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IN MEMORIAM FLORIAN CAJORI
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PLANE AND SOLID
GEOMETEY
BY
WILLIAM J. MILNE, Ph.D., LL.D.
PRESIDENT OP NEW YORK STATE NORMAL COLLEGE, ALBANY N.T-
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NEW YORK:. CINCINNATI.:. CHICAGO
AMERICAN BOOK COMPANY
MILNE'S MATHEMATICS
Milne's Elements of Arithmetic
Milne's Standard Arithmetic
Milne's Mental Arithmetic
Milne's Elements of Algebra
Milne's Grammar School Algebra
Milne's High School Algebra Milne's Plane and Solid Geometry Milne's Plane Geometry — Separate
COPTBIGHT, 1899, BY
WILLIAM J. MILNE. milnk's obom.
B-^ IB
PREFACE
It is generally conceded that geometry is the most interesting of all the mathematical sciences, yet many students have failed to find either pleasure or profit in studying it. The most serious hindrance to the proper understanding of the subject has*been the failure on the part of the student to grasp the geometrical concept which he has been endeav- oring to establish by a process of reasoning. Many attempts have been made by thorough teachers to remedy the difficulty, but there is a very general agreement that the most successful method has been by exercises in " Inventional " Geometry. Students who have been fortunate enough to have the subject presented in that way have usually understood it, and, better still, they have enjoyed it.
While Inventional Geometry has been full of interest to the student, it has often failed to develop that knowledge of the science which is necessary to thorough mastery, because it has not been progressive, and, what is more to be deplored, it has failed to give that acquaintance with the forms of rigid deductive reasoning which is one of the chief objects sought in the study of the science. The student has often been led by this objective method of study to rely upon his visual recognition of the relations of lines and angles in a drawing rather than upon the demon- stration based upon definitions, axioms, and propositions that have been proved.
In this book the effort is made to introduce the student to geometry through the employment of inventional steps, but the somewhat frag- mentary and unsatisfactory result of such teaching is supplemented by demonstrations, in consecutive order, of the fundamental propositions of the science. The desirability of training students to form proper infer- ences from the study of accurately drawn figures has been recognized by the author; such a method awakens keen interest in the subject and develops right habits of investigation, but there is necessity also for the accuracy of statement and the logical training of the older methods to assure the pleasure and profit that belong with both.
Every theorem has been introduced by questions designed to lead the student to discover the geometrical concept clearly and fully before a demonstraticii is attempted. They are not intended to lead to a demon- stration, but rather to a correct and definite idea of what is to be proved.
4 PREFACE,
Many of the exercises at the foot of the page require the student to infer the truth involved in the relations given. The interrogative form is employed for the purpose of compelling the student to obtain the ideas for himself, and the answers he must give to the questions furnish an admirable training in accuracy of expression.
A great abundance of undemonstrated theorems and of unsolved prob- lems is supplied, and teachers will find them quite numerous enough for the needs of any class. The demonstration of original theorems and the solution of original problems are of so great consequence in developing the power to reason that every teacher should insist upon such w^ork.
Much aid in originating demonstrations may be obtained from the Summaries which follow each of the first six books. These summaries are not collections of propositions that have been demonstrated, but are rather groups of the truths established in the book to which they are appended. If the student makes himself thoroughly acquainted with them, much of the difl&culty experienced in demonstrating original theorems, in solving problems, and in determining loci will be removed.
A very small proportion of those who study elementary geometry expect to become mathematicians in any broad sense of the term, and so geometry must serve to give them almost the only training they will get in formal and logical argument in secondary schools and in colleges. For this reason mathematical elegance in demonstrations and in solu- tions has often been sacrificed in the interest of clear and simple steps, even though such a plan has required some expansion of the text. Ele- gant demonstrations are appreciated by mathematicians, but training in formal deductive reasoning is of more consequence to most students.
The author is indebted to many authors, both American and foreign, who have preceded him. Their efforts to present the subject in the best way have aided him very greatly in preparing this work. He has selected large numbers of supplementary theorems and problems from several European authors of renown, and yet he is unable to give credit to any author in particular, because they all seem to have selected their exer- cises from some common source of supply.
WILLIAM J. MILNE.
Albany. N.Y.
SUGGESTIONS TO TEACHERS
1. Thorough teaching and frequent reviews, especially at the beginning of plane and of solid geometry, will be rewarded by intelligent progress and deep interest on the part of the students.
2. Before the assignment of any lesson, the teacher should require the students to draw the figures and answer the questions which are intro- ductory to the propositions that are to be proved at the next lesson.
After the questions have been answered, require the students to express their inferences in the form of a theorem.
3. While the students are answering the introductory questions or stating the inferences suggested by the exercises at the bottom of the page, the inquiry, " How do you know that this is true ? " will often lead to a demonstration.
4. The section numbers are convenient in written demonstrations, but in oral proofs the reason for each step should be given fully and accurately and all why's should be answered,
5. Students may sometimes be allowed to express definitions, axioms, theorems, etc., in their own language, but as a general rule their expres- sions are inaccurate and faulty. The teacher should in such instances call attention to the errors and require concise and accurate statements. It will then be discovered that they approximate very closely those given in the book.
6. The practice of requiring the students to outline, in a general way, the steps they are to take in establishing the truth of a proposition will develop nmch logical power and cause them to look at the argument rather than at its details.
The following are suggestive outlines of steps:
Prop. XXXVI., page 61.
1. Draw the diagonal AC.
2. Prove ^ABC and ADC equal.
3. Prove^5ll2)Cand^i)ll5C. Then, ABCD is a parallelogram.
6
6 SUGGESTIONS TO TEACHERS,
Prop. XII., page 185.
1. Make the required construction, drawing CE^ BJ^ and CK.
2. Prove A AEC and AJB equal.
3. Prove AEKL =<> 2 A AEG, and AGHJ- 2 A ^t/iS.
4. Frove AEKL o^ACHJ.
5. Similarly ^Z>irX=o5(7G^i?'. Then, ABDE<> BCGF+ ACHJ.
7. Demonstrations should never be memorized. If suggestion 6 is observed carefully, students will not be likely to commit to memory the words of the book.
8. Encourage students to prove propositions in their own way, even though the proofs be less elegant than those which are given. Elegant methods will be acquired by practice.
9. Written demonstrations should be required frequently. They serve a double purpose, viz. : they train the eye and develop accuracy in reasoning.
All written work should be done neatly, and all figures should be drawn as accurately as possible.
10. The undemonstrated theorems and unsolved problems are probably more numerous than most classes can prove or solve in the time allotted to the subject, consequently teachers are expected to make selections from the lists given. The exercises are carefully graded so that the more difficult ones come at the end of each list. These may be omitted at the first reading and reserved for a final review.
It is suggested that the exercises in the interrogative form at the foot of the page in Books I and II, except the numerical ones, be employed at first only for the purpose of developing correct geometrical concepts and accuracy in expressing the truth inferred. In review the proofs of the inferences may be required.
11. Particular attention should be given to the Summary at the end of each book. The students should be required to state all the conditions under which the facts given in black-faced type have been shown to be true. They will thus have at immediate command all the facts which can be employed in the demonstration they are attempting.
If the demonstration of the inferences and theorems found at the bottom of the page is required, the students should be referred to the summary. They should understand, however, that they can use no truth given in the summary whose section number indicates that it was estab- lished subsequently to the point in the text where the proposition or exercise is found.
The method of using the summaries is illustrated upon page 78.
CONTENTS GEOMETRY
PAOB
Preliminary Definitions 9
Lines and Surfaces 10
Angles . 12
Measurement of Angles . 15
Equality of Geometrical Magnitudes 15
Demonstration or Proof . . . . 17
Axioms - ... 19
Postulates 19
Symbols . . . • 20
Abbreviations 20
PLANE GEOMETRY
BOOK I
Lines and Rectilinear Figures .21
Parallel Lines 27
Triangles 38
Quadrilaterals 60
Polygons 68
Summary . . , . . .74
Supplementary Exercises 78
BOOK n
Circles . . , . 83
Measurement 98
Theory of Limits 100
Summary 122
Supplementary Exercises 124
BOOK III
Ratio and Proportion 136
BOOK IV
Proportional Lines and Similar Figures 1^7
Summary 167
Supplementary Exercises . 169
I
CONTENTS
BOOK V
PASS
Area and Equivalence , 173
Summary 202
Supplementary Exercises 20^
BOOK VT
Regular Polygons and Measurement of the Circle 213
Maxima and Minima 230
Symmetry 234
Nummary 237
Supplementary Exercises 238
SOLID GEOMETRY
BOOK VII
Planes and Solid Angles 243
Dihedral Angles 257
Polyhedral Angles . 267
Supplementary Exercises 272
BOOK VIII
Polyhedrons 273
Prisms 273
Pyramids 287
Similar and Regular Polyhedrons 299
Formulae. 305
Supplementary Exercises 306
BOOK IX
Cylinders and Cones 309
Formulae 325
Supplementary Exercises 325
BOOK X
Spheres ..... Spherical Angles and Polygons Spherical Measurements .
Formulae
Supplementary Exercises
327
338 352 364 364
Exercises for Review Metric Tables .
369 383
GEOMETRY
PRELIMINARY DEFINITIONS
1. Every material object occupies a limited portion of space and is called a Physical Solid or Body.
2. The portion of space occupied by a physical solid is identical in form and in extent with that solid, and is called a Geometrical Solid.
In this work only geometrical solids are considered, and for brevity they are called simply solids.
3. Any limited portion of space is called a Solid.
A solid has three dimensions, length, breadth, and thickness.
The drawing in the margin is represented as having three dimensions. Fig. 1.
4. The limit of a solid, or the boundary which separates it from all surrounding space, is called a Surface.
A surface has only two dimensions, length and breadth.
A page of a book is a surface, but a leaf of a book is a soUd.
5. The limit or boundary of a surface is called a Line.
A line has only one dimension, length. It has neither breadth nor thickness.
The edges of a cube are lines.
10 GEOMETRY.
6. The limits, or extremities, of a line are called Points.
A point has position only. It has neither length, breadth, nor thickness.
The dots and lines made by a pencil or crayon are not geometrical points and lines, but are convenient representations of them.
7. Lines, surfaces, and solids are called Geometrical Magnitudes, or simply Magnitudes.
8. A line may be conceived of as generated by a point in motion. Hence a line may be considered as independent of a surface, and it may be of unlimited extent.
A surface may be conceived of as generated by a line in motion. Hence a surface may be considered as independent of a solid, and it may be of unlimited extent.
A solid may be conceived of as generated by a surface in motion. Hence a solid may be considered as independent of a material object.
LINES AND SURFACES
9. 1. Select two points upon your paper and draw several lines connecting them.
a. Which is the shortest line you have drawn ? If this line is not the shortest that can be drawn between the points, what kind of a line is the shortest ?
h. What other kinds of lines have you drawn besides a straight line?
2. When a carpenter places a straightedge upon a board and moves it about over the surface, what is he endeavoring to deter- mine regarding the surface ?
3. If the straightedge does not touch every point of the sur- face of the board to which it is applied, what has been discovered about the surface ? ^
4. How does he know whether or not the surface is an even or a plane surface ?
5. If any two points on the surface of a ball or sphere are joined by a straight line, where does the line pass ?
6. How much of the surface of a perfect sphere is a plane sur- face?
GEOMETRY. 11
10. A line which has the same direction throughout its whole extent is called a Straight Line.
A straight line is also called a Bight Line, or simply a Line. In this work the term "line" means a straight line unless otherwise specified.
11. A line no part of which is straight is called a Curved Line.
Consequently, a curved line changes its direc- tion at every point.
12. A line made up of several straight lines which have different directions is called a Broken Line. Fig. a.
13. A line made up of straight and curved lines is called a Mixed Line.
Any portion of a line may be called a segment of that line.
Fig, 4.
14. A surface such that a straight line joining any two of its points lies wholly in the surface is called a Plane Surface, or a Plane.
15. A surface, no part of which is plane, is called a Curved Surface.
16. Any combination of points, lines, surfaces, or solids is called a Geometrical Figure.
A geometrical figure is ideal, but it can be represented to the eye by draw- ings or objects.
17. A figure formed by points and lines in the same plane is called a Plane Figure.
18. A figure formed by straight or right lines only is called a Rectilinear Figure.
19. The science which treats of points, lines, surfaces, and solids, and of the properties, construction, and measurement of geometrical figures, is called Geometry.
20. That portion of geometry which treats of plane figures is called Plane Geometry.
a GEOMETRY.
21. That portion of geometry which treats of figures whose points and lines do not all lie in the same plane is called Solid Geometry.
ANGLES
22. 1. From any point draw two straight lines in different directions. Draw two straight lines from each of several other points, and thus form several angles.
2. How does the angle at the corner of this page compare in size with the angle at the corner of the room? Show your answer to be true by an actual test.
How is the size of any angle affected by the length of the lines which form its sides 9
3. Eorm several angles at the same point; that is, several angles having a common vertex.
4. How many of them have a common vertex and one common side between them and are, at the same time, on opposite sides of the common side; that is, how many angles are adjacent angles?
5. Draw a straight line meeting another straight line so as to form two equal adjacent angles ; that is, two right angles.
6. Draw from a point or vertex two straight lines in opposite directions; that is, form a straight angle. How does a straight angle compare in size with a right angle ?
7. Draw several angles, some greater and some less, than a right angle.
8. Draw a right angle and divide it into two parts, or into two complementary angles.
9. Draw a straight angle and divide it into two parts, or into two supplementary angles.
10. Draw two straiglit lines crossing or intersecting each other, thus forming two pairs of opposite or vertical angles.
23. The difference in direction of two lines which meet is called a Plane Angle, or simply an Angle.
The lines are called the sides of the angle, ^' and the point where they meet is called its vertex.
The lines OA and OB are the sides of the angle formed at the point 0, and 0 is the vertex of the angle. Fig. 6.
GEOMETRY. 18
The size of an angle does not depend upon the length of its sides, but upon the divergence of the sides or upon the opening between them. q.
Compare Figs. 5 and 6. yiq. e.
24. When there is but one angle at a point, it may be desig- nated by the single letter at the vertex, or by three letters.
In Fig. 6 the angle may be called the angle A, or the angle BAG^ or the angle CAB.
When several angles have a common vertex, it is customary to use three letters in designating each, placing the letter at the vertex between the other two.
An angle is sometimes designated by a figure or small letter placed in the opening of the angle. ^ o"
The angles formed by the lines meeting at O may be designated by AOG, the figure 1, and the small letter a.
25. Angles which have a common vertex and a common side, and which are upon opposite sides of the common side, are called Adjacent Angles.
In Fig. 7 angles COA and COB are adjacent angles, having a common vertex O, and a common side CO and lying upon opposite sides of the common side. Also COB and BOD are adjacent angles.
26. When one straight line meets another straight line so as to form two equal adjacent angles, each of the angles is called a Right Angle ; and each line is said to be perpendicular to the other.
The sides of a right angle are therefore perpendicular to each other, and lines per- pendicidar to each other form right angles
with each other. ^^^ ^ ^
27. An angle whose sides extend in opposite directions from the vertex, thus forming one straight line, is called a Straight Angle.
If the sides OA and OB, Fig. 9, extend in
opposite directions from the vertex O, the ^ yi^ ^
angle AOB is a straight angle.
A straight angle is equal to two right angles.
14
GEOMETRY,
28. An angle less than a right angle is called an Acute Angle.
29. An angle greater than a right angle and less than a straight angle is called an Obtuse Angle.
Fig. 10.
Fig. 11.
30. An angle greater than a straight angle and less than two straight angles is called a Reflex Angle.
Acute, obtuse, and reflex angles are called oblique angles in distinction from right angles and straight angles.
31. When two angles are together equal to a right angle, they are called Complementary Angles, and each is said to be the Complement of the other.
If the angle COE is a right angle, the angles COD and DOE are complementary angles ; the angle COD is the complement of the angle ^'^- ^^•
DOE ; and the angle DOE is the complement of the angle COD.
32. When two angles are together equal to two right angles, they are called Supplementary An- gles, and each is said to be the Supplement of the other.
If the angles AOD and DOB are to-
gether equal to two right angles, the -^ angles AOD and DOB are supplemen- tary angles; the angle AOD is the supplement of the angle DOB^ and the angle DOB is the supplement of the angle AOD.
33. When two lines intersect, the opposite angles are called Vertical Angles. a-
The angles AOC and DOB, and the angles AOD and COB are vertical angles.
34. A line or a plane which divides any geometrical magnitude into two equal parts is called the Bisector of that wagnitju.dt^
o
Fig. 14.
Fro. 15.
GEOMETRY. 16
MEASUREMENT OF ANGLES
35. To measure a magnitude is to find how many times it con- tains a certain other magnitude assumed as a unit of measure.
The unit of measure for angles is sometimes a right angle, but very often it is a degree.
Suppose the line OB, having one of its extremities fixed at o, moves from a position coincident with OA to the position indicated by OB. By this motion the angle AOB has been generated.
When the rotating line OB has passed one half the distance from OA around to OA, the lines extend in opposite directions from O, and a straight angle has been gen- erated; and since a straight angle is equal ^^®' ^^' to two right angles (§ 27), when the line has passed one fourth of the distance around to OA, a right angle has been generated, and the lines OB and OA are perpendicular to each other (§ 26). When the line has . rotated entirely around from OA to OA, it has generated two straight angles, or four right angles. Conse- quently : The total angular magnitude about a point in a plane is equal to four right angles.
Inasmuch as it is frequently convenient to employ a smaller unit of angular measure than a right angle, the entire angular mag- nitude about a point has been divided into 360 equal parts, called degrees; a degree into 60 equal parts, called minutes; a minute into 60 equal parts, called seconds.
Degrees, minutes, and seconds are indicated in connection with numbers by the respective symbols °, ', ".
25 degrees, 18 minutes, 34 seconds is written 25° 18' 34".
A right angle is an angle of 90°.
EQUALITY OF GEOMETRICAL MAGNITUDES
36 Geometrical magnitudes which coincide exactly when one is placed upon or applied to the other are equal. Since, however, geometrical magnitudes are ideal they are not actually taken up and placed the one upon the other, but this is conceived to be done.
16 QEOMETRY.
This method of establishing equality is called the Method of Superposition.
If one straight line is conceived to be placed upon another straight line so that the extremities of both coincide, the lines are equal.-
If an angle is conceived to be placed upon another angle so that their vertices coincide and their sides take the same directions, respectively, the angles are equal.
If any figure is conceived to be placed upon any other figure so that they coincide exactly throughout their whole extent, they are equal.
Figures that are superposa-ble are sometimes called congruent.
EXERCISES
37. Draw as accurately as possible the figures which are sug- gested ; study them carefully ; infer the answers to the questions ; state your inference or conclusion in as accurate form as possible ; give the reason for your conclusion when you can.
The student is asked to represent by a drawing any figure that may be required so that it may simply appear to the eye to be accurate. Geometri- cal methods of construction are given at suitable points in the book, but they cannot be insisted upon at this stage.
1. Draw two straight lines intersecting in as many points as possible. In how many points do they intersect ?
Inference : Two straight lines can intersect in only one point.
2. Draw a straight line ; draw another meeting it. How does the sum of the adjacent angles thus formed compare with two right angles ?
Inference : When one straight line meets another straight line, the sum of the adjacent angles is equal to two right angles.
3. Draw a straight line ; from any point in it draw several lines extending in different directions. How does the sum of the consecutive angles formed on one side of the given line compare with a right angle ? With a straight angle ?
4. Draw a straight line ; also another meeting it so as to form two adjacent angles, one of which is an acute angle. What kind of an angle is the other ?
I
GEOMETRY. 17
5. Draw two intersecting lines. How many angles are formed ? How do the opposite or vertical angles compare in size ?
6. Draw two lines intersecting so as to form a right angle. How does each of the other angles formed compare with a right angle ? How do right angles compare in size ? How do straight angles compare in size ?
7. Draw two equal angles. How do their complements com- pare ? How do their supplements compare ?
8. Draw a straight line ; select any point in that line and draw as many perpendiculars as possible to the line at that point. How many such perpendiculars can be drawn on one side of the line ?
DEMONSTRATION OR PROOF
38. The inferences which the student has just made are proba- bly correct, but they must be proved to be true before they can be relied upon with certainty unless their truth is self-evident.
Many truths have been inferred, and used as the basis of im- portant enterprises before they have been logically demonstrated.
Carpenters believe that their squares are true if a line from the 12-inch mark on one side to the 16-inch mark on the other is 20 inches long ; but they may not be capable of giving satisfac- tory reasons for their convictions.
Many valuable facts of geometry may be inferred by observa- tion of figures and objects, but the value of the study to a student consists not so much in the knowledge acquired as in the develop- ment of the logical faculty by the rigid course of reasoning required to prove the truth or falsity of the inference.
Much attention must therefore be given to the demonstration or proof of inferences from known data, and of statements even though they may seem to be true.
39. A course of reasoning which establishes the truth or falsity of a statement is called a Demonstration, or Proof.
40. A statement of something to be considered or done is called a Proposition.
" All men are mortal " and " It is required to bisect an angle " are propo- sitions.
MILNE'S OEOM. 2
18 GEOMETRY,
41. A proposition so elementary that its truth is self-evident is called an Axiom.
An axiom is a self-evident truth to those only who understand the terms employed in expressing it.
Axioms may be illustrated, but they do not require proof.
Axioms have often a general application. Some, however, apply only to geometrical magnitudes and relations.
"A whole is equal to the sum of all itg parts" is a general axiom. It can be employed in demonstrating propositions in arithmetic and algebra as well as in geometry. " A straight line is the shortest distance between two points" is a geometncal axiom. It can be used only in proving propositions which express some geometrical truth.
42. A proposition which requires demonstration or proof is called a Theorem.
" In any proportion the product of the extremes is equal to the product of the means" is an algebraic theorem,
43. A theorem whose truth may be easily deduced from a preceding theorem is often attached to it, and called a Corollary.
The arithmetical theorem, " A number is divisible by 3 when the sum ot its digits is divisible by 3" may be readily deduced from the theorem, "A number is divisible by 9 when the sum of its digits is divisible by 9," and may be attached to it as a corollary,
44. A proposition requiring something to be done is called a Problem. '
" Construct an angle equal to a given angle " is a geometrical problem,
45. A problem so simple that its solution is admitted to be possible is called a Postulate.
*' A straight line may be drawn from one point to another" is a postulate. Postulates are numerous. Some of those employed in geometry may be found in § 50.
46. A remark made upon one or more propositions, and show- ing, in a general way, their extension or limitations, their connec- tion, or their use is called a Scholium.
Thus, after the processes of dividing a common fraction by a common fraction, and a decimal by a decimal, have been taught, a remark showing that precisely the same principles are involved in each process is a scholium.
GEOMETRY, 19
47. The enunciation of a theorem may be separated into the following parts :
1. The things given, or granted, called the Data (singular datum).
2. A statement of what is to be proved, called the Conclusion. The term Hypothesis may be used instead of the term data.
A supposition made in the course of a demonstration is also called an Hypothesis.
48. In proofs, or demonstrations, only definitions, axioms, and propositions which have been proved can be employed to establish the truth of the proposition.
AXIOMS
49. 1. Things which are equal to the same thing are equal to each other.
2. If equals are added to equals, the sums are equal.
3. If equals are taken from equals, the remainders are equal.
4. If equals are added to unequals, the sums are unequal.
5. If equals are taken from unequals, the remainders are unequal.
6. Things which are doubles of equal things are equal. 1. Things which are halves of equal things are equal.
8. Tlie whole is greater than any of its parts.
9. The whole is equal to the sum of all its parts.
10. A straight line is the shortest distance between two points.
11. If two straight lines coincide in two points, they will coincide throughout their whole extent, and form one and the same straight line.
12. Between the same two points but one straight line can be drawn.
POSTULATES
50. 1. A straight line may be produced indefinitely.
2. A straight line may be drawn from any point to any other point.
3. On the greater of two straight lines a part can be laid off equal to the less.
4. A figure can be moved unaltered to a new ptosition.
20
plus, or increased by.
minus, or diminished by.
multiplied by.
multiplied by.
divided by. f equals, I or is (or are) equal to.
is (or are) equivalent to.
is (or are) greater than.
is (or are) less than.
therefore, or hence.
|
GEOMETRY, |
|
|
SYMBOLS |
|
|
A |
triangle. |
|
A |
triangles. |
|
O |
parallelogram. |
|
[EJ |
parallelograms. |
|
O |
circle. |
|
© |
circles. |
II r parallel, I or is (or are) parallel to.
I f perpendicular, I or is (or are) perpendicular to.
Js perpendiculars. " inch or inches.
ABBREVIATIONS
Adj adjacent.
Alt alternate.
Ax axiom.
Circum. . . . circumference.
Corap complement.
Const construction.
Cor corollary.
Def definition.
Ex exercise.
Ext exterior.
Fig figure.
Int. . . . . interior.
Lat. Surf. . . lateral surface.
N note.
Opp Qpposite.
Post postulate.
Prob problem.
Pt point.
Rect rectangle.
Rt right.
Sch scholium.
Sect sector.
Seg segment.
Sim similar.
St straight.
Sup supplement.
The letters q.e.d. are placed at the end of a proof; they are the initial letters of the Latin words qtiod erat demonstrandum, meaning which was to be proved.
The letters q.e.f. are placed at tlie end of a solution of a, problem for quoa erat faciendum, meaning which was to be done.
PLA^E GEOMETRY
BOOK I
LINES AND RECTILINEAR FIGURES
Proposition I
51. Draw a straight line and as many perpendiculars as possible to the line at one point. How many can be drawn ? (§ 37 )
Theorem. At any point in a straight line one perperir- dicular to the line can be drawn, and only one.
Data : Any straight line, as AB, and any point in that line, as O. i"
To prove that a perpendicular to AB ' \ ^.p
can be drawn at the point O, and that I ,.-'''
only one can be drawn. j x'
Proof. Suppose a line DO to rotate a o b
about the point 0 as a pivot, from the position BO to AO.
As DO rotates from the position BO toward the position AO, the angle DOB will, at first, be smaller than the angle DO A.
As DO continues to rotate, the angle DOB will increase continu- ously, and will eventually become larger than angle DOA.
Therefore, since angle DOB is at first smaller than angle DOA, and afterwards larger than angle DOA, there must be one position of DO, as, for example, CO, in which the two angles are equal.
By § 26, CO is then perpendicular to AB.
Since there is but one position in which the line DO makes equal angles with the line AB, there can be but one perpendicular.
Therefore, at any point in a straight line one perpendicula,r to the line can be drawn, and only one. q.e.d.
21
22 PLANE GEOMETRY. — BOOK L
Proposition II
62. 1. Draw two lines intersecting so as to form a right angle. How does each of the other angles formed compare in size with a right angle ? How do right angles compare in size? How'do straight angles compare? (§ 37 )
2. Draw two equal angles and their complements. How do their com- plements compare in size? How do their supplements compare? (§ 37 )
Theore^n, All right angles are equal.
Data : Any right angles, as ABC and DEF.
To prove angles ABC and DBF equal.
A B D ■ E
Proof. Suppose that Z DBF is placed upon Z ABC in such a way that the point B falls upon the point B and the line BD takes the direction of the line BA.
Since by § 26, BC is perpendicular to BA and BF is perpendicu- Jar to BD and on the same side of the line,
line BF must take the same direction as line BC, for otherwise there would be two perpendiculars to BA at the point B and by § 51, this is impossible.
Consequently, the line BF falls upon the line BC,
and Z DEF coincides with Z ABC. Hence, § 36, A ABC and DEF are equal.
Therefore, all right angles are equal. q.e.d.
53. Cor. I. All straight angles are equal.
54. Cor. II. The complements of equal angles are equal, also the supplements of equal angles are equal.
Ex. 1. Find the complement of an angle of 15° ; 27° ; 35° ; 49°.
Ex. 2. Find the supplement of an angle of 38° ; 96° ; 114°.
Ex. 3. The complement of an angle is 63°. What is the angle ?
Ex. 4. The supplement of an angle is 103°. What is the angle ?
Ex. 5. Find the complement of the supplement of an angle of 165° ; 140° ; 122°; 113°; 108°; 99°.
Ex. 6. Find the supplement of the complement of an angl' of 48°; 84°; 27° ; 16° ; 31° ; 64° ; 39°.
PLANE GEOMETRY. — BOOK I. 23
Proposition III
55. 1- Draw a straight line, and another meeting it. How does the sum of the adjacent angles thus formed compare with a right angle? With a straight angle ? (§ 37 )
2. Draw a straight line, and from any point in it draw several lines extending in different directions. How does the suin of the consecutive angles formed on one side of the line compare with a right angle? With a straight angle? How does the sum of the consecutive angles on both sides of the line compare with a right angle ? With a straight angle? (§37)
Theorem, If one straight line meets another straight line, the sum of the adjacent angles is equal to two right angles.
Data : Any straight line, as AB, and -P
any other straight line, as GO, meeting it in the point 0.
To prove the sum of the adjacent
«,ngles, AOC and t, equal to two right j
angles.
Proof. When CO is perpendicular to AB, by § 26, each of the A AOC and t is a rt. Z,
and their sum is two rt, A.
When CO is not perpendicular to AB,
draw DO perpendicular to AB at the point O.
Then, by § 26, A r and DOB are rt. A, and Ar-{- A DOB = 2 it. A;
by Ax. 9, A DOB = As-{-At,
.'. by substitution,
Ar-^As-{-At = 2Ti A.
By Ax. 9, A AOC =Ar + As,
and by substitution, A AOC -[- At = 2 vt. A.
Therefore, if one straight line meets another straight line, the sum of the adjacent angles is equal to two right angles. q.e.d.
24
PLANE GEOMETRY. — BOOK L
56. Cor. I. The sum of aU the consecutive angles which hcuve a common vertex in a line, and which lie .-^ on one side of it, is equal to two right angles, or a straight angle.
57. Cor. II. The sum of all the
consecutive angles thai can be formed about a point is equal to four right angles, or two straight angles.
Ex. 7. One line meets another, making two angles with it. One angle contains 87°. How many degrees are there in the other ?
Ex. 8. Four of the five consecutive angles about a point contain 17®, 36°, 89°, and 110° respectively. How many degrees are there in the fifth angle ?
Ex. 9. K two lines meet a third line at the same point, making with the third line angles of 27° and 63° respectively, what is the angle between the two lines?
Proposition IV
58. Constmct two angles which are adjacent such that their sum is equal to two right angles. What kind of a line do their exterior sides form?
Theoi^ew,. If the sum of two adjacent angles is equal to tivo right angles, their exterior sides form one straight line.
Data : Any two adjax3ent angles, as AOC and COB, whose sum is equal to two right angles.
To prove that the exterior sides, AO 2
and OB, form one straight line.
Proof.
From data, Z AOC -f- Z COB = 2 rt. ^ ;
.-. by § 27, Z AOC+Z COB = a st. Z.
But by Ax. 9, Z AOC + Z COB = Z ^OF, .-. by Ax. 1, Z AOB = a st Z.
Hence, by % 27, AO and OB, the sides of Z AOB extending in opposite directions from the point G, iiorm one straight line.
Therefore, etc. q.b.d.
PLANE GEOMETRY. — BOOK L Proposition V
25
59. Draw two intersecting lines. How many angles are formed? How do the opposite or vertical angles compare in size ?
Theorem. If two straight lines intersect, the vertical angles are equal.
Data : Any two intersecting straight lines, as AB and CD.
To prove tlie vertical angles, as v and t, equal.
Z r + Z V = 2 rt. A
Proof.
By § 55,
and Z.r + Z.t = 2Tt.A)
hence, by Ax. 1, Zr-\-/.v = /.r-\-Zt.
Subtracting Z r from both sides of this equality,
by Ax. 3, /.v = Z. t.
In like manner it may be proved that /.r = Z.s. Therefore, etc.
Q.E.D.
Ex. 10. The complement of an angle is 43°. What is the supplement of the angle ?
Ex. 11. The supplement of an angle is 125°. What is the complement of the angle ?
Ex. 12. How many degrees are there in the supplement of the comple- ment of an angle of 60° ? Of 43° 25' 50" ?
Ex. 13. How many degrees are there in the complement of the supple- ment of an angle of 159° ? Of 133° 15' 25" ?
Ex. 14. How many degrees are there in the angle formed by the bisectors of two supplementary adjacent angles ?
Ex. 15. If a line drawn through the vertex of two vertical angles bisects one angle, how does it divide the other ? (§ 37)
Ex. 16. If one of the vertical angles formed by the intersection of two straight lines is 37°, what is the value of each of the other angles ?
Ex. 17. Lines are drawn to bisect the two pairs of vertical angles fonned by two intersecting straight lines. What is the direction of these bisectors with reference to each other? (§37)
26 PLANE GEOMETRY.^BOOK L
Proposition VI
60. 1. Draw a straight line ; from a point not in the line draw as many perpendiculars to the line as possible. How many can be drawn ?
2. How does the perpendicular compare in length with any other line drawn from the point to the given line ?
Theorem, From a point without a straight line only one perpendicular can be drawn to the line.
Data : Any straight line, as AB, any point without it, as P, and the per- pendicular PC drawn from the point P to the line AB. [1
To prove that PC is the only per- pendicular that can be drawn from / \ the point P to the line AB.
G
F
Proof. Prolong PC to F, making CF=PC; from P draw any other line to AB, as PD ; and draw DF.
Then, PCF is a straight line.
§ 26, ^ ^ 3,nd s are rt. A,
and, § 52, Zr=Zs.
Revolve the figure PCD about AB as an axis and apply it to the
figure FCD.
DC being in the axis remains fixed,
and since Zr=Zs,
CP will take the direction of CF,
Since, const., CP = CF,
P and F will coincide ;
.*. Ax. 11, DP and DF will coincide,
and, § 36, Zt = Zv.
Revolve PCD back to its original position and prolong PD to G.
Then, § 56, Zt-\- Zv -\- Zw = 2 vt A,
PLANE GEOMETRY.-^ BOOK L
27
and since /.t = /.Vj
2Zt-\-Zw = 2Yt.A', Z f + i Z w = 1 rt. Z ; that is, Z ^ is less than a rt. Z.
PD is not perpendicular to AB. But since PD represents any line from P to AB other than PC^ PC is the only perpendicular that can be drawn to AB from P. Therefore, etc. q.e.d.
61. Cor. A perpendicular is the shortest line that can be drawn from a j^oint to a line.
1. Since PCF is a straight line, is PDF a straight line ? Ax. 12.
2. Which line, then, is the shorter, PCF or PDF? Ax. 10.
3. What part of PCF is PC ? Of PDF is PD ?
4. Then, how do PC and PD compare in length ?
5. Since PD represents any line from P to AB other than the perpendicular PC, what is the shortest line that can be draw» from a point to a line ?
62. The distance from a point to a line is always understood to be the perpendicular or shortest distance.
PARALLEL LINES
63. Lines which lie in the same plane, and ~
which cannot meet however far they may be
extended, are called Parallel Lines.
64. A straight line which crosses or cuts two or more straight lines is called a Trans- versal.
EF is a transversal of AB and CD.
Eight angles are formed by the transversal EF with the lines AB and CD.
65. The angles above AB and those below CD, or those without the two lines cut by the trans* versal, are called Exterior Angles.
Angles r, 5, y, and z are exterior angles.
28 PLANE GEOMETRY. — BOOK I.
66. The angles between, or within the two lines cut by the transversal, are called Interior Angles.
Angles t, r, w, and x are interior angles.
67. Non-adjacent angles without the two lines, and on opposite sides of the transversal, are called Alternate Exterior Angles.
Angles r and sr, or s and y, are alternate exterior angles.
68. Non-adjacent angles within the two lines, and on opposite sides of the transversal, are called Alternate Interior Angles.
Angles t and x, or v and w, are alternate interior angles.
69. Non-adjacent angles, which lie one without and one with- in the two lines, and on the same side of the transversal, are called Corresponding Angles.
Angles r and w, s and aj, t and y, or v and z, are corresponding angles. Corresponding angles are also called Exterior Interior Angles.
70. Ax. 13. TJirough a given point but one straight line can be drawn parallel to a given straight line.
Proposition VII
71. Draw a straight line; also two other lines each perpendicular to the first line. In what direction do the perpendiculars extend with reference to each other ?
Theorem, If two straight lines are perpendicular to the same straight line, they are parallel.
Data: Any straight line, as AB, and any two straight lines each per- pendicular to AB, as CD and EF.
To prove CD and EF parallel.
i) F
Proof. Since, by data, both CD and EF are perpendicular to AB, they cannot meet, for, if they should meet, there would then be two perpendiculars from the same point to the line AB, which is impossible. § 60.
Hence, § 63, CD W EF.
Therefore, etc. q.e.d.
PLANE GEOMETRY. — BOOK I. • 29
Proposition VIII
72. Draw two parallel lines ; also a transversal perpendicular to one of them. What is the direction of the transversal with reference to the other parallel line?
Theorem* If a straight line is perpendicular to one of two parallel straight lines, it is perpendicular to the other.
Data: Any two parallel straight lines, as AB and CD, and any straight a- line perpendicular to AB, as EF, cut- ting CD at the point J. q.
To prove EF perpendicular to CD. G
Proof. If EF is not perpendicular to CD at the point J, it will be perpendicular to some other line drawn through that point.
Suppose GH is that Jine.
Then, hyp., EF ± GH.
But, data, EF A. AB,
then, § 71, GH II AB.
But, data, ^ CD W AB,
then, § 70, GH and CD passing through J cannot both be parallel to AB.
Hence, the hypothesis that EF is not perpendicular to CD is untenable.
Consequently, EF A. CD.
Therefore, etc., q.e.d.
Ex. 18. Two lines are drawn each parallel to AB^ and another line mak- ing an angle of 90° with AB. What is the direction of this line with refer- ence to each of the other two lines ?
Ex. 19. State and illustrate the differences between a plumb line, a per- pendicular line, and a vertical line.
Ex. 20. Two parallel lines are cut by a third line making one interior angle 35". What is the value of the adjacent interior angle ?
30 • PLANE GEOMETRY. — BOOK I.
Proposition IX
73. 1. Draw two parallel lines ; also a transversal. How many pairs of vertical angles are formed? How many pairs of supplementary adjacent angles ? How many sizes of angles are formed '? How many angles of each size ? When may they all be of the same size ?
2. Name a pair of angles whose sum is equal to two right angles. Name seven other pairs. Name a set of four angles whose sum is equal to four right angles. Name three other sets.
3. Name the pairs of alternate interior angles. How do the angles ol any pair compare in size ?
Theorem. If two parallel straight lines are cut hy a transversal, the alternate interior angles are equal.
Data: Any two parallel straight lines, as AB and CD, cut by a trans- versal, as EF, in the points H and J.
To prove the alternate interior an- gles, as AHJ and DJH, equal.
Proof. Through L, the middle point of HJ, draw GK ± CD.
Then, § 72, GK i. AB.
Revolve the figure JLK about the point L and apply it to the figure HLG, so that LJ coincides with LH.
Then, since, § 59, Z JLK = Z HLG,
LK takes the direction of LO.
Const., JK ± GK and HG ± GK,
and since the point J falls upon the point H, § 60, JK must fall upon HG.
Since LJ coincides with LH,
and JK takes the same direction as HQ^
§ 36, Z GHL = Akjl.
Therefore, etc. q.e.d.
PLANE GEOMETRY. — BOOK I. 31
74. If two theorems are related in such a way that the data and conclusion of one become the conclusion and data, respec- tively, of the other, the one is said to be the converse of the other.
Thus, the converse of the theorem just proved is, " Two straight lines cut by a transversal are parallel, if the alternate interior angles are equal."
Converse propositions cannot be assumed to be true. They may be true, but their truth must be established by proof.
Thus, the truth of the proposition, "The product of two even numbers is an even number," can be established readily, but its converse, "An even number is the prodjict of two even numbers," is evidently false.
Proposition X
75. Draw two lines ; also a transversal. In what direction do the lines extend with reference to each other, if the alternate interior angles are equal?
Theorefn, Two straight lines cut by a transversal are parallel, if the alternate interior angles are equal. (Con- verse of Prop. IX.)
/E
Data : Two straight lines, as AB and /
CD, such that when cut by any trans- a --'/'" ^
versal, as EF, the alternate interior ^ /
angles, as ahf and EJD, are equal. /
To prove AB and GB parallel. /
Proof. If AB is not parallel to CZ), then some other line, as KL, drawn through the point H is parallel to CD.
Then, hyp. and § 73, Z KUF = Z EJD ;
but, data, Zahf = /. ejd ;
hence, Ax. 1, Z KHF = Z AHF,
which is absurd, since a part cannot be equal to the whole.
Hence, the hypothesis, that some other line, as KL, drawn through the point H is parallel to CD, is untenable.
Consequently, AB II CD.
Therefore, etc. q.e.d.
32
PLANE GEOMETRY. — BOOK 1.
Proposition XI
76. Draw two parallel lines ; also a transversal. Name the pairs of 'corresponding angles. How do the angles of any pair compare in size?
Theorem, If two parallel straight lines are cut by a transversal, the corresponding angles are equal.
Data: Any two parallel straight lines, as AB and CD, cut by any- transversal, as EF.
To prove the corresponding angles, as t and s, equal.
Proof. § 59,
§73,
hence. Ax. 1, Therefore, etc.
Z.s = Zr; Zt = Zs.
Q.E.D.
Proposition XII
77. Draw two lines; also a transversal. In what direction do the lines extend with reference to each other, if the corresponding angles are equal?
Theorem. Two straight lines cut hy a transversal are parallel, if the corresponding angles are equal. (Converse of Prop. XI.)
Data : Two straight lines, as ^5 and CD, such that when cut by any transversal, as EF, the correspond- ing angles, as t and s, are equal.
To prove AB and CD parallel.
Proof. § 59,
data,
then. Ax. 1, Hence, § 75, Therefore, etc.
Zr = Zt, Zs = Zt; Zr = Zs. AB II CD.
Q.E.D.
PLANE GEOMETRY. — BOOK I. 33
Proposition XIII
78. Draw two parallel lines; also a transversal. How does the sum of the two interior angles on the same side of the transversal compare with a right angle ?
Theorem, If two parallel straight lines are cut hy a transversal, the suin of the two interior angles on the same side of the transversal is equal to two right angles.
Data: Any two parallel straight lines, as AB and CD, cut by a trans- /
versal, as EF. ^
To prove the sum of the two in- terior angles on the same side of the transversal, as t and s, equal to two /
right angles. f
Proof. § 73, Z.r = /.s.
Adding Z ^ to each member of this equation, Ax. 2, Z r + Z ^ = Z s 4- Z ^.
But, § 55, Zr + Z^ = 2rt. ^;
.•. Ax. 1, Zs + Z^ = 2rt. A.
Therefore, etc. q.e.d.
Ex. 21. If two parallel lines are cut by a transversal, what is the sum of the two exterior angles on the same side of the transversal ?
Ex. 22. The straight lines AB and CD are cut hy EF in G and H respectively ; angle EHD — 38°. What must he the value of the angle EGB in order that AB and CD may be parallel ?
Ex. 23. A transversal cutting two parallel lines makes an interior angle of 50°. What is the value of the other interior angle on the same side of the transversal ?
Ex. 24. TWO parallel lines are cut by a third line making one interior angle 35°. What is the value of each of the other interior angles ? How many degrees are there in the sum of the interior angles upon the same side of the transversal ?
Ex. 25. How do lines bisecting any two alternate interior angles, formed by two parallel lines cut by a transversal, lie with reference to each other ?
Ex. 26. The straight lines AB and CD are cut by EF in G and H re- spectively ; angle EHD = 40°. What must be the value of the angle A GF, if AB and CD are parallel ? milne's geom. — 3
34 PLANE GEOMETRY. — BOOK i.
Proposition XIV
79. Draw two lines; also a transversal. In what direction do th« lines extend with reference to each other, if the sum of the two interior angles on the same side of the transversal is equal to two right angles ?
Theorein, Two straight lines cut by a transversal are parallel, if the sum of the two interior angles on the same side of the transversal is equal to two right angles. (Con- verse of Prop. XIII.)
Data: Two straight lines, as AB and CZ), such, that when cut by any- transversal, as EF, the sum of the two interior angles on the same side of the transversal, as t and s, is equal to two right angles.
To prove AB and CD parallel.
Proof. §55, Zr + Z^ = 2 rt. zi;
data, Z^ + Zs = 2 rt. A]
.'. Ax. 1, Z.r + At = /.t + /.s.
Taking Z t from each member of this equation,
Ax. 3, Zr = /.s.
Hence, § 75, AB II CD.
Therefore, etc. q.e.d.
Ex. 27. AB and CD are two lines cut in O and JST, respectively, by EF-, A BCF = 123*, and Z GHD = 62°. Are the lines AB and CD parallel ?
Ex. 28. If two lines are cut by a transversal and the sum of the two exterior angles on the same side of the transversal is equal to 180°, are the lines parallel ?
Ex. 29. Two parallel lines are cut by a transversal so that one exterior angle is 105°. How many degrees are there in the sum of each pair of alter- nate interior angles ?
Ex. 30. The bisectors of two adjacent angles are perpendicular to each other. What is the relation of the given angles to each other ?
Ex. 31. Two lines are cut by a transversal. In what direction do they extend with reference to each other, if the alternate exterior angles are equal ?
PLANE GEOMETRY. — BOOK L 35
Proposition XV
80. Draw a straight line ; also two other lines each parallel to the given line. In what direction do these two lines extend with reference to each other?
Theorem, Straight lines which are parallel to the same straight lUie are parallel to each other.
/K
Data: Any straight lines, as AB pG"
and CD, each parallel to another q
straight line, as EF.
To prove AB parallel to CB. e
ryn
'L
Proof. Draw any transversal, as KL, cutting the lines AB, CD, and EF.
|
Since, data, |
CD II EF, |
|
73, |
Z.r=As. |
|
Since, data, |
AB II EF, |
|
73, |
At = Z.s. |
|
Then, Ax. 1, |
/.r = /.t. |
|
Hence, § 77, |
AB II CD. |
|
Therefore, etc. |
Q.E.p.
Ex. 32. The straight lines AB, CD, and EF are cut in Cr, H, and J respectively, by ifZ;. angle KGB = S1°; angle i^JTC = 149° ; angle FJL = 143°. Are the lines AB and CD parallel ? AB and EF ? CD and EF ?
Ex. 33. Can two intersecting straight lines both be ptiallel to the same straight line ?
Ex. 34. How many degrees are there in the angle formed by the bisec- tors of two complementary adjacent angles ?
Ex. 35. If the line BD bisects the angle ABC, and EF is drawn through B perpendicular to BD, how do the angles CBE and ABF compare in size ?
Ex. 36. If a straight line is perpendicular to the bisector of an angle at the vertex, how does it divide the supplementary adjacent angle formed by producing one side of the given angle through the vertex ?
36 PLANE GEOMETRY. — BOOK L
Proposition XVI
81. 1. Construct two angles whose corresponding sides are parallel. How do the angles compare in size, if both corresponding pairs of sides extend in the same direction from their vertices ? If both pairs extend in opposite directions from their vertices ?
2. Discover whether it is possible for the angles to have their sides parallel and yet not be equal.
Theorem, Angles whose corresponding sides are parallel are either equal or supplementary.
Data: AB parallel to DE, and BC parallel to HF, forming the angles r, s, tf s', and t'.
To prove 1. Z.r = Zs, ot Z s'.
2. Zr and Zi or ZV supplemen- ^ V^' ^
tary. ^ ^
Proof. 1. Produce BG and EB, if necessary, to intersect as at G.
§76, Zr = Zv, and Zs = Z'y; .'. Ax. 1, Zr = Zs.
§59, Zs^Zs^', .-.Ax. 1, Zr = Zs'.
2. § 55, Zs + Z^ = 2rt.Z;
but Zr = Z8\
Zr-\-Zt = 2Tt.A.
Hence, § 32, Zr and Z t are supplementary; also, sincCj § 59, Z ^ = Z ^', Zr and Z t' are supplementary.
Therefore, etc. q.e.d.
82. Scholium. The angles are equals if both corresponding pairs of sides extend in the same or in opposite directions from their vertices ; they are supplementary, if one pair extends in the sa^ne and the other in opposite directions.
Ex. 37. If two straight lines are perpendicular each to one of two par- allel straight lines, in what direction do they extend with reference to each other ?
Ex. 38. How do lines bisecting any two corresponding angles, formed by parallel lines, cut by a transversal, lie with reference to each other ?
PLANE GEOMETRY. — BOOK L
87
Proposition XVII
83. 1. Construct two angles whose corresponding sides are perpen- dicular to each other. How do the angles compare in size, if both are acute ? If both are obtuse ?
2. Discover whether it is possible for the angles to have their sides perpendicular and yet not be equal.
Theore^n, Angles whose corresponding sides are perpen- dicular to each other are either equal or supplementary.
Data : AB perpendicular to DB, and CE perpendicular to FB, forming the angles r, s, and t.
To prove 1 / r = Z s.
2. Z t and Z s supplementary.
Proof. 1. § 26, Aabd and GBF are vt. A\ .'. § 52, Z ABD = Z CBF,
and. Ax. 9, Zr + Zv = Zv + Zs.
Taking Z v from each member of this equation, Ax. 3, Zr = Z 8.
2. §§ 55, 32, Z t and Z r are supplementary ; but Zr = Zsy
hence, Z ^ and Z s are supplementary.
Therefore, etc. q.e.d.
84. Sch. The angles are equal, if both are acw^e or if both are obtuse ; they are supplemerdaryy if OTie is acwie and the other obtuse.
Ex. 39. The bisectors of two adjacent angles form an angle of 45°. What is the relation of the given angles to each other ?
Ex. 40. Two angles are supplementary, and the greater is five times the less. How many degrees are there in each angle ?
Ex. 41. Two angles are complementary, and the greater is five times the less. How many degrees are there in each angle ?
Ex. 42. Two parallel straight lines are cut by a transversal so that one of the two interior angles on one side of the transversal is eleven times the other. How many degrees are there in each of the exterior angles ?
88
PLANE GEOMETRY. — BOOK I.
TRIANGLES
85. A portion of a plane bounded by three straight lines is called a Plane Triangle, or simply a Triangle.
The straight lines which bound a triangle are called its sides, their sum is called its perimeter, and the vertices of the angles of a triangle are called the vertices of the triangle.
86. The angle formed by any side of a triangle and the prolongation of another side is called an Exterior Angle of the triangle.
Angle s is an exterior angle.
87. An angle formed within a triangle by any two of its sides is called an Interior Angle of the triangle.
Whenever the angles of a triangle or other enclosed figure are mentioned, the interior angles are referred to unless otherwise specified.
Angles ^, C, and r are interior angles.
88. The interior angles which are not adjacent to the exterior angles are called Opposite Interior Angles.
Angles A and C are opposite interior angles when s is the exterior angle.
89. A triangle whose three sides are unequal is called a Scalene Triangle.
90. A triangle two of whose sides are equal is called an Isosceles Triangle.
91. A triangle whose three sides are equal is called an Equilateral Triangle.
PLANE GEOMETRY. — BOOK I.
39
92. The side upon which a triangle is assumed to stand is called the Base of the triangle.
See figure accompanying § 95.
93. The angle opposite the base of a triangle is called the Vertical Angle, and its vertex is called the Vertex of the triangle.
94. The perpendicular distance from the vertex of a triangle to its base, or its base produced, is called the Altitude of the triangle.
Since any side of a triangle may be considered as its base, it is evident that a triangle may have three altitudes and that they will be unequal, if the sides of the triangle are unequal. If the triangle is equilateral, then all three altitudes will be equal ; if the triangle is isosceles, only two of the altitudes will be equal.
95. A triangle, one of whose angles is a right angle, is called a Right Triangle.
In a right triangle, the side opposite the right angle is called the hypotenuse.
96. A triangle, one of whose angles is an obtuse angle, is called an Obtuse Triangle.
97. A triangle, each of whose angles is an acute angle, is called an Acute Triangle.
Obtuse triangles and acute triangles are called oblique triangles.
98. A triangle whose three angles are equal is called an Equl angular Triangle.
See figure accompanying § 91.
99. A line drawn from any vertex of a tri- angle to the middle of the opposite side is called a Median, or Median Line of the triangle.
40 PLANE GEOMETRY. — BOOK L
Proposition XVIII
100. 1. Make two triangles such that two sides of one, and the angle formed by them, shall be equal to the corresponding parts of the other. How do the triangles compare ? How do the third sides compare ? How do the angles of one compare with the corresponding angles of the other?
2. Under what conditions are two triangles equal?
Theorem, Two triangles are equal, if two sides and the included angle of one are equal to two sides and the in- cluded angle of the other, each to each.
A B D E
Data : Any two triangles, as ABC and DEF, in which AB = DE^ AC = DF, and angle A = angle D.
To prove triangles ABC and BFF equal.
Proof. Place A ABC upon A DEF, AB coinciding with DE.
Data, Za — Zb,
hence, AC will take the direction of BF;
and since AC = BF,
the point C will fall upon the point F.
Since the point B falls upon E and the point C upon Fj BC will coincide with EF.
Then, A ABC and BEF coincide in all their parts.
Hence, § 36, Aabc = A BEF.
Therefore, etc. q.e.d.
Prove Prop. XVIII
(1) when AB — BEy BC ~ EF, and angle B = angle E.
(2) when AC = BF, BC = EF, and angle C = angle F.
101. Sch. Every triangle has six parts or elements ; namely, three sides and three angles. Two equal triangles may be made to coincide in all their parts. Therefore, each part of one is equal to the corresponding part of the other.
PLANE GEOMETRY.— BOOK I. 41
Proposition XIX
102. 1. Make two triangles such that a side of one, and the angles formed at its extremities, shall be equal to the corresponding parts of the other. How do the triangles compare? What parts are equal?
2. Under what conditions are two triangles equal ?
Theorem, Two triangles are equal, if a side and two adjacent angles of one are equal to a side and two adja- cent angles of the other, each to each.
A B D E
Data: Any two triangles, as ABG and DEF, in which AB = DE, angle A = angle D, and angle B = angle E.
To prove triangles ABC and DEF equal.
Proof. Place A ABC upon A DEF, AB coinciding with DE.
Data, /.A=^Ad;
hence, AC will take the direction of DF,
and the point C will fall upon DF, or upon DF produced.
Also, data, /.B = /.E\
hence, BG will take the direction of EF,
and the point C will fall upon EF, or upon EF produced.
Since the point C falls upon each of the lines DF and EF, it must fall upon their point of intersection, F.
Then, A ABC and D^i^ coincide in all their parts.
Hence, § 36, A ^^C = A DEF.
Therefore, etc. q.e.d.
Prove Prop. XIX
(1) when angle C = angle F, angle B = angle E, and BC = EF.
(2) when angle A = angle D, angle C = angle F, and AC = DF.
Ex, 43. Are two triangles equal, if the three angles of one are equal to the three angles of the other, each to each ?
Ex. 44. Can two triangles, having two sides and an angle of one respec- tively equal to two sides and an angle of the other, be unequal ?
42
PLANE GEOMETRY.^ BOOK I.
Proposition XX
103. 1. Draw a straight line and a perpendicular to that line at its middle point ; select any point in the perpendicular and from that point draw straight lines to the extremities of the given line. How do these lines compare in length ? How do the angles made by these lines with the perpendicular compare in size ? How do the angles made by these lines with the given line compare ?
2. Select any point not in the perpendicular and from that point draw straight lines to the extremities of the given line. How do they compare in length ?
3. Draw a straight line and find a point equidistant from its ex- tremities; find another point equidistant from its extremities; con- nect these points by a line and if necessary extend it until it intersects the given line. At what point does it intersect the given line ? What kind of angles does it make with the given line ?
4. What line contains every point that is equidistant from the ex- tremities of a straight line?
Theorem, If a perpendicular is drawn to a straight line at its middle point,
1, Any point in the perpendicular is equidistant from the extremities of the line.
2, Any point not in the perpendicular is unequally dis- tant from the extremities of the lint*
Data : Any straight line, as ^5 ; a per- pendicular to it at its middle point, as CD ; any point in CD, as E ; and any point not in CD, as F,
To prove
1. E equidistant from A and 5.
2. F unequally distant from A and B,
Proof. 1. Draw AF, BE, AF, and BF.
Data and § 26, Z r and Z s are rt z1^ and, §52, /Lr Z«.
In A ABE and BBE, AD = J57),
ED is common, and Z r = Z 5 ;
.-.§100, ^ADE=^l^BDE,
PLANE GEOMETRY. — BOOK I. 43
and, § 101, AE = BE.
That is, E is equidistant from A and B,
2. From the point G, where AF cuts CD, draw GB,
Ax. 10, BF <BG + GF]
but BG = AG'j . Why?
.-. substituting AG for its equal BG,
BF <AG + GF,
or ^i^ < AF.
That is, 2^ is unequally distant from A and B»
Therefore, etc. q.e.d.
104. Cor. I. Every point that is equidistant from the extremi- ties of a straight line lies in the perpendicular at the middle point of that line.
105. Cor. II. If a perpendicular is erected at the middle point of a straight line, the lines joining the extremities of this line with any point in the perpendicular make equal angles with the line and also with the perpendicular. § 101
106. Cor. III. Two points each equidistant from the extremities of a straight line determine the perpendicular at the middle point of that line.
Ex. 45. How does the distance between two parallel lines at a given point compare with tlie distance between them at any other point ?
Ex. 46. Can two angles which are not adjacent have a common vertex and a common side ?
Ex. 47. If in an equilateral triangle a line is drawn from the vertex to the middle point of the base, how do the triangles thus formed compare in size ?
Ex. 48. If two lines bisect each other, in what direction do the lines joining their opposite extremities extend with reference to each other ?
Ex. 49. If two sides of a triangle are equal, and a line is drawn bisecting their included angle and intersecting the third side, how do the segments of the third side compare in length ?
Ex. 50. Perpendiculars are erected at the extremities of a line and termi- nate in any bisector of the line that is not perpendicular to the line. How do the perpendiculars compare in length ?
Ex. 51. If through the middle point of a straight line terminating in two parallel lines, a second straight line is drawn also terminating in the parallels, how do the parts of the second line compare in length ?
44
PLANE GEOMETRY. — BOOK L
Proposition XXI
107. 1. Make two triangles such that the sides of one shall be equal to the corresponding sides of the other. How do the triangles compare ? How do the corresponding angles compare?
2. Under what conditions , are two triangles equal ?
Theorem, Two triangles are equal, if the three sides of one are equal to the three sides of the other, each to each.
Data : Any two triangles, as ABC and DEF, in which AB = DE^ AC = BF, and BC = EF. To prove triangles ABC and DEF equal.
Proof. Place A DEF in the position ABF so that the «qual sides, BE and AB, coincide, and the vertex F falls oppo&ite C. Draw CF.
Data, AF = AC, and BF = BC;
A and B are each equidistant from F and C; hence, § 106, AB ± CF Sit its middle point,
and, § 105, Zr = Zs.
In A ABC Sind ABF, AC = AF,
AB is common,
Zr = Zs; A ABC = A ABF. AABC = ABEF,
and
.-. § 100,
That is,
Therefore, etc.
Prove by placing the triangle so that
(1) BF will coincide with AC.
(2) EF will coincide with BC.
Q.E.D
PLANE GEOMETRY.^ BOOK L 45
108. Sch. It is evident that in equal triangles the parts which are similarly situated are equal ; that is, the angles included by the equal sides are equal; the angles opposite the equal sides are equal, the sides included between equal angles are equal, and the sides opposite the equal angles are equal.
109. In equal figures, the parts which are similarly situated are called Homologous parts.
Ex. 52. Draw two parallel lines intersecting two parallel lines, and draw a line joining two opposite points of intersection. How do the triangles thus formed compare ?
Ex. 53. Perpendiculars are drawn from the extremities of a line to any line that bisects it and is not perpendicular to it. How do the perpendiculars compare in length ?
Ex. 54. The line BD is the bisector of the angle ABC whose sides are equal. Lines are drawn from any point of BD^ as ^, to A and G. How do AE and CE compare in length ?
Ex. 55. In a triangle ABC angle A equals angle B ; a line parallel to AB intersects AC m D and BC in E. How do the angles ADE and BED com- pare ?
Ex. 56. If D is the middle point of the side BC of the triangle ABC, and BE and CF are perpendiculars from B and 0 to AD^ or AD produced, how do BE and GF compare in length ?
Proposition XXII
110. 1. Cut out a paper triangle ABC. Cut off the corners and place the vertices A, B, and C together. To how many right angles is the sura of the three angles equal ?
2. If one angle of a triangle is a right angle, how does the sura of the other two angles compare with a right angle?
3. What is the greatest nuraber of obtuse angles that a triangle raay have? The greatest nuraber of right angles?
4. If there are two triangles such that the sum of two angles of one is equal to the sum of two angles of the other, how do the third angles com- pare in size ?
5. If there are two right triangles such that a side and an acute angle of one are equal to the corresponding parts of the other, how do the triangles compare?
6. Extend one side of a triangle through a vertex ; through the same vertex draw a line parallel to the opposite side of the triangle. Since the figure thus formed contains two parallel lines and a transversal, M'hat angles of the figure are equal? How do3s the exterior angle of the tri- angle compare with the sum of the two opposite interior angles?
46 PLANE GEOMETRY. — BOOK 1.
Theorem, The sum of the angles of a triangle is equal to two right angles.
o
A B ~ ~~~ "d
Datum : Any triangle, as ABC.
To prove Zr -{- Z.s-\- Zt = two right angles.
Proof. Produce AB to D and draw BE W AC.
§ m, Zr-\-Zs' + Zt' = 2it.A',
but, § 73, Zs' = Zs,
and, § 76, . Zt' = Zt',
.'. substituting Zs and Zt for Zs' and Z ^' in the first equation, Zr + Zs + Zt = 2Tt. A.
Therefore, etc. q.e.d,
111. Cor. I. In a right triangle the sum of the two acute angles is equal to a nght angle.
112. Cor. II. A triangle cayinot have more than one right angl^, nor more than one obtuse angle.
113. Cor. III. If two angles of one triangle are equal to tn'^i angles of another, the third angles are equal.
114. Cor. IV. Two right triangles are equal, if a side and an acute angle of one are equal to a side and an acute angle of the of^^er, each to each.
115. Cor. V. Any extenor angle of a triangle is equal to the sum of the tico opposite interior angles.
1. What interior angle is equal to Zt'? Why ?
2. What interior angle is equal to Z s' ? Why ?
3. To what, then, is the whole exterior angle equal ?
Ex. 57. May a triangle be formed whose angles are 93°, 40°, and OP re- spectively ? 08°, 24°, and 58° ? 57°, 49°, and 74° ?
Ex. 58. Two angles of a triangle are together equal to 76°. What ia the value of the third angle ?
PLANE GEOMETRY. — BOOK I
47
Ex. 59. Show by each of the-foUowing figures that the sum of the three angles of a triangle is equal to two right angles, assuming that the construc- tion lines are drawn as they appear to be drawn.
Proposition XXIII
116. 1. Draw an isosceles triangle. How do the angles opposite the aqxial sides compare in size ?.
2. How do the angles of an equilateral triangle compare?
Theorem, In an isosceles triangle the angles opposite the equal sides are equal.
Data : Any isosceles triangle, as ABCy in which To prove angle A = angle B.
Proof. Draw CD bisecting Z C. Then, in A ADC and BDC,
data,
and, const., .-. § 100, and, § 108, Therefore, etc.
AC=BOy
CD is common,
Aadc = Abdc, Aa = /.b,
Q.E.D.
117. Cor. An equilateral triangle is also equiangular.
48
PLANE GEOMETRY. — BOOK L
Proposition XXIV
118. 1. Draw a triangle such that two of its angles are equal. How do the sides opposite these angles compare in length ? What kind of a triangle is it?
2. How do the sides of an equiangular triangle compare in length ?
Theorein, If two angles of a triangle are equal, the sides opposite the equal angles are equal and the triangle is isos- celes.
c
Data: Any triangle, as ABC, having
angle A = angle B. To prove AC = BC, and triangle ABC isosceles.
Proof. Draw CD bisecting Z C. Then, in A ADC and BDC,
data,
const.,
.-. § 113,
and, since
§ 102,
and, § 108, Hence, § 90, Therefore, etc.
Za = Zb, Zr^Zs;
Zt = Zv;
CD is common,
AADC = ABDC,
AC = BC.
A ABC is isosceles.
Q.S.P,
119. Cor. An equiangular triangle is also equilateral
Ex. 60. If equal distances from the vertices of an equilateral triangle are laid off on its sides in the same order, what kind of a triangle do the lines joining these points form ?
Ex. 61. From the extremities of the base of an isosceles triangle perpen- diculars are drawn to the opposite sides ; the points where the perpendiculars meet the opposite sides are joined by a straight line. What is the direction of this hne with reference to the base ?
Ex. 62. The base of an isosceles triangle is 6 inches and the opposite angle is 60°. How many degrees are there in each of the base angles? What is the length of each of the other two sides?
PLANE GEOMETRY. — BOOK I
49
Proposition XXV
120. 1. Draw an isosceles triangle and a line bisecting its vertical angle. How does this line divide the base ? What kind of angles does it form with the base ?
2. Draw a line perpendicular to the base of an isosceles triangle at its middle point. How does it divide the triangle ? How does it divide the vertical angle ?
3. Draw a line from the vertex of an isosceles triangle and perpen- dicular to the base. How does this line divide the base of the triangle ? How does it divide the vertical angle?
Theorem. The bisector of the vertical angle of an isos- celes triangle is perpendicular to the hose at its middle point.
Data: Any isosceles triangle, as ABC^ in which AG — BGy and CD bisects the angle C.
To prove CB perpendicular to J 5 at its mid- dle point.
Proof. Data, AC—BC^
AA — ^B\
Aadc=i^bdc,
AD^BD'y D is the middle point of AB j
CD ± AB,
and, § 116, .-. § 102, and, § 108, that is, also, hence, § 26,
Therefore, etc. * Q.B.D.
121. Cor. I. A perpendicular which bisects the base of an isos- celes triangle bisects the vertical angle,
122. Cor. II. A line perpendicular to the base of an isosceles triangle and passing through the vertex bisects both the base and the vertical angle of the triangle.
Ex. 63. How do the lines joining the extremities of the bases of two opposite, or vertical, isosceles triangles compare in length ?
MILNE^S GEOM. 4
50
PLANE GEOMETRY. — BOOK /.
Proposition XXVI
123. Draw two right triangles such that the hypotenuse and a side of one shall be equal to the corresponding parts of the other. How do the triangles compare ?
Theorem, Two right triangles are equal, if the hypote- nuse and a side of one are equal to the hypotenuse and a side of the other, each to each.
Data: Any two right triangles, as ABC and bef^ in which the hypotenuse AC = the hypotenuse BF, BC = EF, and angles r and 5 are the right angles.
To prove triangles ABC and BEF equal.
Proof. Place A BEF in the position BBC so that the equal sides EF and BC coincide, and the vertex B falls opposite A.
Data, Z r and Z s are rt. A ;
then, Z r + Z s = 2 rt. ^,
and, § 58, AB and BB form one straight lina
Data, AC = BC',
.'. §90, A ^DC is isosceles.
Hence, in ^ ABC and BBC,
AG^BCy
Z.A=ABy
§116, and, § 113, .-. § 102,
That is.
Therefore, etc.
Aabc = Abbc. aabc = abef.
Q.E.D.
Ex. 64. The median line from the vertex to the base of a certain triangle is equal to one half the base. What kind of an angle is the vertical angle ?
PLANE GEOMETRY.— BOOK L 51
Proposition XXVII
124. 1. Draw any triangle. How does any side compare in length
with the sum of the other two sides?
2. How does the sum of any two sides compare with the third side?
Theorem, Any side of a triangle is less than the sum of the other two sides.
Data : Any triangle, as ABC, and any side,
as AC.
To prove AC less than AB 4- BC.
A B
Proof. By Ax. 10, the straight line AC, which is a side of the triangle, is the shortest distance between the points A and C.
Hence, AC is less than the broken line ABC which joins the points A and C.
That is, AC is less than AB -f- BC.
Therefore, etc. « q.e.d.
125. Cor. The sum of any two sides of a triangle is greater than the third side.
Ex. 65. May a triangle he formed with lines 4, 2, and 3 inches long ? With lines 6, 1, and 2 inches long ? 5, 2, and 3 inches long?
Ex. 66. If a line is drawn joining the middle points of the equal sides of an isosceles triangle, what kind of a triangle is formed ?
Ex. 67. If the bisectors of the base angles of an isosceles triangle are produced to the opposite sides, how do they compare in length ?
Ex. 68. The sum of the two angles at the base of an isosceles triangle is 64°. What is the value of each angle of the triangle ?
Ex. 69. If the straight line which joins the vertex of a triangle with the middle point of the base is perpendicular to the base, what liind of a tri- angle is it ?
Ex. 70. The perpendicular distance between two parallel lines is 20 inches, and a line is drawn across the parallels making an angle of 45° with the perpendicular at its upper extremity. What distance does this line cut off from the foot of the perpendicular ?
Ex. 71. If from a point within a right angle perpendiculars are drawn to the sides containing the right angle and each perpendicular is produced its own length, what kind of a line will join the extremities of the produced lines and the vertex of the right angle ?
62 PLANE GEOMETRY. — BOOK L
Proposition XXVIII
12^6. Draw a scalene triangle. Where is the smaller angle situated with reference to the shorter side ? Where is the greater angle situated with reference to the greater side ?
Theorem, If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is op- posite the greater side.
Data: Any triangle, as ABC, in which AB is greater than BC.
To prove angle ACB, opposite AB, is greater than angle A, opposite BC.
Proof. On BA take BF equal to BC, and draw CF. Ax. 8, /.ACB is greater than Z BCF\
but, § 116, Z BCF=Z. BFC;
Z ACB is greater than Z BFC. But since, § 115, Z BFC = Za + Zacf,
Z BFC is greater than Z A. Then, Zacb, the angle opposite AB, is greater than Z A, the angle opposite BC.
In like manner, ii AB is greater than AC, Z ACB may be proved greater than Z B,
Therefore, etc. q.e.d.
Prove Z ACB greater than Z B when AB is greater than AC.
Ex. 72. How do the angles of a scalene triangle compare ?
Ex. 73. What is the value of the angle formed by the bisectors of the acute angles of a right triangle ?
Ex. 74. How many degrees are there in an angle of an equilateral triangle ?
Ex. 75. How many degrees are there in each of the equal angles of an isosceles triangle, the angle at the vertex being 35° 50' ?
Ex. 76. In an isosceles triangle one base angle is 35°. What is the value of the vertical angle ?
Ex. 77. AD is the bisector of a base angle of the isosceles triangle ABC, the bisector meeting the side BC in 7> ; the vertical angle C is 28°. How many degrees are there in angle ADC ?
PLANE GEOMETRY.— BOOK I 53
Proposition XXIX
127. 1. Draw any triangle. Where is the shorter side situated with reference to the smaller angle ? Where is the greater side situated with reference to the greater angle ?
2. Which side of a right triangle is the greatest ?
Theorem, If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle. (Converse of Prop. XXVIII)
Data: Any triangle, as ABC, in which ^
angle ABC is greater than angle A.
To prove AC^ opposite angle ABC, greater than BC, opposite angle A.
Proof. Draw BD so that Z ABB = Z BAD. Then, § 118, AD = BD.
In A BCD, % 125, BD-\-DC>BG.
Substituting AD for its equal BD,
AD-\-DC>BC, or AC>BC.
That is, AC, opposite angle ABC, is greater than BC, opposite angle A.
In like manner, if angle ABC is greater than angle C, AC may- be proved greater than AB.
Therefore, etc. q.e.d.
Prove that ^c is greater than AB when Zabc is greater than Zc.
128. Cor. The hypotenuse is the greatest side of a right triangle.
Ex. 78. The angles A, B, and C of the triangle ABC &re 40°, 60°, and 80° respectively. How do ^O and AB compare in length ? AB and BC? ^C and 50?
Ex. 79. CD bisects the base of an isosceles triangle ABC, Si base angle of which is 55°. How many degrees are there in angle ADC ? In angle CDB ? In angle ACD? In angle DCB?
6i
PLANE GEOMETRY. — BOOK 1.
Proposition XXX
129. Construct two triangles having two sides of one equal to two sides of the other, and the angles included between the sides unequal. How do the third sides compare in length? Which triangle has the greater third side?
Theorem. If two sides of one triangle are equal to two sides of another, each to each, and the included angles are unequal, the remaining sides are unequal, and the greater side is in the triangle which has the greater included angle.
Data: Any two triangles, as ABC and DBF, in which AC — DF^ BC = EF, and angle ACB is greater than angle F. To prove AB greater than DE.
Proof. Of the two sides DF and EF, suppose that EF is the side which is not greater. Place Abef in the position BBC so that the equal sides, EF and BC, coincide. Draw CH bisecting Z ACB and draw BH. In A AHC and BHC, AC = BC,
CH is common,
Zt = Zs; A AHC = A BHC,
AH = BH. BH + HB > BB.
and, const., .-. § 100, and, § 108,
In A BHB, § 125,
Substituting AH for its equal BH,
AH+ HB>BB\ that is, AB > BB or BE.
Therefore, etc.
Q.E.D.
PLANE GEOMETRY. — BOOK I. 55
Proposition XXXI
130. Construct two triangles that have two sides of one equal respec« tively to two sides of the other, but the third sides unequal. How do the angles opposite the third sides compare in size?
Theorem, If two sides of one triangle are equal to two sides of another, each to each, and the third sides are un- equal, the angles opposite the third sides are unequal, and the greater angle is in the triangle which has the greater third side. (Converse of Prop. XXX.)
Data: Any two triangles, as ABC and DEF, in which AC = DF, BC = EF, and AB is greater than DE.
To prove angle c greater than angle F.
Proof. Since AC = DF,
and BC = EF,
if Zc = Zf,
then, § 100, A ABC = A DEFy
and, § 108, AB = DE,
which is contrary to data.
If Z (7 is less than Z F,
then, Z i^ is greater than Z C,
and, § 129, DE> AB,
which is also contrary to data.
Therefore, both hypotheses, namely, that Z.C — /.F, and that Z c is less than Z F, are untenable.
Consequently, Z C is greater than Z F.
Therefore, etc. q.e.d.
Ex. 80. If one angle of a triangle is equal to the sum of the other two, what is the value of that angle ? What kind of a triangle is the triangle ?
Ex, 81. AT) is perpendicular to 50, one of the equal sides of the isosceles triangle ABC whose vertical angle is 30°. How many degrees are there in each of the angles CAB, DAB, and ABC^
66
PLANE GEOMETRY. — BOOK L
Proposition XXXII
131. Choose some point within any triangle and from it draw lines to the extremities of one side. How does the sum of these lines com- pare with the sum of the other two sides of the triangle?
Theorem., The sum of two lines drawn from a point within a triangle to the extremities of one side is less than the sum of the other two sides.
Data: Any triangle, as ABC-, any point within it, as D ; and the two lines, AD and BB, drawn from D to the extremities of AB.
To prove AD -\- BD less than AC -\-BC.
Proof. Produce AD to meet BC in E.
In A AEC, § 124, AE<AC+ CE.
Adding BE to both members of this inequality, Ax. 4, AE -\-BE<AC-\-CE -\- BE,
or AE -\-BE <AC -^BC.
In A DBEj BD <DE -\- BE.
Adding AD to both members of this inequality, Ax. 4, AD -^ BD < AD -\- DE ■\- BE,
or AD-^BD<AE-\- BE.
It has been shown that
ae-\-'be<ac + bc\ hence, AD + BD < AC -[■ BC.
Therefore, etc. q.e.d.
Why?
Proposition XXXIII
132. 1. Draw a straight line and a perpendicular to it; select a point in the perpendicular, and from that point draw two oblique lines meet- ing the given line at equal distances from the foot of the perpendicular. How do the oblique lines compare in length ?
2. Draw oblique lines from that point to points unequally distant from the foot of the perpendicular. How do they compare in length ? Which is the greater ?
PLANE GEOMETRY. — BOOK I. 57
3. Draw two unequal lines from that point to the given line. Which one meets the line at the greater distance from the foot of the perpen- dicular ?
4. How many equal straight lines can be drawn from a point to a straight line?
Theorem. If from a point in a perpendicular to a given straight line, oblique lines are drawn to the given line,
1. The ohlique lines which meet the given line at equal distances from the foot of the perpendicular are equal.
2. Of oblique lines which meet the given line at unequal distances from the foot of the perpendicular the more remote is the greater.
Data : Any straight line, as AB ; any perpendicular to AB, as PB ; and any point in PB, as C, from
which oblique lines, as CE, CF, y^
and (JG, are drawn meeting AB so // that BE = BF, and /) G is greater ^ gx '\e —
than BE. ^^<\ I
To prove 1. CE = CF. ' <\ j
2. CG greater than CE. ''"-^H
Proof. 1. Data, CB X EF at its middle point.
Then, § 103, CE = CF.
2. Produce CB to H, making BH = CB ; draw EH and GH.
Then, data and const., AB J. CH Sit its middle point.
.-. § 103, EH= CE, and GH= CG\
hence. Ax. 2, CE + EH = 2 CE, and CG -\- GH = 2 CG.
But, § 131, CE -\-EH<CG-\-GH',
2CE <2 CG, ov CE <CG\ that is, CG > CE.
Therefore, etc. q.e.d.
133. Cor. Only tivo equal straight lines can he drawn from a point to a straight line; and of two unequal lines the greater cuts of the greater distance from the foot of a perpendicular drawn to the line from the given point.
68 PLANE GEOMETRY. — BOOK I.
Proposition XXXIV
134. Bisect any angle ; from any point in the biseclor draw lines pe? pendicular to the sides of the angle. How do the perpendiculars com- pare in length ? How do the distances of the point from the sides ot the angle compare ?.
Theorem, Every point in the bisector of an angle is equidistant froin the sides of the angle-,
/A
Data: Any angle, as ABC, and any point in e/ ^
its bisector BB, as .F. / :^<p
To prove F equidistant from AB and CB. >^^^^^ ■
B Q c
Proof. Draw the perpendiculars fe and FG representing the distances of the point F from AB and CB respectively.
§ 26, Z r and Z s are rt. A.
Then, in the rt. A BFE and BFG,
BF is common, and, data, /.t = /.v',
.-. § 114, A BFE = A BFQ,
and, § 108, fe = fG',
that is, F is equidistant from AB and CB.
Therefore, etc. q.e.d.
Ex. 82. The perpendicular let fall from the vertex to the base of a tri- angle divides the vertical angle into two angles. How does the difference of these angles compare with the difference of the base angles of the triangle ?
Ex. 83. ^^a is a triangle. Angle A = 60°, angle B = 40°. The bisector of angle A is produced until it cuts the side BC. How many degrees are there in each angle thus formed ?
Ex. 84. A perpendicular is let fall from one end of the base of an isos- celes triangle upon the opposite side. How does the angle formed by the perpendicular and the base compare with the vertical angle ?
Ex. 85. If an angle of a triangle is equal to half the sum of the other two, what is tlie value of that angle ?
Ex. 86. How does the sum of the lines from a point within a triangle to the vertices of the triangle compare with the sum of the sides of the tri- angle ? With half that sum ?
PLANE GEOMETRY.-^BOOK 1, 69
Proposition XXXV
135. Within an angle select any number of points that are each equi- distant from its sides. Will the lines joining these points form a straight line? How will it divide the angle?
Theorem, Every point within an angle and equidistant from its. sides lies in the bisector of the angle. (Converse of Prop. XXXIV.)
/A
Data: Any angle, as ABC, and any point /
within the angle equidistant from AB and GB, r^^^-- ^^^
as i^. / ^^^
To prove F is in the bisector of the angle A> si
ABC. * ° "
Proof. Through the point F draw BB\ also draw the perpen- diculars FE and FG representing the distances of the point F from AB and GB respectively.
Then, § 26, Z r and Z s are rt. A
In the rt. A BFF and BGF,
BF is common, and, data, FE = FG ;
.-. § 123, A BEF = A BGFf
and, § 108, Zt = Zv',
that is, ^D is the bisector of Z ABC.
Hence, F is in the bisector of Z ABC.
Therefore, etc. q.e.d.
Ex. 87. ABC is an isosceles triangle having a vertical angle of 30°. From each extremity of the base perpendiculars are drawn to the opposite sides. What angles are formed at the intersection of these perpendiculars ?
Ex. 88. The exterior angle at the vertex of an isosceles triangle is 110°. How many degrees are there in each angle of the triangle ?
Ex.89. The exterior angle at the base of an isosceles triangle is 110°. How many degrees aa-e there in each angle of the triangle ?
Ex. 90. The angle C at the vertex of the isosceles triangle ABC is one fourth of the exterior angle at C. How many degrees are there in angle A f In the exterior angle at B?
Ex. 91. How does the angle formed by the bisectors of the base angles of an isosceles triangle compare with an exterior angle at the base ?
60
PLANE GEOMETRY. — BOOK L
QUADRILATERALS
136. A portion of a plane bounded by four straight lines is called a Quadrilateral.
137. A quadrilateral which has no two sides parallel is called a Trapezium.
138. A quadrilateral which has only two sides parallel is called a Trapezoid.
The parallel sides of a trapezoid are called its bases.
139. A trapezoid whose non-parallel sides are equal is called an Isosceles Trapezoid.
140. A quadrilateral whose opposite sides are parallel is called a Parallelogram.
141. A parallelogram whose angles are right angles is called a Rectangle.
142. A parallelogram whose angles are oblique angles is called a Rhomboid.
143. An equilateral rectangle is called a Square.
144. An equilateral rhomboid is called a Rhombus.
145. The straight lines which join the vertices of the opposite angles of a quadrilateral are called Diagonals.
146. The side upon which a figure is assumed to stand is called the Base.
The side upon which a trapezoid or a parallelogram is assumed to stand is called its lower hose, and the side opposite is called its upper base.
147. The perpendicular distance between the bases of a trape- zoid or of a parallelogram is called its Altitude.
PLANE GEOMETRY. — BOOK L 61
Proposition XXXVI
148. 1. Draw a quadrilateral whose opposite sides are equal. What kind of a quadrilateral is it ?
2. How do the opposite angles of a parallelogram compare in size?
Theorem, If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.
D
Data: Any quadrilateral, as ABCD, in which AB = DC and AD = EC.
To prove ABCD di. parallelogram.
Proof. Draw AC.
Then, in the A ABC and ADC, ^
data, AB =DC, BC = AD,
and ^(7 is common;
.-.§107, AABC = l\ADC,
§108, Zr=Z^, andZs = Zv;
.-. § 75, AB II DC, and AD II BC.
Hence, § 140, ABCD is a parallelogram.
Therefore, etc. q.e.d.
149. Cor. The opposite angles of a parallelogram are equal.
Z.r=^Z.t and Z <; = Z s ;
Ex. 92. If lines are drawn joining in succession the middle points of the sides of a square, what figure will be formed ?
Ex. 93. To how many right angles is the sum of the angles of a parallelo- gram equal ? To what is the sum of any two angles of a parallelogram, which 'are not opposite, equal ?
Ex. 94. If medians are drawn from two vertices of a triangle and each is produced its own length, what kind of a line will join the extremities of the produced medians and the other vertex of the triangle ?
62 PLANE GEOMETRY.— BOOK /.
Proposition XXXVII
150. 1. Draw a quadrilateral having two of its sides equal and pa^ allel to each other. What kind of a quadrilateral is it ?
2. Draw two parallel lines and two parallel transversals. How do the segments of the transversals between the parallel lines compare in length?
Theorem, If two sides of a quadrilateral are equal and parallel, the figure is a parallelogram.
Data: Any quadrilateral, as ABCD^ in which two of the sides, <is AB and DC, are equal and parallel. .
To prove ABCD Si parallelogram.
Proof. Draw AC.
In the A ABC and ADC, data, AB — DC,
^C is common, and, § 73, Zr = /.t',
.-.§100, AABC = AADC,
and Zs = Zv.
Hence, § 75, BCWAD,
and since, data, AB \\DC,
§ 140, ABCD is a parallelogram.
Therefore, etc. q.e.d.
151. Cor. Parallel lines intercepted betiveen parallel lines are equal, and parallel lines are everywhere equally distant.
Ex. 95. If the sides of a parallelogram are bisected and these middle points joined in succession, what figure is formed by the connecting lines ?
Ex. 96. If the four interior angles formed by a transversal crossing two parallel lines are bisected and the bisectors produced until they meet, what figure will be formed ?
Ex. 97. If a line is drawn through the vertices of two isosceles triangles on the same base, how does it divide the base ?
Ex. 98. If two equal straight lines are drawn from a point to a line, how do the angles formed with the given line compare ?
Ex. 99. If lines are drawn from the vertex of an isosceles triangle to points in the base equally distant from its extremities, how do they compare in length ?
PLAkE GEOMETRY. —BOOK L
63
Proposition XXXVIII
152. 1. Draw a parallelogram and either diagonal. How do the tri- angles thus formed compare in size ?
2. How do the opposite sides of the parallelogram compare in length ?
Theorem. The diagonal of a parallelogram divides the figure into two equal triangles.
Data: Any parallelogram, as ABCD^ and one of its diagonals, as ^C.
To prove triangles ABC and ADC equal.
Proof. To be given by the student.
153. Cor. Tlie opposite sides of a parallelogram are equal.
Proposition XXXIX
154. Draw a parallelogram and its diagonals. How do the segments of each diagonal compare in length?
Theoretn. The diagonals of a parallelogram bisect each other.
•Data: Any parallelogram, as A BCD, and its diagonals, AO and BB, inter- secting at E.
To prove AE = CE and BE == BE.
Proof. In the A ABE and CDE, § 153, AB = DC,
§73, Z.r=/.t,
and Z s = Z v ;
.-. § 102, A ABE = A CDE,
§ 108, AE = CEj and BE = DE.
Therefore, etc. q.e.I).
Ex. 100. If the diagonals of a quadrilateral are equal and bisect each other, what kind of a figure is the quadrilateral ?
Ex. 101. From a figure representing a parallelogram and its diagonals, select four pairs of equal triangles.
64 PLANE GEOMETRY. — BOOK I.
Proposition XL
165. 1. Draw two parallelograms such that two sides of one, and the angle between them, shall be equal to the corresponding parts of the other. How do the parallelograms compare?
2. How do two rectangles compare, if the base and altitude of one are equal to the corresponding parts of the other?
Theorem, Two parallelograms are equal, if two sides and the included angle of one are equal to two sides and the included angle of the other, each to each.
Data: Any two parallelograms, as abqb and efoh, in which AB = EF, AD = EH, and angle A = angle E.
To prove parallelograms ABCD and EFGH equal.
Proof. Place EJ EFGH upon EJ ABCD so that EF coincides with its equal AB and Z E with its equal Z A.
Then, EH coincides with, its equal AD,
§ 70, HG takes the direction of DC,
and O falls upon DC, or upon DC produced.
Also, FG takes the direction of BC,
and O falls upon BC, or upon BC produced.
Since G falls upon both DC and BC, it must fall upon their point of intersection c, which is the only point common to DC and BC.
Hence, § 36, O ABCD = EJefgh.
Therefore, etc. q.e.d.
156. Cor. Two rectangles are equal, if the base and altitude of one are equal to the base and altitude of the other, each to each.
Ex. 102. From any point in the base of an isosceles triangle lines are drawn parallel to the equal sides and produced until they meet the sides of the triangle. How does the sum of these two lines compare with one of the equal sides of the triangle ?
PLANE GEOMETRY.— 'BOOK I.
65
Proposition XLI
157. Draw three or more parallel lines intercepting equal parts on a transversal; draw any other transversal. How do the parts which the parallels intercept on the second transversal compare in length ?
Theorefu, If three or more parallel lines intercept equal parts on any transversal, they intercept equal parts on every transversal.
Data : Any parallel lines, as AH, CM, EK, and GP, intercepting the equal parts AC, CE, and ^G^ on the transver- sal A G, and the parts HM, MK, and KP on any other transversal, as HP.
To prove HM = MK — KP.
Proof. Draw AB, CD, and EF each parallel to HP. Then, § 140, ABMH, CDKM, and EFPK are parallelograms, and, § 153, HM = AB, MK = CD, and KP = EF.
Now, in A ABC, CDE, and EFG, § 80, AB II CD WEF;
hence, § 76, Zr = Zs =Zt,
Z.v = Zw = Z.X, AC=CE = EG; A ABC = A CDE = A EFG, AB = CD = EF.
and, data, .-. § 102, and
But since HM = AB, MK = CD, and KP = EF, Ax. 1, HM — MK = KP.
Therefore, etc.
Q.E.D.
Ex. 103. In the triangle ABC angle A is double angle B and the ex- terior angle at C is 105°. How many degrees are there in angles A and B respectively ?
Ex. 104. If one angle of a parallelogram is a right angle, what is the value of each of the other angles ?
Ex. 105. One angle of a parallelogram is three times its supplement What is the value of each angle of the parallelogram ?
MTT.-VK'S aHOTW-
66
PLANE GEOMETRY. — BOOK I
Proposition XLII
158. 1. Draw a triangle and a line parallel to the base, bisecting one of the sides. How does it divide the other side? How does the part of this line intercepted by the sides of the triangle compare in length with the base of the triangle ?
2. Draw a triangle and a line connecting the middle points of two of its sides. What is the direction of this line with reference to the third side of the triangle?
Theorem, If a straight line drawn parallel to the base of a triangle hisects one of its sides, it bisects the other side, and is equal to one half of the base.
Data: Any triangle, as ABC, and a straight line DE drawn parallel to AB bisecting ^(7 at Z). d^ /^
To prove 1. BE =^ EC.
2. DE = i AB,
Proof. 1. Draw FD II BC. Then, in A DEC and AFD,
|
data, |
DC = AD, |
|
§76, |
Zdce = Zadf, |
|
and |
Zcde = Zdaf', |
|
.-.§102, |
A DEC = A AFD, |
|
and |
EC = FD. |
|
But, § 151, |
BE = FD; |
|
.-. Ax. 1, |
BE = EC. |
|
2. §108, |
AF = DE, |
|
and, § 151, |
FB =DE; |
|
.*. |
AF = FB=^AB; |
|
but |
FB =DE; |
|
hence, |
DE = \ AB, |
|
Therefore, etc. |
Why?
Q.E.D.
159. Cor. TJie line joining the middle points of two sides of a triangle is parallel to the third side.
PLANE GEOMETRY. — BOOK I. 67
For if the line is not parallel to the third side, suppose a line drawn through 2), the middle point of ACy parallel to AB. By § 158, it will pass through E, the middle point of BCj and we shall have two straight lines drawn between the same two points, which by Ax. 12 is impossible. Consequently, the line joining the mid- dle points ot two sides of a triangle is parallel to the third side.
Proposition XLIII
160. Draw a trapezoid and a line connecting the middle points of the non-parallel sides. What is the direction of this line with reference to iihe bases of the trapezoid? How does it compare in length with the sum of the bases ?
Theorem, The line which joins the middle points of the non-parallel sides of a trapezoid is parallel to the bases and is equal to one half their sum.
Data: Any trapezoid, as A BOB, and R ^r^c
the line EF joining the middle points / ^.'"''
of the non-parallel sides AB and B C. eA -^^
To prove EF parallel to AB and BC /.'-''
and equal to one half AB + BC. ^ ^
Proof. Draw AC intersecting EF at K, and from H, the middle point of AC, draw HE and HF.
Data, AE = EBy
and, const., AH = HC ;
.-. § 159, HE II BC,
and, § 158, ' HE — ^BC,
In like manner, HF II AB and HF = ^AB,
Then, § 80, hfWbc;
but HEWbC;
:, § 70, EHF is a straight line parallel to AB and DC,
But, data, EKF is a straight line,
then, EHF and EKF coincide,
and the point H coincides with the point K,
Now, HE, or KE = lBC,
and HF, or KF = ^AB -,
hence. Ax. 2, EF — ^ (AB -\- BC).
Therefore, etc. qM.D.
68 PLANE GEOMETRY. -BOOK L
POLYGONS
161. A portion of a plane bounded by any number of straight lines is called a Polygon.
The sum of the straight lines which bound a polygon is called its perimeter.
The term polygon is usually applied to figures of more than four sides.
162. A polygon of three sides is called a trigon or triangle ; one of four sides, a tetragon or quadrilateral ; one of five sides, a pentagon ; one of six sides, a hexagon ; one of seven sides, a heptagon ; one of eight sides, an octagon ; one of ten sides, a decagon; one of twelve sides, a dodecagon; one of fifteen sides, a pentadecagon.
163. A polygon such that none of its sides, if produced, extend within it is called a Convex Polygon.
164. A polygon such that two or more of its sides, if produced, extend within it is called a Concave Polygon.
The reflex angle ABCis called a re-entrant angle. Unless otherwise stated, polygons considered hereafter will be understood to be convex.
165. A straight line joining the vertices of two non-adjacent angles of a polygon is called a Diagonal of the Polygon.
Proposition XLIV
166. 1. Draw convex polygons, each having a different number of sides, and from any vertex of each draw its diagonals. How does the number of triangles into which each polygon is divided compare with the number of sides of the polygon?
To how many right angles is the sum of the angles of a triangle equal ? To how many times two right angles is the sum of the interior angles of a polygon equal ?
PLANE GEOMETRY. — BOOK I. 69
2. Produce the sides of any polygon in succession. To how many right angles is the sum of all the exterior and i:.terior angles equal? To how many right angles is the sum of the exterior angles of a polygon equal ?
Theorem, The sum of the angles of any eonvejc polygon is equal to twice as many right angles as the polygon has sides less two. .
D
Data: A convex polygon of any number (n) of sides, as ABODE.
E<
To prove the sum of the angles, A^ B, o, D, and E equal to twice as many right angles as the polygon has sides less two.
Proof. From any vertex, as A, draw the diagonals, AC and AD.
The number of triangles thus formed is two less than the num- ber of sides of the polygon, or {n — 2) triangles.
By § 110, the sum of the angles of each triangle is equal to two right angles, therefore, the sum of the angles of all the triangles ; that is, the sum of the angles of the polygon is equal to(w — 2)2rt. A
Therefore, etc, q.e.d.
167. Cor. The sum of the exterior angles of any convex polygon formed by producing the Bides of the polygon in succession is equal to four right angles.
Ex. 106. If from the extremities of the shorter base of an isosceles trapezoid lines are drawn parallel to the equal sides, two triangles are formed. How do they compare ?
Ex. 107. If in a parallelogram any two points in a diagonal equally dis- tant from its extremities are joined to the vertices of the opposite angles, what kind of a figure is thus formed ?
Ex. 108. How many degrees are there in each angle of an ecjuiangular polygon of five sides ?
Ex. 109. How many sides has a polygon the sum of whose mterior angles ^s double the sum of its exterior angles ^
70 PLANE GEOMETRY. — BOOK I.
Proposition XLV
168. Draw any triangle and its three medians. Do the medians intersect in a point ? Measure the distance from this point to each ver- tex. How do these distances compare with the medians of which tihey are a part ?
Theorem, The medians of a triangle pass throiCgh a point which is two thirds of the distance from each vertex to the middle of the opposite side.
c Data: Any triangle, as ABC, and yf\
its medians, AD, BE, and CF. /I \
To prove that AB, BE, and CF pass / / \
through a point, which is two thirds ^^^^^/--'^A
of the distance from A, B, and C to /^Z^^^^J^^\/ \ the middle of the opposite sides re- /^-^^ J ^^"^^ spectively. a f b
Proof. Since two of the medians will intersect, if sufficiently produced, it needs to be shown only that the third median passes through the point of intersection, to prove that the three pass through the same point.
Let any two of the medians, as AD and BE, intersect at H.
Draw KG, joining K and G, the middle points of AH and BH respectively ; also draw KE, ED, and GD.
|
Then, § 159, |
KG Wab, and ED WAB-, |
|
.-. § 80, |
KG II ED ; |
|
also, § 158, |
KG = ^AB, and ED^^AB |
|
.•. |
KG— ED', |
|
and, § 150, |
KGDE is a parallelogram. |
|
Hence, § 154, |
KH=HD, and GH=HE. |
|
But since, const.. |
AK = KH, and BG = GH, |
|
AH = I AD, 2^Ti^ BH=lBE, |
Then, since the medians from any two vertices intersect in a point which is two thirds of the distance from each vertex to the middle of the opposite side, the median from C intersects AD at if.
That is, CF passes through H, and CH = ^ CF.
Therefore, etc. q.e.d.
PLANE GEOMETRY. — BOOK I. 71
Proposition XLVI
169'. Draw any triangle and lines bisecting its angles. Do these lines intersect in a point ? How do the distances of the point from the sides of the triangle compare?
Theorem, The bisectors of the three angles of a triangle pass through a point' which is equidistant from the sides of the triangle.
Data: Any triangle, as ABC, and the lines AD, BE, and CF, bisecting the angles A, B, and C respectively.
To prove that AD, BE, and CF pass through a point which is equidistant from AB, BC, and AC.
Proof. Since two of the bisectors will intersect, if sufficiently produced, it needs to be shown only that the third bisector passes through the point of intersection to prove that the three pass through the same point.
Let any two of the bisectors, as AD and BE, intersect in H.
Then, § 134, H is equidistant from AB and AC, and also from AB and BC.
Hence, H is equidistant from AC and BC\
.'. § 135, H lies in the bisector of angle C.
That is, CF passes through the point H, which is equidistant from AB, BC, and AC.
Therefore, etc. q.e.d.
Ex. 110. How does the angle formed by the diagonals of a square com- pare with a right angle ?
Ex. 111. How does the angle formed by the diagonals of a rhombus compare with a right angle ? How do the diagonals divide each other ?
Ex. 112. How do the diagonals of a rectangle compare in length ?
Ex. 113. If from any point in the bisector of an angle straight lines are drawn parallel to the sides of the angle and are produced to meet the sides, what figure is thus formed ?
Ex, 114. The difference between two angles of a parallelogram which have a common side is 60°. What is the value of each angle of the parallelogram ?
Ex. 115. If the middle points of any two opposite sides of a quadrilateral are joined to each of the middle points of the diagonals, what kind of a figure will the four joining lines form ?
72 PLANE GEOMETRY. — BOOK I.
Proposition XLVII
170. Draw any triangle and Jines perpendicular to its sides, bisecting them. Do these lines intersect in a point ? How do the distances of the point from the vertices of the triangle compare?
Theorem, The perpendicular bisectors of the sides of a triangle pass through a point which is equidistant from the vertices of the triangle.
Data: Any triangle, as ABC, and FG, DK, and EJ, the perpendicular bi- sectors of AB, BC, and AC respectively.
To prove that FG, DK, and EJ pass through a point which is equidistant from A, B, and C. a f b
Proof. Since tvs^o of the perpendiculars will intersect, if sufh- ciently produced (why ?), it needs to be shown only that the third - perpendicular passes through the point of intersection, to prove that the three pass through the same point.
Let any two of the perpendiculars, as FG and DK, intersect at H.
Then, § 103, H is equidistant from A and B, and also from B and C.
Hence, H is equidistant from A and C ;
.*. § 104, H lies in the perpendicular bisector of AC.
That is, EJ passes through the point H, which is equidistant from A, B, and C.
Therefore, etc. q.e.d.
Ex. 116. The middle points of the sides of an equilateral triangle are joined. What kind of triangles are formed ?
Ex. 117. How do the lines drawn from the middle points of the equal sides of an isosceles triangle to the opposite extremities of the base compare in length ?
Ex. 118. Tiie parallel sides of a trapezoid are 35 and 55 feet respectively. What is the length of the line joining the middle points of the non-parallel sides ?
Ex. 119. If from the extremities of the shorter base of rn isosceles trape. zoid perpendiculars are drawn to the longer base, two triangles are formed. How do they compare ?
PLANE GEOMETRY,-^ BOOK I. 73
Proposition XLVIII
171. Draw any triangle and lines from the vertices perpendicular to the opposite sides. Do these lines intersect in a point?
Theorem, The perpendiculars from the vertices of a tri- angle to the opposite sides pass through the same point.
H c K
Data: Any triangle, as ABC, \ ^.
and the lines AD, BE, and CF \ /
drawn from the vertices A, B, \ /^
and C respectively, perpendicu- 2\
lar to the opposite sides. \
To prove that AD, BE, and CF pass through the same point.
Q Proof. Through the vertices A, B, and G draw GH, GK, and EK parallel to BC^ ACy and AB respectively, and intersecting in G, H, and K,
§153, AG=BCf
and AH = BC;
.♦. Ax. 1, AG = AH.
§72, AD±GH',
AD is the perpendicular bisector of GH, In like manner,
BE is the perpendicular bisector of GK^ and CF is the perpendicular bisector of HK.
Hence, § 170, AD, BE, and CF pass through the same point. Therefore, etc. q.e.d.
Ex. 120. How many sides has a polygon the sum of whose exterior angles is double the sum of its interior angles ?
Ex. 121. How many sides has a polygon the sum of whose interior angles is equal to the sum of its exterior angles ?
Ex. 122. The perimeter of an isosceles triangle is 176 feet, and the base is 1^ times one of the equal sides. What is the length of each side of the triangle ?
Ex. 123. How many sides has an equiangular polygon which can be divided into equile-teral triangles by lines drawn from a point within to the vertices of the polygon ?
T4 PLANE GEOMETRY -^ BOOK L
SUMMARY 172. Truths established in Book L 1. Two lines are equal,
0. If they can be made to coincide. § 36 6. If they are sides of an equilateral triangle. § 91
c. If they represent the distances from the extremities of a straight line to any point in the perpendicular erected at its middle point. § 103
d. If they are homologous sides of equal triangles. § 108 «. If they are sides of a triangle opposite equal angles. § 118 /. If they are sides of an equiangular triangle. § 119 g. If they are drawn fronj any point in a perpendicular to a line and cut
off equal distances on that line from the foot of the perpendicular. § 132
h. If they represent the distances of any point in the bisector of an angle
from its sides. § 134
i. If they are the non-parallel sides of an isosceles trapezoid. § 139
j. If they are the sides of a square. § 143
k. If they are the sides of a rhombus. § 144
1. If they are parallel and are intercepted between parallel lines. § 151 w>. If they are opposite sides of a parallelogram. § 153 w. If they are parts intercepted on one transversal by parallel lines which
mtercept equal parts on another transversal. § 157
0. If one is half a side of a triangle and the other is drawn parallel to it
and bisecting one of the other sides. § 158
p. If one joins the middle points of the non-parallel sides of a trapezoid
and the other is equal to half the sum of the parallel sides. § 160
2. Two lines are parallel,
a. If both are perpendicular to the same line. § 71
6. If when cut by a transversal the alternate interior angles are equal.
§75 c If when cut by a transversal the corresponding angles are equal. § 77
d. If when cut by a transversal the sum of the two interior angles on the same side of the transversal is equal to two right angles. § 79
e. If both are parallel to a third line. » § 80 /. If they are the bases of a trapezoid. § 138 g. If they are opposite sides of a parallelogram. § 140 h. If one is a side of a triangle and the other joins the middle points
of the other two sides. § 159
i. If one is either base of a trapezoid and the other joins the middle
points of the non-parallel sides. § 160
8. Two lines are perpendicular to each other,
a. If they form equal adjacent angles with each other. § 26
^. If one is perpendicular to a line which is parallel to the other. § 72
PLANE GEOMETRY. — BOOK I. 75
c. If any two or more points in one are each equidistant from the extremi- ties of the other. §§ 106, 104
d. If one is the base of an isosceles triangle and the other is the bisector of the vertical angle. § 120
4. Two lines form one and the same straight line,
a. If they are the sides of a straight angle. § 27
h. If they are the exterior sides of adjacent supplementary angles. § 58
6. Two lines are unequal,
a. If one is a perpendicular from a point to a straight line and the other is any other line from that point to the straight line. § 61
h. If they represent the distances from the extremities of a straight line to any point without the perpendicular erected at its middle point. § 103
c. If they are sides of a triangle and lie opposite unequal angles. § 127
d. If they are the third sides of two triangles whose other sides are equal, each to each, and include unequal angles. ' § 129
e. If they are drawn from any point in a perpendicular to a line and cut off unequal distances on that line from the foot of the perpendicular. § 182
/. If they are distances cut off on a line from the foot of a perpendicular to it by unequal lines from any point in the perpendicular. § 133
g. If one is any side of a triangle and the other is equal to the sum of the other two sides. §§ 124, 125
h. If one is equal to the sum of two lines from a point within a triangle to the extremities of one side, and the other is equal to the sum of the other two sides. § 131
6. A line is bisected,
a. If it is the base of an isosceles triangle, by the bisector of the vertical angle. § 120
h. If it is the base of an isosceles triangle, by a perpendicular from the vertex. § 122
c. If it is either diagonal of a parallelogram, by the other diagonal. § 154
d. If it is the side of a triangle, by a straight line drawn parallel to the base and bisecting the other side. § 158
7. Lines pass through the same point,
a. If they are the medians of a triangle. § 168
6. If they are the bisectors of the three angles of a triangle. § 169
c. If they are the perpendicular bisectors of the sides of a triangle. § 170
d. If they are perpendiculars from the vertices of a triangle to the oppo- site sides. § 171
8. A perpendicular, and only one, can be drawn to a straight line,
a. At a point in the line. § 51
6. From a point without the line. § 60
T6 PLANE GEOMETRY. — BOOK L
9. Two angles are equal,
a. If tliey can be made to coincide. § 36
h. If they are right angles. § 52
c. If they are straight angles. § 53
d. If they are complements of equal angles. § 54
e. If they are supplements of equal angles. § 54 /. If they are vertical angles. § 59 g. If they are alternate interior angles formed by a transversal and paral- lel lines. § 73
h. If they are corresponding angles formed by a transversal and parallel lines. § 76
i. If their sides are parallel and both pairs extend in the same or in opposite directions from their vertices. § 81
j. If their sides are perpendicular to each other and both angles are acute or both are obtuse. § 83
k. If they are angles of an equiangular triangle. § 98
I. If they are formed adjacent to a straight line by lines joining the ex- tremities of that line with any point in its perpendicular bisector. § 105 TO. If they are formed by the perpendicular bisector of a straight line and lines from any point in it to the extremities of the straight line. § 105 n. If they are homologous angles of equal triangles. § 108 0. If they are the third angles of two triangles whose other angles are equal, each to each. § 113 p. If they are opposite the equal sides of an isosceles triangle. § 116 q. If they are angles of an equilateral triangle. § 117 r. If they are the opposite angles of a parallelogram. § 149
10. Two angles are supplementary,
a. If their sum is equal to two right angles. § 32
h. If their corresponding sides are parallel and one pair extends in the
same direction and the other in opposite directions from their vertices. § 81 c. If their corresponding sides are perpendicular and one angle is acute
and the other obtuse. § 83
11. Two angles are unequal,
a. If they are angles of a triangle and lie opposite unequal sides. § 126
h. If they are the angles opposite unequal sides of two triangles whose
other two sides are equal, each to each. § 130
12. An angle is bisected,
a. If it is the vertical angle of an isosceles triangle, by the perpendicular bisector of the base. § 121
h. If it is the vertical angle of an isosceles triangle, by a line from the vertex perpendicular to the base. § 122
c. By a line every point of which is equidistant from the sides of the angle. § 136
PLANE GEOMETRY.^ BOOK L . 77
13. An angle is equal to the sum of two angles,
a. If it is an exterior angle of a triangle, and the two angles are the opposite interior angles. § 115
14. The sum of angles is equal to a right angle,
a. If they are complements of each other. § 31
h. If they are the acute, angles of a right triangle. § 111
15. The sum of angles is equal to two right angles,
a. If they are supplements of each other. § 32
6. If they are adjacent angles formed by one straight line meeting
another. § 55
c. If they are all the consecutive angles which have a common vertex in a line and lie on the same side of the line. § 56
d. If they are the interior angles formed by a transversal and parallel lines and lie on the same side of the transversal. § 78
e. If they are the angles of a triangle. § 110
16. The sum of angles is equal to four right angles,
a. If they are all the consecutive angles that can be formed about a point. § 57
6. If they are the exterior angles of any convex polygon formed by pro- ducing the sides in succession. § 167
17. The sum of angles is equal to (/; — 2) 2 rt. A^
a. If they are the angles of any convex polygon. § 166
18. Two triangles are equal,
a. If two sides and the included angle of one are equal to the correspond- ing parts of the other. § 100
h. If a side and two adjacent angles of one are equal to the correspond- ing parts of the other. § 102
c. If the three sides of one are equal to the three sides of the other. § 107
d. If they are right triangles, and a side and an acute angle of one are equal to the corresponding parts of the other. § 114
e. If they are right triangles, and the hypotenuse and a side of one are equal to the corresponding parts of the other. § 12?
/. If they are formed by dividing a parallelogram by one of its diagonals.
§152
19. Two parallelograms are equal,
a. If they can be made to coincide. '' ^ § 36
6. If two sides and the included angle of one are equal to the correspond- ing parts of the other. § 155
20. A quadrilateral is a parallelogram,
a. If its opposite sides are parallel. § 14C
h. If its opposite sides are equal. § 148
c. If two of its sides are equal and parallel. § 150
78 PLANE GEOMETRY.— BOOK L
SUPPLEMENTARY EXERCISES
Ex. 124. If through a point halfway between two parallel lines two transversals are drawn, they intercept equal parts on the parallel lines.
Suggestions for Demonstration. 1. What are the data of the propo- sition ? •
2. What lines and point in the figure in ^ ^^ ^
the margin represent the data of the propo- sition ?
3. What parts of the figure are to be C -f ^^ 1>
proved equal ?
4. How may two lines be proved equal ? Summary, § 172, 1.
5. Since FH and JG, which are to be proved equal, are parts of tri- angles, what propositions might we expect to employ in the proof ?
6. In what ways may two triangles be proved equal ? Summary, § 172, 18.
7. What facts in the data suggest aid in determining the equality of angles ? Ans. Parallel lines.
8. What homologous angles in the two triangles are equal ?
9. What other homologous elements of the two triangles must also be equal before the triangles can be proved to be equal ?
10. By careful examination of the given figure discover whether any two homologous sides can be proved equal.
11. Since the homologous sides cannot be proved equal from the given figure, if they can be proved equal at all, what expedient must be resorted to ?
Ans. Construction lines must be drawn which will enable us to prove a side of one of the triangles equal to an homologous side of the other.
12. What fact in the data has not yet been considered which might suggest aid in drawing the construction lines ?
13. What kind of a line measures the distance between two parallel lines ? If such a line be drawn through the given point, how is it divided at the given point ? Then, what line may aid in the proof ?
14. Drawing the figure as in the margin,
with LK perpendicular to the parallel lines, A L JJ ^^
and passing through the point E, which is halfway between the parallel lines, discover .
how the triangles FEL and GEK compare ; C J ^ ^
also, how FE and GE compare.
15. Since the homologous angles of the original triangles have been dis- covered to be equal, and since the equality of two homologous sides, FE and GE, has also been shown, how do the original triangles compare ? How do the sides FH and JG compare ?
Write out the demonstration.
PLANE GEOMETRY. — BOOK I. 79
General Suggestions. I. Study the theorem carefully to discover the data.
II. Construct a figure, or figures, to correspond ivith the data.
III. Discover what parts of the figure correspond to the conclusion given in the theorem.
IV. Study the theorem and the figure to discover as many truths as possible regarding the lines, angles, or other parts.
V. Keeping in mind the truths just discovered and the facts to he proved, consult the Summary and find which truth will best aid in establishing the proposition.
VI. If none of the truths in the Summary seem to be directly ap- plicable to the demonstration sought, draw construction lines which may aid in applying some one of the truths.
Ex. 125. A straight line cutting the sides of an isosceles triangle and parallel to the base makes equal angles with the sides.
Ex. 126. If the base of a triangle is divided into two parts by a perpen- dicular from the vertex, each part of the base is less than the adjacent side of the triangle.
Ex. 127. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
Ex. 128. The perpendiculars to the diagonal of a parallelogram from the opposite vertices are equal.
Ex. 129. If one side of a quadrilateral is extended in both directions, the sum of the exterior angles formed is equal to the sum of the two interior angles opposite the side produced.
Ex. 130. If in an isosceles triangle perpendiculars are drawn from the middle point of the base to the equal sides, the perpendiculars are equal.
Ex. 131. A straight line drawn from any point in the bisector of an angle to either side and parallel to the other side makes, with the bisector and the side it meets, an isosceles triangle.
Ex. 132. The difference between two sides of a triangle is less than the third side.
Ex. 133. Any straight line through the middle point of a diagonal of a parallelogram, and terminated by the opposite sides, is bisect^d at that point,
Ex. 134. If either of the equal sides of an isosceles triangle is produced through the vertex, the line bisecting the exterior angle thus formed is parallel to the base of the triangle. *
Ex. 135. If the bisector of one of the angles of a triangle meets the opposite side, the lines from the point of meeting parallel to the other sides and terminated by them are equal.
80 PLANE GEOMETRY. — BOOK I.
Ex. 136. If each of the angles at the base of an isosceles triangle is one fourth the vertical angle, every line perpendicular to the base forms an equi- lateral triangle with the other two sides, produced when necessary.
Ex. 137. If the straight line bisecting an exterior angle of a triangle is parallel to a side, the triangle is isosceles. *
Ex. 138. If the non-parallel sides of a trapezoid ar^ equal, the base angles are equal, and the diagonals are equal.
Suggestion^. Through one extremity of the shorter parallel side draw a line parallel to the opposite non-parallel side.
Ex. 139. If the angles adjacent to one base of a trapezoid are equal, those adjacent to the other base are also equal.
Suggestion. Produce the non-parallel sides.
Ex. 140. If the upper base of an isosceles trapezoid equals the sum of the non-parallel sides, lines drawn from the middle point of the upper base to the extremities of the lower divide the figure into three isosceles triangles.
Ex. 141. The opposite angles of an isosceles trapezoid are supplements of each other.
Ex. 142. The segments of the diagonals of an isosceles trapezoid form with the upper and lower bases two isosceles triangles.
Ex. 143. The triangle formed by joining the middle points of the sides of an isosceles triangle is isosceles.
Ex. 144. If the two angles at the base of an isosceles' triangle are bisected, the line which joins the intersection of the bisectors with the vertex of the triangle bisects the vertical angle.
Suggestion. " Refer to § 172, 9, n.
Ex. 145. ABCD is a parallelogram ; E and F are the middle points of AD and BO respectively. Show that BE and FD trisect the diagonal AG.
Suggestion. Refer to § 172, 6, d.
Ex. 146. The exterior angle of an equiangular hexagon is equal to the interior angle of an equiangular triangle.
Ex. 147. If one diagonal of a quadrilateral bisects two of the angles, it is perpendicular to the other diagonal.
Ex. 148. If one triangle has two sides and a median to one of them equal respectively to the corresponding parts of another triangle, the triangles are equal.
Ex. 149. The diagonals of a rhombus are unequal.
Ex. 150. If one angle of a triangle is equal to the sum of the other two, the triangle can be divided into two isosceles triangles.
Suggestion. From the vertex oi ZB which is equal to the sum of the other two angleS, draw BD to meet AO a,t D, making ZABD = ZBAD.
>Ex. 151. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram.
Ex. 152. The bisectors of two adjacent angles of a parallelogram inter- sect each other at right angles.
PLANE GEOMETRY. — BOOK I, 81
Ex. 153. If the bisectors of the equal angles of an isosceles triangle are produced until they meet, they form with the base an isosceles triangle.
Ex. 154. The diagonals of a rhombus bisect the opposite angles.
Ex. 155. If two equal straight lines bisect each other at right angles, the lines joining their extremities form a square.
Ex. 156. If the base of any triangle is produced in both directions, the sum of the two exterior angles diminished by the vertical angle is equal to two right angles.
Ex. 157. In a quadrilateral, if two opposite sides which are not parallel are produced to meet, the perimeter of the greater triangle thus formed is greater than the perimeter of the quadrilateral.
Ex. 158. If from any point in the base of an isosceles triangle lines paral- lel to the sides are drawn, a parallelogram is formed whose perimeter is equal to the sum of the equal sides of the triangle.
Ex. 159. ABCD is a quadrilateral having angle ABC equal to angle ADC] AB and DG produced meet in E ', AD and BG in F. Show that angle AED equals angle AFB.
Ex. 160. ABO is an isosceles triangle having AG equal to BC^ and AG is' produced through G its own length to D. Then, ABD is a right angle.
Ex. 161. ABG is a triangle, and through i>, the intersection of the bisec- tors of the angles B and C, EDF is drawn parallel to BG^ meeting AB in E and ^C in i^. Then EF = EB+ FG.
Suggestion. Prov5 ED = EB and FD = FG.
Ex. 162. ABGD is an isosceles trapezoid and AG and BD its diagonals intersecting at 0. Prove that the following pairs of triangles are equal: ABG and ABD ; ADG and BDO; AOD and BOG.
Ex. 163. In any right triangle the line drawn from the vertex of the right angle to the middle of the hypotenuse is equal to one half the hypotenuse.
Suggestion. From the middle point of the hypotenuse draw a line paral- lel to one of the other sides.
Ex. 164. If through each of the vertices of a triangle a line is drawn parallel to the opposite side, a new triangle is formed equal to four times the given triangle.
Ex. 165. Two equal lines, AB and CZ), intersect at E, and the triangles GAE and BDE are equal. Show that OB is parallel to AD.
Ex. 166. ABG and ABD are two equilateral triangles on opposite sides of the same base ; BE and BF are the bisectors respectively of the angles ABG and ABD, meeting AG and AD in E and i?* respectively. Then, the triangle BEF is equilateral.
Suggestion. Eefer to § 172, 1, f.
Ex. 167. ABO is any triangle, and on AB and AC equilateral triangles ADB and AEG are constructed externally. Show that GD equals BE. Suggestion. Refer to § 172, 18. a kilns^s obom. — 6
82 PLANE GEOMETR Y. — BOOK I.
Ex. 168. If through the extremities of each diagonal of a quadrilateral lines parallel to the other diagonal are drawn, a parallelogram double the given quadrilateral will be formed.
Ex. 169. ABCD is a parallelogram ; E and F are points on AC, such that AE = FC ; G and H are points on BD, such that BG- HD. Then, EGFH is a parallelogram.
Ex. 170. The lines joining the middle points of the sides of a rhombus taken in order form a rectangle.
Ex. 171. The bisector of the vertical angle of a triangle and the bisectors of the exterior angles at the base formed by producing the sides about the vertical angle meet in a point which is equidistant from the base and the sides produced.
Suggestion. Use the method of proof employed in Prop. XL VI.
Ex. 172. If in a right triangle one of the acute angles is twice the other, the hypotenuse is equal to twice the side opposite the smaller acute angle.
Suggestion. From the vertex of the right angle drav/ a line to the hy- potenuse, making with one side an angle equal to the acute angle adjacent to that side.
Ex. 173. A parallelogram is bisected by any straight line passing through the middle point of one of its diagonals.
Ex. 174. If two quadrilaterals have three sides. and the two included angles of one equal, each to each, to three sides and the two included angles of the other, the quadrilaterals are equal.
Suggestion. Draw homologous diagonals.
Ex. 175. If two quadrilaterals have three angles and the two included sides of one equal, each to *^ach, to three angles and the two included sides of the other, the quadrilaterals are equal.
Ex. 176. ' The bisectors of the exterior angles of a rectangle form a square.
Ex. 177. The bisectors of the interior angles of a parallelogram form a rectangle.
Ex. 178. The bisectors of the exterior angles of a quadrilateral form a quadrilateral whose opposite angles are supplementary.
Ex. 179. The bisectors of the interior angles of a quadrilateral form a quadrilateral whose opposite angles are supplementary.
Ex. 180. The straight line drawn from any vertex of a triangle to the middle point of the opposite side is less than half the sum of the other two sides.
Suggestion. Draw lines from the middle point of the side opposite the given vertex, parallel to each of the other two sides.
Ex. 181. The lines which join the middle points of the sides of a quad- rilateral successively form a parallelogram whose perimeter is equal to the sum of the diagonals of the quadrilateral.
BOOK II
CIRCLES
173. A plane figure bounded by a curved line, every point ol which is equally distant from a point within, is called a Circle ; the point within is called the Center ; and the bounding line is called the Circumference.
174. Circles which have the same center are called Concentric Circles.
175. A straight line from the center to the circumference of a circle is called a Radius of the circle.
OB, Fig. 1, is a radius of the circle AEDC.
176. A "straight line which passes through ihe center of a circle, and whose extremities '^xe in the circumference, is called a Diameter of the circle.
AD and EG, Fig. 1, are diameters of the circle AEDC.
177. Any part of a circumference is called an Arc.
The curved lines between A and B, and between A and E, Fig. 1, are arcs of the circumference AEDC.
178. An arc which is one half a circumference is called a Semi- circumference.
179. A straight line which joins any two points in a circum- ference is called a Chord of the circle.
A chord which joins the extremities of an arc is said to subtend that arc, and 'an arc is said to be subtended by its chord.
83
84
PLANE GEOMETRY. — BOOK IL
Every chord of a circle subtends two arcs. The chord AB, Fig. 2, subtends the arc ADB, and also the arc ACB. •
When a chord and its subtended arc are men- tioned, the less arc is meant unless otherwise specified.
180. The part of a circle included between an arc and its chord is called a Segment of the circle.
ADB, Fig. 2, is a segment of the circle ADBC.
181. A segment which is one half a circle is called a Semi- circle.
182. The part of a circle included between an arc and the radii drawn to its extremities is called a Sector of the circle.
OABB, Fig. 2, is a sector of the circle ADBC.
183. An arc which is one fourth of a circumference, or a sector which is one fourth of a circle, is called a Quadrant.
184. A straight line which touches a circle and does not cut the circumference if produced is called a Tangent of the circle.
The circle is then said to be tangent to the line.
The point where a tangent touches a circle is called the point of tangency, or the point of contact.
AB, Fig. 3, is a tangent of the circle ; and P is the point of tangency.
185. Two circles are said to be tangent to ^' each other, if they are both tangent to the fw. 8. same straight line at the same point.
They are tangent internally or externally according as one circle lies within or without the other.
186. A straight line which cuts a circumference in two points is called a Secant of the circle.
CD, Fig. 3, is a secant of the circle.
PLANE GEOMETRY. — BOOK IL
85
187. An angle whose vertex is at the center of a circle, and whose sides are radii of the circle, is called a Central Angle.
Angle AOB is a central angle.
188. An angle whose vertex is in the circumference of a circle, and whose sides are chords, is called an Inscribed Angle.
Angle AGE is an inscribed angle.
An angle whose vertex is in the arc of a segment, and whose sides pass through the extremities of the arc, is said to be inscribed in the segment.
189. A polygon is said to be inscribed in a circle, if the vertices of its angles are in the circumference.
The circle is then said to be circumscribed about the polygon.
190. A polygon is said to be circum- scribed about a circle, if each side is tangent to the circle.
The circle is then said to be inscribed in
the polygon.
191. Ax. 14. All radii of the same cirde^ or of equal circles, are equal.
15. All diameters of the same circle, or of equal circles, are equal.
16. Turn circles are equal, if their radii or diameters are equal.
17. A tangent has only one point in common with a circle.
Proposition I
192. 1. Draw a circle and one of its diameters; draw several chords of the circle. How does the diameter compare in length with any other chord? How does the diameter divide the circle? The circumference?
2. How do two arcs of the same circle, or of equal circles, compare, if their extremities can be made to coincide?
86 PLANE GEOMETRY,— BOOK II.
Theorem. A diameter of a circle is greater than any other chord, and bisects the circle and its circumference.
' Data : A circle whose center is 0 ; any diameter, as AB ; and any other chord, as EF.
To prove 1. AB greater than EF.
2. AB bisects the circle and its circumference.
Proof. 1. Draw the radii OE and OF, § 125, OE + OF > EF\
but. Ax. 14, OE = AO, and OF = OB]
AO -\- OB > EFj or AB > EF. 2. § 176, AB passes through the center 0.
Eevolve the segment ACB upon ^B as an axis until it comes into the plane of the segment ADB.
Then, the arc ACB must coincide with the arc ABB ; for, if the arcs do not coincide, some points in the two arcs are unequally distant from the center.
But, § 173, every point in the arcs is equally distant from 0.
Hence, the arcs ACB and ADB coincide and are equal.
That is, AB bisects the circle and its circumference.
Therefore, etc. q.e.d."
193. Cor. In the same circle, or in equal circles^ arcs whose extremities can be made to coincide are equal.
Proposition II
194. 1. Draw a circle and divide a part of its circumference into a number of equal arcs ; from the points of division draw lines to the center. How do the central angles thus formed compare in size?
2. In a circle construct two equal central angles. How do the arcs which subtend them compare in size?
3. Draw a circle and take two unequal arcs ; from their extremities draw lines to the center. How do the angles at the center compare in size ? Which arc subtends the larger angle ? Which angle at the center is subtended by the smaller arc ?
PLANE GEOMETRY. — BOOK II. 87
Theorem, In the same circle, or in equal circles, equal arcs subtend equal central angles; conversely, the sides of equal central angles intercept equal arcs.
Data: The equal circles whose centers are 0 and P, and any equal arcs, as AB and DE.
To prove angle 0 = angle P.
Proof. Place the circle whose center is 0 upon the equal circle whose center is P so that arc AB coincides with the equal arc DE, and the point 0 with the point P,
then, OA coincides with PD, and OB with PE.
Hence, § 36, Zo = Z.P. . q.e.d.
Conversely : Data : Any equal central angles in these circles, as Z 0 and Z P.
To prove arc AB = arc DE.
Proof. Place the circle whose center is 0 upon the circle whose center is P so that the point 0 coincides with the point P, and OA takes the direction oi PD.
Data^ Zo =:^ ZP)
OB takes the direction of PE)
and, since. Ax. 14, OA = pd, and OB = PE ;
A coincides with D, and B with E.
Hence, § 193, arc AB = arc DE.
Therefore, etc. q.e.d.
195. Cor. In the same circle, or in equal circles, the greater of two arcs subtends the greater central angle; conversely, the sides of the greater of two central angles intercept the greater arc.
88 PLANE GEOMETRY. — BOdk II.
Proposition III
196. 1. Draw two equal circles and two equal chords, one in each circle. . How do the subtended arcs compare in length ?
2. Draw two chords, one in each of two equal circles, subtending equal arcs. How do the chords compare in length ?
3. Draw two equal circles and two unequal chords, one in each circle. Which chord subtends the larger arc ? Which arc is subtended by the less chord ?
Theorem, In the same circle, or in equal circles, equal chords subtend equal arcs; conversely, equal arcs have equal chords.
Data : The equal circles whose centers are 0 and P, and any equal chords, as ^B and DE.
To prove arc AB = arc DE.
Proof. Draw the radii OA, OB, PD, and PE,
In the A OAB and PDE, AB = DE, Ax. 14, OA = PD, and OB = PE',
A OAB = A PDE, Why ?
and Zo=:Zp.
Hence, § 194, arc AB = arc DE. Q.e.d.
Conversely : Data : Any two equal arcs in these circles, as AB and. DE, and the chords subtending them.
To prove chord AB = chord DE.
Proof. By the student.
197. Cor. In the same circle, or in equal circles, the greater of two chords subtends the greater arc; conversely, the greater of two arcs has the greater cliord.
PLANE GEOMETRY. — BOOK 11.
89
Proposition IV
198. Draw a circle and a chord of the circle ; draw a radius perpen- dicular to the chord. How does this radius divide the chord? How does it divide the arc subtended by the chord ?
Theorem. A radius which is perpendicular to a chord bisects the chord and its subtended arc.
Data: A circle whose center is O; any chord, as AB ; and a radius, as OD, perpen- dicular to AB at E.
To prove AE = BE, and arc AD = arc BD.
|
Proof. Di |
aw |
the radii OA and OB. |
|
|
In the rt. Aaeo and |
BEO, |
||
|
OA — OB, |
|||
|
and |
OE is common ; |
||
|
.-. § 123, |
Aaeo = Abeo, |
||
|
and |
AE = BE. |
||
|
Also, |
Z.t = Av', |
||
|
hence, § 194, |
arc AD = arc BD. |
||
|
Therefore, |
etc |
/ |
Why?
Why?
Q.E.D.
Ex. 182. If two circumferences intersect, how does the distance between their centers compare with the difference of their radii ?
Proposition V
199. 1. Draw a chord of any circle and a perpendicular to that chord at its middle point. Determine whether the perpendicular passes through the center of the circle.
2. Draw a line through the center of a circle perpendicular to a chord; How does it divide the chord ? How does it divide the subtended arc ?
3. Draw a circle and two chords which are not parallel. Erect per- pendiculars at their middle points and produce these perpendiculars until they intersect. At what point in the circle do the perpendicular bisectors of the chords npie^t ?
90
PLANE GEOMETRY. — BOOK II.
Theorem. A line perpendicular to a chord at its mid- dle point passes through the center of the circle.
Data: A circle whose center is 0; any chord, as ^5 ; and CD a perpendicular to AB 3i,t its middle point.
To prove that CD passes through 0.
Proof. § 173, A and B are equally distant from 0, data, CD is the perpendicular bisector of AB ;
hence, § 104, 0 must lie in CD ;
tfiat is, CD passes through 0.
Therefore, etc. q.e.d.
200. Cor. I. A line passing through the center of a circle and perpendicular to a chord bisects the chord and its subtended arc.
201. Cor. II. The point of intersection of the perpendicular bi- sectors of two non-parallel chords is the center of the circle.
Proposition VI
202. 1. Draw a circle and two equal chords. How do their distances from the center compare ?
2. Draw a circle and two chords equally distant from the center. How do the chords compare in length ?
Theorem, In the same circle, or in equal circles, equal chords are equally distant from the center; cotiversely^ chords equally distant from the center are equal.
Data : Any two equal chords, as ^£ and DEy in the circle whose center is 0.
To prove AB and DE equally distant from 0.
Proof. Draw the radii OA and OD ; also draw the perpendicu-
PLANE GEOMETRY. — BOOK II. 91
lars OG and OH representing the distances from o to AB and DEj
respectively.
Then, § 200, AG = ^AB, and BH = ^BE ; .-. Ax. 7, AG =BH.
In the rt. Aaog and pOH, Ax. 14, OA = OBf
AG = BH',
.'. § 123, Aaog = Aboh,
and OG = OH', . Why?
that is, AB and i)j& are equally distant from O. q.e.d.
Conversely: Data: Any two chords, as AB and BE, equally distant from 0, the center of the circle. • To prove AB = BE.
Proof. In the rt. ^AOG and BOH, data, OG = OH,
Ax. 14, O^ = 02) ;
.-. § 123, Aaog = Aboh,
and AG = BH. Why 9
But, § 200, AG = \AB, and BH = ^BE;
hence. Ax. 6, ^£ = D^.
Therefore, etc. q.e.d
Ex. 183. In how many points can a straight line cut a circumference ?
Ex. 184. How many centers may a circle have ?
Ex. 185. If a straight line bisects a chord and its subtended arc, what is its direction with reference to the chord ?
Ex. 186. Do the perpendicular bisectors of the sides of an inscribed quadrilateral meet in a common point?
Ex. 187. If a diameter of a circle bisects a chord, how does it divide the subtended arc ? In what direction does it extend with reference to the chord ?
Ex, 188. If a diameter of a circle bisects an arc, how does it divide the chord of the arc ? What is its direction with reference to the chord ?
Ex. 189. If the distance from the center of a circle to a straight line is less than the radius, will the line cut the circumference ? If the distance is greater than the radius, will the line cut the circumference ?
92
PLANE GEOMETRY. — BOOK IL
Proposition VII
203. Draw two unequal chords in the same circle, or in equal circles. Which chord is nearer the center, the longer or the shorter one?
Theorem, In the same circle, or in equal circles, the less of two unequal chords is at the greater distance from the center ^
Data : In the equal circles whose centers are 0 and P, any two ciiords, as AB and BE^ of which BE is the less.
To prove BE at a greater distance from P than AB is from 0.
Proof. Draw the radii OA^ OB, PB, and PE\ also draw the perpendiculars OG and PH representing the distances from 0 to AB, and from P to BE, respectively.
Data, AB > BE,
§200, AG=\AB,2iTidLBH = ^BE\
AG>BH.
Then, take AK equal to BH, and draw OK.
In the isosceles A ABO and BEP, § 130, /.AOBi^ greater than Z BPE'^
Z PBE is greater than Z OAB,
Then, in Abhp and AKO, BP = AO, const., BH = AK,
and Z PBH is greater than ZOAK;
PH>OK', but OK>OG',
PH>OG', that is, BE is at a greater distance from P than ^P is from 0.
Therefore, etc. Q.e.d
Why? Why?
Why? Why?
PLANE GEOMETRY. — BOOK IL 93
Prbposition VIII
204. In the same circle, or in equal circles, draw two chords unequally distant from the center. Which one is the shorter ?
Theorem. In the same circle, or in equal circles, of two chords unequally distant from the center, the one at the greater distance is the less. (Converse of Prop. VII.)
D
Data: In the circle whose center is 0, /^ \.h\ any two chords, as AB and DE, of which / /^^\\
DE is at the greater distance from 0. of Ne
To prove DE less than AB. ^V g~ y^
Proof. Draw the perpendiculars OG and OH representing the distances from 0 to AB and DE, respectively.
Now, if DE = ABj
then, § 202, OH = OG, which is contrary to data.
If DE > AB,
then, AB < DE,
and, § 203, OG > OH, which is also contrary to data.
Then, since DE is neither equal to, nor greater than AB,
DE is less than AB. Therefore, etc. q.e.d.
Ex. 190. Where does the line drawn through the middle points of two parallel chords in a circle pass with reference to the center ?
Ex. 191. If two circles are concentric, how do any two chords of the greater, which are tangent to the less, compare in length ?
Ex. 192. If an isosceles triangle is constructed on any chord of a circle as its base, where does the vertex lie with reference to the diameter that is perpendicular to the chord, or to that diameter produced ?
Ex. 193. If two chords of a circle cut each other and make equal angles with the straight line which joins their point of intersection with the center, how do the chords compare in length ?
Ex. 194. If from any point within a circle two equal straight lines are drawn to the circumference, where will the bisector of the angle thus formed pass with reference to the center of the circle ?
y4 PLANE GEOMETRY. — BOOK II.
Proposition IX
205. 1. Draw a circle and one of its radii; also a line perpendicular to the radius at its extremity. Is this line a tangent or a secant ?
2. Draw a tangent to a circle and a radius to the point of contact. What kind of an angle is formed by these lines?
Theorem,' A line perpendicular to a radius at its ex- tremity is tangent to the circle; conversely, a tangent is perpendicular to the radius drawn to the point of contact.
Data : A circle whose center is O ; any radius, as 0D\ and a line AB perpendicular to OD at D.
To prove AB tangent to the circle.
Proof. From 0 draw any other line to AB^ as OE.
Then, OD < OE. Why ?
Since every point in the circumference is at a distance equal to oj) from the center, and jK is at a greater distance, E is without the circumference.
Therefore, every point of AB, except D, is without the circum- rerence.
Hence, § 184, AB is tangent to the circle at D. q.e.d.
Conversely : Data : Any tangent to this circle, as AB, and the fadius drawn to the point of contact, as OD.
To prove AB perpendicular to OD.
Proof. Ax. 17, every point of AB, except D, is without the circumference.
.*. OD is the shortest line that can be drawn between 0 and AB.
Hence, § 61, OD ± AB.
Therefore, etc. q.e.d.
Ex. 195. If in a circle a chord is perpendicular to a radius at any point, how does it compare in length with any other chord which can be drawn through that point ?
Ex. 196. If tangents are drawn through the extremities of a diameter, what is their direction with reference to each other ?
PLANE GEOMETRY. — BOOK II, Proposition X
95
206. 1. Draw a circle, a tangent to it, and a chord parallel to the tan- gent. How do the arcs intercepted between the point of tangency and the extremities of the chord compare?
2. Draw a circle and two parallel secants "or chords. How do the intercepted arcs compare? *
Theorem. Parallel lines intercept equal arcs on a cir- cumference.
Data : A circle whose center is O, and any a two parallel lines, as AB and CDj intercepting arcs on the circumference.
To prove that the arcs intercepted hj AB and OD are equal.
Proof. Case I. WJien AB is a tangent and CD is a chord.
Draw to the point of tangency the radius OE.
Then, § 205, OE±AB;
.'. § 72, oi:±CD',
hence, § 198, arc CE = arc DE.
Case II. When both AB and CD are chords.
Draw EF II AB, and tangent to the circum- ^. g
ference at G.
Then, § 80, EF II CD,
Case I, arc CG = arc DG,
ind SiTC AG — arc BG ;
.-. Ax. 3, arc CA = arc DB.
Case III. When AB and CD are tangents as at G and H respectively.
Draw the chord EF II AB.
Then, § 80, EF II CD,
Case I, arc EH = arc FH,
and arc Ji^G' = arc FG ;
.-. Ax. 2, arc HEG = arc HFG,
Therefore, etc. O.B.D.
96 PLANE GEOMETRY. — BOOK II.
Proposition XI
207. Select three points not in the same straight line. How many circumferences can be passed through them ?
Theorem. Through three points not in the same straight line one circumference can he drawn, and only one.
Data: Any three points not in the same straight line, as A, B, and C. y^' ~~""n
To prove that one circumference can be / N
drawn through A^ B, and C, and only one. ; ^^^Jp ^.^-fp
Proof. Draw AB, BC, and AC-, and at their \ j-V'' >/'''
middle points, E, F, and G, respectively, erect \ ,.-'''' ; perpendiculars. ^ ^^--~.£l""^ "^
§ 170, these perpendiculars meet in a point, as 0, which is equidistant from A, B, and C;
.'. § 173, a circumference described from O as a center, and with a radius equal to the distance OA, passes through the points Ay B, and C; and, since the perpendiculars intersect in but one point, there can be but 07ie center, and consequently but one cir- cumference passing through the points A, B, and C. q.e.d.
208. Cor. I. Circles circumscribing equal triangles are equal. Cor. II. Two circumferences can intersect in only two points. If
two circumferences have three points in common, they coincide and form one circumference.
Proposition XII
209. Draw a circle and from a point outside draw two tangents. How do the tangents compare in length?
Theorem. The tangents drawn to a circle from a point without are equal.
Data: A circle whose center is 0 ; any point without it, as A ; and AB and AC the tangents to the circle at the points B and C, respec- tively.
To prove AB = AC.
PLANE GEOMETRY. — BOOK II - 97
Proof. Draw OAy OB, and OC.
§ 205, A B and c are rt. A
Then, in the rt. A OAB and OAC,
OB = 0(j, Why?
and . OA is common;
A OAB = A OAC, Why?
and AB = ^C»
Therefore, etc. ' q.e.d.
210. The line which joins the centers of two circles is called their line of centers.
211. A common tangent to two circles which cuts their line of centers is called a common interior tangent; one which does not cut their line of centers is called a common exterior tangent.
Proposition XIII
212. Draw two intersecting circles and a chord that is common to both. What kind of an angle does a line joining their centers make with this common chord ?
Theorem. If the circumferences of two circles intersect, the line of centers is perpendicular to the common chord at its middle point.
Data : Two circles whose centers are O and P, whose circumferences intersect at A and 5; AB the common chord ; and OP the line of centers.
To prove OP perpendicular to AB at its middle point.
Proof. § 173, P is equidistant from A and B, and also, 0 is equidistant from A and B ;
.*. § 104, both P and 0 lie in the perpendicular bisector of AB.
Hence, Ax. 11, OP coincides with this perpendicular bisector i that is, OP ± AB at its middle point.
Therefore, etc. Q.B.D
milne's geom. — 7
98
PLANE GEOMETRY. — BOOK IL
Proposition XIV
213. Draw two circles tangent to each other, and a line joining their centers. Through what point will this line pass?
Theorem, If two circles are tangent to each other, their line of centers passes through the point of contact.
Data : The two tangent circles whose centers are 0 and P ; OP their line of centers; and C their point of contact.
To prove that OP through C
Proof. At C draw the common tangent AB.
Ax. 17,
.'. § 205, also,
hence, § 68, that is, Therefore, etc.
C lies in both circumferences;
if radius PC is drawn, PC A.ABf
if radius OC is drawn, OC l.AB\
A OCA +ZPCA =2rt. ^;
OCP is a straight line;
OP passes through C.
MEASUREMENT
Why
Q.E.D.
214. The theorems thus far presented and proved have usually- established only the equality or inequality of two magnitudes, but it is sometimes desirable to measure accurately the magni- tudes that are given.
A magnitude is measured when we find how many times it contains another magnitude of the same kind, called the unit of measure.
The number which expresses how many times a magnitude con- tains a unit of measure is called its numerical measure.
215. The relation of two magnitudes which is determined by finding how many times one contains the other, or what part one is of the other, is called their Ratio.
PL4NE GEOMETRY.— BOOK 11. 99
The ratio of two magnitudes is the ratio of their numerical measures. It may be either an integer or a fraction.
The ratio of a line 12 ft. long to one 4 ft. long is 3 ; that is, the 12 ft. line is three times the 4 ft. line. Also the ratio of an angle of 10° to an angle of 60° is ^ ; that is, an angle of 10° is one sixth of an angle of 60°.
216. Two magnitudes of the same kind which contain a com- mon unit of measure an integral number of times are called Com- mensurable Magnitudes.
217. Two magnitudes of the same kind which have no common unit of measure are called Incommensurable Magnitudes.
218. The ratio of incommensurable magnitudes is called an mcommensurahle ratio ; that is, it cannot be expressed exactly by numbers; and yet we can approximate to the exact numerical value as nearly as we please.
Thus, if the side of a square is 1 ft. in length, the diagonal is V2 ft. in length, as will be shown later, and the ratio of the
diagonal to the side of the square is —— •
Now, no integer or mixed number can be found which .is ex- actly equal to V2^ but by expressing the square root of 2 in a decimal form, the ratio can be determined within a fraction as small as we please.
Thus, V2 = 1.414213+ ; that is, V2 lies between 1.414213 and 1.414214; therefore, the ratio of the diagonal of a square to its
«5iflp llP<5 hptwPPTl 1414213 o-pfl 1414214.
blue neb uetweeu yTycroTO^TT '*'""• T"o7orTyTo
That is, if the one-millionth part of a foot be assumed approxi- mately as the common unit of measure, the side of the square will be equal to 1,000,000, and the diagonal will be equ^l to be- tween 1,414,213 and 1,414,214 such units.
By continuing the process of finding the square root we can approximate as closely to the actual ratio as we please, or until the fraction contains an error so small that it may be disregarded, though it cannot be eliminated.
It is evident, therefore, that the ratio of two magnitudes of the same kind, even when they are incommensurable, may be obmined to any required degree of precision.
100 PLANE GEOMETRY. — BOOK II.
To generalize : suppose Q to be divided into n equal parts, and that P contains m of these parts with a remainder less than one
of the parts ; then, - = — within less than — ^ Q 71 n
Since n may be taken as large as we please, - may be made
n less than any assigned measure of precision, and, consequently,
the value of — may be regarded as the approximate value of the
. p . .^ ratio — within any assigned degree of precision.
THEORY OF LIMITS
219. A magnitude which remains unchanged throughout the same discussion is called a Constant.
220. A magnitude which, under the conditions imposed upon it, may have an indefinite number of different successive dimen- sions is called a Variable.
221. When a variable increases or decreases so that the dif- ference between it and a constant may be made as small as we please, but cannot be made absolutely equal to zero, the constant is called the Limit of the variable, and the variable is said to approach this limit.
Suppose, for example, that a point moves from ^ to P under
the condition that it shall ^ , , \ \ b
move one half the distance ^ d e f
during the first interval of time ; one half the remaining distance the second interval; one half the distance still remaining the next interval; and so on. The distance from A to the moving point is an increasing variable, which approaches the distance AB as a limit, though it cannot actually reach it. Also, the distance from B to the moving point is a decreasing variable, which ap- proaches zero as a limit, though it cannot actually reach it.
Again, the sum of the descending series 1, \, \, \, -jV) ^V' "^V? etc., approaches 2 as a limit, but never quite reaches it. The sum of the first two terms is 1|, of the first three terms If, of the first four terms IJ, etc. It is evident that the sum approaches 2, and that, by taking terms enough, it may be made to differ from 2 by
PLANE GEOMETRY. — BOOK 11. 101
as small a quantity as we please, but it cannot be actually equal to 2. That is, the sum of the series approaches the limit 2 as th6 number of terms is increased.
For further illustration, if, in the right triangle ABC, the ver- tex C indefinitely approaches the vertex B, the angle A diminishes and indefinitely approaches zero as its limit, and if it should actually become zero, the triangle would vanish and become the straight line AB. Again, if the vertex C indefinitely moves away from the vertex B, the angle A increases and indefinitely approaches a right angle as its limit, and if it should actually become a right angle the triangle would vanish and AC and BC would be two parallel lines, each perpendicular to AB. Hence, the value of angle A lies between the limits zero and a right angle, but it can never actually reach either limit so long as the triangle exists.
Proposition XV
222. What is a variable? What is the limit of a variable? If two variables are always equal, how do their limits compare ?
Theorem. If, while approaching their respective limits, two variables are always equal, their limits are equal.
Data : Two equal variables,
as ^a; and Cy, which approach a- + -^ ^
the limits AB and CD, respec- .
tively. c + ^
To prove AB and CD equal.
Proof. Suppose that AB is greater than CD ; then some part of AB, as Az, is equal to CD.
Then, the variable Ax may have values between Az and AB ; that is, the variable Ax may have values greater than CD, while the variable Cy cannot have values greater than CD.
This is contrary to the condition that the variables are equal.
Hence, AB cannot be greater than CD.
In like manner it can be proved that AB cannot be less than CD,
Consequently, AB = CD.
Therefore, etc. Q-e.d.
102
PLANE GEOMETRY.^ BOOK IL
Proposition XVI
223. In the same circle, or in equal circles, draw two central angles such that the arc intercepted by the sides of the first shall be three times the arc intercepted by the sides of the second. How does the first angle compare in size with the second? How does the ratio of the central angles compare with the ratio of the arcs intercepted by their sides ?
Theorem, In the same circle, or in equal circles, two central angles have the same ratio as the arcs intercevted by their sides.
M
Data: In the equal circles whose centers are 0 and P, any tjcntral angles, as AOB and DPE, whose sides intercept the arcs AB and DE respectively.
To prove
ratio
Z AOB
ratio
arc ^5
Z DPE arc DE
Proof. Case I. When the arcs are commensurable. Suppose that M is the common unit of measure for the two arcs, and that it is contained in arc AB 7 times and in arc DE 4 times.
Then,
tatio
arc AB
ratio
arc DE 4
Divide the arcs AB and DE into parts each equal to the common measure M, and to each point of division draw a radius.
By § 194, each of these central angles formed by any two adjacent radii is equal to every other central angle so formed.
The number of equal parts into which the angles A OB and DPE are divided by these radii is equal to the number of times M is contained in the arcs AB and DE respectively.
ratio
Z AOB Z DPE
ratio
Hence, Ax. 1,
, . Z AOB ,.
ratio — = ratio
^DPE
4
arc AB
SiVC DE
PLANE GEOMETRY. — BOOK II. Case II. When the arcs are incommensurable.
103
M
bmce arcs AB and DE are incommensurable, take these com- mensurable arcs AG and DE, and suppose that GB is less than M.
Then, Case I, ratio
Z AOG
= ratio
arc A G
Z DPE arc DE
1 • if is indefinitely diminished, angle GOB and arc GB decrease,
and the ratios — and — remain equal, and indefinitely
Z DPE arc DE "^
approach the limiting ratios ^ ^__ and ^^^ _ respectively.
Z DPE arc DE
Hence, § 222, ratio ^^^ = ratio ^^^^. Z DPE arc DE
Q.E.D.
J224. It was stated in § 35 that the total angular magnitude about A point in a plane is divided into degrees, minutes, and seconds. In the same way the circumference of a circle is divided into 360 equal arcs, called arc degrees, or simply degrees; each arc degree is divided into 60 equal parts, called minutes; and each arc minute into 60 equal parts, called seconds.
Thus it will be seen that the angle degree and the arc degree, the angle minute and the arc minute, the angle second and the arc second correspond, each to each. The sides of a central angle of one degree therefore intercept an arc of one degree, the sides of a central angle of ten degrees intercept an arc of ten degrees, and, in general, the sides of a central angle of any number of degrees intercept an arc of an equal number of degrees.
Therefore, a central angle is measured by its intercepted arc; that is, the central angle contains between its sides the same propor- tion of the total angular magnitude about a point m a plane, that the arc intercepted by its sides is of the whole circumference.
104 PLANE (JEOMETRY.-^BOOK IL
Proposition XVII
225. 1. Draw a circle and an inscribed angle one of whose sides is a diameter; draw a radius to the extremity of the other side. How does the inscribed angle compare in size with the central angle subtended by the same arc? Since the central angle is measured by the arc which subtends it, by what part of the arc is the inscribed angle measured?
2. Draw other inscribed angles no one of whose sides is a diameter and draw a diameter from the vertex of each. By what part of its arc is each inscribed angle measured?
3. If inscribed angles are subtended by the same arc or chord or are inscribed in the same segment, how do they compare in size?
4. How many degrees are there in a semicircumference? Whai^ then, will be the size of all angles inscribed in a semicircle ?
5. How does an angle inscribed in a segment greater than a semicircle compare with a right angle? How, if in one less than a semicircle?
Theorem. An inscribed angle is measured hy one half the arc intercepted hy its sides.
Data : A circle whose center is O, and any- inscribed angle, as ABO, ^ To prove angle ABC measured by ^ arc AC.
Proof. Case I. When one side of the angle is a diameter of the circle.
When AB is a, diameter.
Draw the radius 00,
|
Then, |
OB = OOy |
Why? |
|
§90, |
A 5 oc is isosceles, |
|
|
and |
Zb = Zc. |
Why? |
|
§115, |
ZAOC— Zi? + Zc = 2Z5; |
|
|
or |
Zb = \Z AOC. |
|
|
But, § 224, |
Z AOC is measured by arc AO\ |
|
|
hence^ |
Z jB is measured by \ arc AC, |
PLANE GEOMETRY. — BOOK IL
106
Case II. When the diameter from the vertex of the angle lies between the sides.
Draw the diameter BB.
Case I., Z ABB is measured by ^ arc AD, and Z CBB is measured by i arc CB ;
but, Ax. 9, Z ABB + Z CBB =Z ABC,
and arc JZ) + arc CB = arc ^C;
hence, Z ^4^8 (7 is measured by |- arc ^C.
Case III. When both sides of the angle are on the same side of the diameter from the vertex.
Draw the diameter BB.
Case I., Z ABB is measured by ^ arc AD, and Z CBB is measured by ^ arc CB ;
but Z ABB — Z C5D = Z ^5C,
and arc ^D — arc CD = arc AC,
hence, Z ^5C7 is measured by ^ are AC.
Therefore, etc.
Q.E.D.
226. Cor. I. Angles inscribed in the same segment of a circle, or in equal segments of the same circle, or of equal circles are equal.
227. Cor. II. An angle inscribed in a semi- circle is a right angle.
228. Cor. III. An angle inscribed in a segment greater than a semicircle is less than a right angle.
229. Cor. IV. An angle inscribed m a segment less than a semi- circle is greater than a right angle.
106 PLANE GEOMETRY. — BOOK II.
Proposition XVIII
230. Draw a circle and two intersecting chords ; construct an inscribed angle equal to one of the vertical angles thus formed, by drawing from an extremity of one chord a chord parallel to the other. How does the arc subtending the inscribed angle compare with the sum of the arcs intercepted by the sides of the vertical angles ? What, then, is the meas- ure of the angle formed by the two intersecting chords ?
Theorem, An angle formed hy two intersecting chords
is measured hy one half the sum of the arc intercepted by
its sides and the arc intercepted by the sides of its vertical
an^le.
c
Data: Any two intersecting chords, as X^ \ ^\ AB and CD, forming the angle r. a/- ^ — \b
To prove angle r measured by I \ I |(arc^CH-arc£i)). Jv i)^
Proof. Draw BE II AB.
Then, § 76, Zs^Zr.
§ 225, Zs is measured by ^ arc CAE ;
but ' arc CAB = arclc + arc AB,
and, § 206, arc AB = arc BB ;
* arc CAB = arc ^C + arc BB ;
hence, Z 5 is measured by ■i-(arc ^C -h arc BB).
Consequently, Z r is measured by ^(arc^C + arc^D).
Therefore, etc. ^ q.e.d.
Ex. 197. Prove by Prop. XVII that the sum of the angles of a triangle is equal to two right angles.
Ex. 198. The opposite angles of an inscribed quadrilateral are supple- mentary.
Ex. 199. If two chords of a circle intersect at right angles, to what is the sum of any pair of opposite arcs equal ?
Ex. 200. If one of the equal sides of an isosceles triangle is the diameter of a circle, the circumference bisects the base.
Ex. 201. If one side of an angle of a quadrilateral inscribed in a circle is produced, the exterior angle is equal to the opposite angle of the quadrilateral.
PLANE GEOMETRY. — BOOK 11. 107
Proposition XIX
231. 1. At any point in the circumference of a circle form an angle by a tangent and a chord; construct vertical angles equal to this by drawing through the given chord a chord parallel to the tangent. How does the sum of the arcs subtending these vertical angles compare with the arc intercepted by the sicies of the given angle ? What, then, is the measure of the given angle ?
2. How does an angle between a tangent and a diameter compare with a right angle ? What is its arc measure ?
Theorem, An angle formed hy a tangent and a chord is measured hy one half the intercepted arc.
Data : Any tangent, as EB, and any X"^^ \^\
chord, as AGj forming with EB the ^/ ^__\ Xi)
angle r. i \ I
To prove angle r measured by \ 7^
iarc^C. \ /
Proof. Draw any chord, as HD^ parallel to EB and cutting AO in F.
Then, § 76, Z.s = /.r.
§ 230, Z s is measured by \ (arc ^iT + arc i) (7) ; but, § 206, arc ^jH" = arc ^D ;
hence, Z s is measured by \(diVQ, AD + arc DC), or ^arc^C
Consequently, Z r is measured by ^ arc AC.
Therefore, etc. * q.e.d.
232. Cor. A right angle is measured by one half a semicircum- ference.
Ex. 202. The angle between a tangent to a circle and a chord drawn from the point of contact is half the angle at the center subtended by that chord.
Ex. 203. A line which is tangent to the inner of two concentric circles, and is a chord of the outer circle, is bisected at the point of tangency.
Ex. 204. The diagonals of a rectangle inscribed in a circle are diameters ' of the circle.
Ex. 205. The bisector of the vertical angle of an inscribed isosceles . triangle passes through the center of the circle.
108 PLANE GEOMETRY. — BOOK IL
Proposition XX
233. At any point without the circumference of a circle form an angle between a tangent and a secant ; construct an angle equal to this by drawing from the point of tangency a chord parallel to the secant. How does the arc intercepted by the sides of this angle compare with the diiference of the arcs intercepted by the sides of the given angle ? Whatj then, is the measure of the given angle ?
Theorem, An angle formed by a tangent and a secant which meet without a circumference is measured hy one half the difference of the intercepted arcs.
Data: Any tangent, as AB, and any- secant, as AC, meeting AB without the ^ circumference and forming witli it the angle A.
To prove angle A measured by ^ (arc DF — arc DE),
c
Proof. From D, the point of tangency, draw DM 11 AG. Then, § 76, Zr=ZA.
§ 231, Z r is measured by ^ arc DH;
but arc DH = arc DF — arc HF,
and, § 206, arc HF=SiVC de ;
arc DH = arc DF — arc DE ;
hence, Z r is measured by | (a^c DF — arc DE).
Consequently, Z ^ is measured by ^ (arc DF — arc DE), Therefore, etc. q.e.d.
Ex. 206. Two chords perpendicular to a third chord at its extremities are equal.
Ex. 207. If a quadrilateral is inscribed in a circle and its diagOKdls are drawn, the diagonals will divide the angles of the quadrilateral so that there will be four pairs of equal angles.
Ex. 208. If from the center of a circle a perpendicular is drawn to either side of an inscribed triangle, and a radius is drawn to one end of this side, the angle between the radius and the perpendicular is equal to the opposite angle of the triangle.
PLANE GEOMETRY. — BOOK II.
109
Proposition XXI
234. At any point without the circumference of a circle form an angle between two secants ; construct an inscribed angle equal to this by draw- ing from the point of intersection of either secant and the circumference a chord parallel to the other secant. How^ does the arc subtending the inscribed angle compare with the difference of the arcs intercepted by the secants ? What, then, is the measure of the given angle ?
Theorem, An angle formed hy two secants which meet without a cirjcumference is measured hy one half the differ- ence of the intercepted arcs.
Data : Any two secants, as AB and AC, meeting without the circumference and forming the angle A.
To prove angle A measured by ^ (arc ^C — arc DE).
Proof. Draw the chord DF W AC.
Then, § 76, Zr = /.A.
§ 225, Z r is measured by \ arc BF ;
but arc BF = arc BC — arc FC,
and, § 206, arc FC = arc DE ;
arc 5i^ = a-rc 5(7 — arc Z)^ ; hence, Z r is measured by ^ (arc BC — arc DE).
Consequently, Z ^ is measured by ^ (arc BC — arc DE).
Therefore, etc. q.e.d.
Ex. 209. The tangent at the vertex of an inscribed equilateral triangle forms equal angles with the adjacent sides.
Ex. 210. The angle between two tangents from the same point is 24° 15'. How many degrees are there in each of the intercepted arcs ?
Ex. 211. Ond angle of an inscribed triangle is 38°, and one of its sides subtends an arc of 124°. What are the other angles of the triangle ?
Ex. 212. AB^ a chord of a circle, is the base of an isosceles triangle whose vertex C is without the circle, and whose equal sides intersect the circumference at D and E. Prove that CD is equal to CE.
110 PLANE GEOMETRY. — BOOK IL
Proposition XXII
235. At any point without the circumference of a circle form an angle between two tang'ents; construct an angle equal to this by drawing from the point of contact of either tangent a chord parallel to the other. How does the arc intercepted by the sides of this angle compare with the difference of the arcs intercepted by the tangents? What, then, is the measure of the given angle?
Theorem. An angle formed by two tangents is -measured by one half the difference of the intercepted arcs.
Data : Any two tangents, as AB and AC, forming the angle A. To prove angle A measured by i (arc DFE — arc DE),
Proof. From D, the point of tangency, draw the chord DF\\ AC.
Then, § 76, /.r= Z.A. .
§ 231, Z r is measured by ^ arc DF ;
but arc DF — arc BFE — arc FE^
and, § 206, arc FE = arc DE ;
arc BF — arc BFE — arc bE\ hence, Z r is measured by J (arc BFE - sltcBE).
Consequently, Z ^ is measured by ^ (arc BFE — arc T)E).
Therefore, etc. q.e.d.
Ex. 213. In any quadrilateral circumscribing a circle any pair of opposite sides is equal to half the perimeter of the quadrilateral.
Ex. 214. If three circles touch each other externally, ahd the three com- mon tangents are drawn, these tangents meet in a point equidistant from the points of contact of the circles.
Ex. 215. If two triangles are inscribed in a circle, and two sides of one are parallel, each to each, to two sides of the other, the third sides are equal.
PLANE GEOMETRY.-- BOOK IL 111
Ex. i216. Every parallelogram inscribed in a circle is a rectangle.
Ex. 217. Two sides of an inscribed triangle subtend \ and \ of the cir« cumference, respectively. What are the angles of the triangle ?
Ex. 218. If a circle is circumscribed about the triangle ABC, and a line is drawn bisecting angle A and meeting the circumference in Z>, angle BCB is equal to one half angle BA C.
Ey 219. AB is an arc of 65°, DC an arc of 75° in a circle whose center is 0. ^C is a diameter. How many degrees are there in each angle of the triangles A OD and BOG^
Ex 220. If an equilateral triangle is inscribed in a circle, and a diameter is drawn from one vertex, the triangle formed by joining the other extremity of the diameter and the center of the circle with one of the other vertices of the inscribed triangle is also equilateral.
Ex. 221. If tangents aie drawn to a circle from a point without, the line joining that point with the center of the circle bisects (1) the angle formed by the tangents ; (2) the angle formed by the radii drawn to the points of tangency ; and (3) the arc intercepted by these radii.
Proposition XXIII 236. Problem,* To bisect a straight line.
>P
Datum : Any line, as AB. Required to bisect AB,
yp
Solution. From A and B as centers, with equal radii each ^eater than one half AB, describe arcs intersecting at C and D. Draw CD intersecting AB Sit E. Then, CD bisects AB at J^. Q.E.F.
Proof. Const., C and D are each equidistant from A and B,
Hence, § 106, cn is the perpendicular bisector of AB.
* 1. The student is urged to attempt to solve each problem before he studies the solution given in the book. He will very likely discover for him- self the same method of solution or perhaps another one equally good.
2. The only implements used in solving problems in plane geometry are the straightedge and compasses.
112
PLANE GEOMETRY.-^BOOK IL
a-
Proposition XXIV 237. Problem, To bisect an arc of a circle.
Datum : Any arc of a circle, as AB. Required to bisect arc AB.
Solution. Draw the chord AB.
From A and B as centers, with equal radii each greater than one half chord AB, describe arcs intersecting at C and D. Draw CD intersecting arc AB d^t E. Then, CD bisects arc AB at Z. q.e.p.
Proof. Const., C and D are each equidistant from A and B ; .*. § 106, CD is perpendicular to the chord AB at its middle point, and, § 199, CD passes through the center of the circle.
Hence, § 200, CD bisects the arc AB 2i\, E.
Proposition XXV 238. Brohlem, To bisect an angle.
Datum: Any angle, as ^BC. Required to bisect the angle ABC.
Solution. From 5 as a center with any radius, as BE, describe an arc intersecting AB 'm. D and CB in E.
From D and E as centers, with equal radii each greater than one half the distance DEj describe arcs intersecting at F, and draw BF.
Then, BF bisects the angle ABC, q.e.f
Proof. Draw DF and EF.
Then, in A BDF and BEF^
PLANE GEOMETRY. — BOOK IL 113
const., BD — BE,
DF = EF, and BF is common ;
.-.§107, ABBF = ABEFy
and /.'DBF = Zebf; Why?
that is, BF bisects the angle ABC.
Proposition XXVI
239. Problem, At a given point in a line to erect a perpendicular to the line.
%
/o
E: ^ HX
I
Case I. When the given point is between the extremities of the line.
Data : Any line, as AB, and any point in AB, as C.
Required to erect a perpendicular to AB at C.
Solution. From C as a center, with any radius, describe arcs intersecting AB 3it D and E.
From D and E as centers, with equal radii, describe arcs inter- secting at F. Draw CF.
Then, CF is the perpendicular required. q.e.f.
Proof. By the student. Suggestion. Refer to § 106.
Case II. When the given point is at the extremity of the line.
Data : Any line, as AB, and the point at either extremity, as B.
Required to erect a perpendicular to AB at 5.
Solution. From 0, any point without AB, as a center, with OB as a radius, describe an arc intersecting AB in H.
From H, draw a line through 0 intersecting this arc in K, and draw KB.
Then, KB is the perpendicular required. q.e.f.
Proof. By the student. Suggbstion. Refer to § 227. milne's geom. — 8
114 PLANE GEOMETRY. — BOOK 11.
Proposition XXVII
240. Problem, To draw a perpendicular to a line from a point without the line.
Data: Any line, as AB, and any point without the line, as C. /ff
Required to draw a perpendicular from c to AB.
Solution. From C, as a center, with a radius greater than the distance from C to AB, describe an arc intersecting AB at D and E.
From D and E, as centers, with equal radii each greater than one half of DE, describe arcs intersecting at F.
Draw CF and produce it to meet AB as at G.
Then, CG is the perpendicular required. q.e.f.
Proof. By the student. Suggestion. Refer to § 106L
Proposition XXVIII
241. Problem, To construct an angle equal to a given angle.
Datum : Any angle, 2^s ABC.
Required to construct an angle equal to ABC.
Solution. Draw any line, as DE.
From 5 as a center, with any radius, describe an arc intersect- ing BA and BC in F and G respectively.
From D as a center, with the same radius, describe an arc inter- secting DE in H.
PLANE GEOMETRY. — BOOK II. 115
From fi- as a center, with a radius equal to the distance GF^ describe a second arc intersecting the first at J", and draw DJ.
Then, Z JDH is the angle required. q.e.f.
Proof. Draw GF and HJ.
In Abgf and DHJ, . const., BG= DH, BF = DJ, and GF = JEfJ ;
.-. § 107, Abgf = Abhj,
and Zb = Zb.
Proposition XXIX
242. rroblem. Through a given point to draw a line parallel to a given line. ,e
c!
Data: Any line, as AB, and any point not in AB, as C.
Required to draw a line /""\^ '
through a parallel to ^^. /
A i \ -B
Solution. Through G draw any line meeting AB, as EB. Construct Z ECF equal to Z EBB.
Then, CF is parallel to AB. q.e.f.
Proof. Const., Z ECF = Z EBB ;
.-. f 77, CF II AB.
Ex. 222. Two angles of a triangle being given to construct the third.
Ex. 223. Two secants cut each other without a circle ; the intercepted arcs are 72° and 48°. What is the angle between the secants ?
Ex. 224. A tangent and a secant cut each other without a circle ; the intercepted arcs are 94° and 32°. What is the angle between the tangent and the secant ?
Ex. 225. Two chords of a circle intersect and two opposite intercepted arcs are 88° and 26°. What are the angles between the chords ?
Ex. 226. A tangent of a circle and a chord from the point of contact intercept an arc of 110°. What is the angle between the tangent and the chord ?
Ex. 227. If the radii of two intersecting circles are 4*'" and 9^™, respec- tively, what is the greatest and the least possible distance between their centers ?
116 PLANE GEOMETRY. — BOOK II.
Proposition XXX
243. Problem, To construct a triangle when two sides and the included angle are given.
yQ
Tb--^
Data : Two sides of a triangle, as m and n, and the included angle, as r.
Required to construct the triangle.
Solution. Draw any line, as AD, and on it measure AB equal to n.
Construct the /, A equal to Z r, and on AG measure AC equal to m.
Draw CB.
Then, Aabc'ib, the A required. q.e.f.
Proof. By the student. .
Proposition XXXI
244. Prohlem, To construct a triangle when a side and two adjacent angles are given.
rj,--D
Data : A side of a triangle, as m, and the angles, as r and s, ad- jacent to it.
Required to construct the triaugle.
PLANE GEOMETRY. — BOOK II. 117
Solution. Draw any line, as AD, and on it measure AB equal to m.
Construct Z A equal to Z r, and Z 5 equal to Z s.
Produce the' sides of these two angles until they intersect, as at C.
Then, AABCi^ the A required. . q.e.f.
Proof. By the student.
245. Sch. The problem is impossible when the sum of the given angles equals or exceeds two right angles. Why ?
Proposition XXXII
246. Problem. To construct a triangle when the three sides are given.
P
1^ ^.. .^—-D
Data : The three sides of a triangle, as m, n, and o.
Required to construct the triangle.
Solution. Draw any line, as AD, and on it measure AB equal to 0.
From ^ as a center, with a radius equal to n, describe an arc.
From J5 as a center, with a radius equal to m, describe a second arc intersecting the first in C. Draw AC and BC.
Then, AABCis the A required. q.e.f.
Proof. By the student.
247. Sch. The problem is impossible when any one side is equal to or greater than the sum of the other two sides. Why ?
Ex. 228. Construct an equilateral triangle.
Ex, 229. Prove that the radius of a circle inscribed in an equilateral triangle is equal to one third the altitude of the triangle.
Ex. 230. In an inscribed trapezoid how do the non-parallel sides com- pare ? How do the diagonals compare ?
118 PLANE GEOMETRY. — BOOK 11.
Proposition XXXIII
248. Problem, To construct a parallelogram when two sides and the included angle are given.
c/. z-Ad
'tK
m.
Data: Two sides of a parallelogram, as m and n, and tlie included angle, as r.
Required to construct the parallelogram.
Solution. Draw any line, as AE, and on it measure AB equal to n.
Construct the Z A equal to Z r, and on AG measure A 0 equal to m.
From c as a center, with a radius equal to n, describe an arc.
From 5 as a center, with a radius equal to m, describe a second arc intersecting the first in D.
Draw CD and BD.
Then, ABDC is the parallelogram required. q.e.f.
Proof. By the student. Suggestion. Refer to § 148.
Proposition XXXIV 249. Problem, To inscribe a circle in a given triangle.
Datum : Any triangle, as ABC. Required to inscribe a circle in A ABC.
A E B
Solution. Bisect A A and B, produce the bisectors to intersect at 0, and draw OE _L AB.
PLANE GEOMETRY. — BOOK IT. 119
From 0 as a center, with OE as a radius, describe the circle EFG.
Then, EFG is the circle required. q.e.f.
Proof. Const., 0 lies in the bisectors oi Aa and B ; .-. § 134, 0 is equidistant from AB, AC, and BC.
Hence, a circle described from 0 as a center, with a radius equal to OE, touches AB, AC, and BC.
That is, § 190, the circle EFG is inscribed in A ABC.
Proposition XXXV 250. Problem, To divide a straight line into equal parts.
Datum : Any straight line, as AB.
Required to divide AB into equal parts.
Solution. From A draw a line of indefinite length, as AD, making any convenient angle with AB.
On AB measure off in succession equal distances corresponding in number with the parts into which AB is to be divided.
From the last point thus found on AD, as C, draw CB, and from each point of division on AC draw lines II CB and meeting AB.
These lines divid-e AB into equal parts. q.e.f.
Proof. By the student. Suggestion. Refer to § 157.
Ex. 231. If the sides of a central angle of 35° intercept an arc of 75"", what will be the length of an arc intercepted by the sides of a central angle of 80° in the same circle ?
Ex. 232. AB and CD are diameters of the circle whose center is 0 ; BD is an arc of 116°. How many degrees are there in each angle of the triangles ^OC and i)05?
Ex. '233. If a circle is circumscribed about a triangle ABC, and perpen- diculars are drawn from the vertices to the opposite sides and produced to meet the circumference in the points D, E, and jP, the arcs EF^ FD^ and DE are bisected at the vertices.
J> •
120 PLANE GEOMETRY. — BOOK II.
Proposition XXXVI 251. Problem, To find the center of a circle.
Datum : Any circle, as ABD. Required to find the center oi ABD.
Solution. Draw any two non-parallel chords, as AB and CD. Draw the perpendicular bisectors of AB and CD, and produce them until they intersect, as in 0. Then, 0 is the center of the circle. Proof. By the student. Suggestion. Refer to § 201.
Ex. 234. To circumscribe a circle about a given triangle.
Ex. 235. AB is a chord of a circle and J.C is a tangent at J. ; a secant parallel to AB, as EFD, cuts ^C in ^ and the circumference in F andD ; the lines AF, AD, and BD are drawn. Prove that tite triangles AEF and ADB are mutually equiangular.
Proposition XXXVII
252. Problem. Through a given point to draw a tangent to a given circle.
Data: A circle whose center is 0, and any point, as Ji.
Required to draw through A a tangent [ ^/ ] "^ j?.
to the circle.
Solution. Case I. When A is on the circumference.
Draw the radius OA.
At A dT3iW EF ± OA.
Then, EF is the tangent required.
Proof. Bj the student.
PLANE GEOMETRY, — BOOK II. 121
Case II. When A is without the circumference.
Draw OA.
From B, the middle point of OA, as y^^'^^^^Tc^ ^^ \ a center, with -B 0 as a radius, describe / / nT"^-^^ \
a circle intersecting the given circle / J \ | -"--:^J^
at C and D. ' \ ^^ / '^ -^^^ '
Drsiw AC Siiid AD. \ \dv^""" /'
Then, AO ot AD is the tangent re- T^^^^^^^-^-J-iL-'""'
quired. q.e.f.
Proof. By the student.
Suggestion. Draw the radii OC and OB, and refer to § 227 and § 205.
Proposition XXXVIII
253. Problem, To describe upon a given straight line a segment of a circle which shall contain a given angle.
Data : Any straight line, as AB. / / \ \
and any angle, as r. / / xjq\ \
Required to describe a segment \ / j '^^v\^ h
of a circle upon AB which shall i*r -f y^
contain Z r. """- ^^7
Solution. Construct Z ABD equal to Z r.
Draw FE 1. AB at its middle point.
Erect a perpendicular to DB at B, and produce it to intersect FE at 0.
From 0 as a center, with a radius equal to OB, describe a circle.
Then, AMB is the segment required. q.e.f.
Proof. Inscribe any angle in segment AMB, as Z s.
Const., Z ABD = /.r,
§ 231, Z ABD is measured by ^ arc ^5 ;
but, § 225, Z s is measured by \ arc AB,
Z.S — /. ABD = Zr.
Hence, AMB is the segment required.
122 ' PLANE GEOMETRY. — BOOK 11.
SUMMARY 254. Truths established in Book II.
1. Two lines are equal,
a. If they are radii of the same circle, or of equal circles. Ax. 14
b. If they are diameters of the same circle, or of equal circles. Ax. 15
c. If they are chords which subtend equal arcs in the same circle, or in equal circles. § 196
d. If they represent the distances of equal chords in the same circle, or in equal circles, from the center. § 202
e. If they are chords equally distant from the center of the same circle, or of equal circles. § 202
/. If they are tangents drawn to a circle from a point without. § 209
g. If they are the limits of two variable lines which constantly remain
equal and indefinitely approach their respective limits. § 222
2. Two lines are perpendicular to each other,
a. If one is a tangent to a circle and the other is a radius drawn to the point of contact. § 205
b. If one is the common chord of two intersecting circles and the other is their line of centers. § 212
3. Two lines are unequal,
a. If one is a diameter of a circle and the other is any other chord of that circle. § 192
b. If they are chords of the same circle, or of equal circles, subtending unequal arcs. § 197
c. If they represent the distances of unequal chords in the same circle, or in equal circles, from the center. § 203
d. If they are chords of the same circle, or of equal circles, unequally distant from the center. § 204
4. A line is bisected,
, a. If it is a chord of a circle, by a radius perpendicular to it. § 198
b. If it is a chord of a circle, by a line perpendicular to it and passing through the center. § 200
c. If it is the common chord of two intersecting circles, by their line of centers. § 212
5. A line passes through a point,
a. If it is the perpendicular bisector of a chord and the point is the center of the circle. § 199
b. If it is the line of centers of two tangent circles, and the point is their point of contact. § 213
PLANE GEOMETRY.— BOOK II. 123
6. Two angles are equal,
a. If they are central angles subtended by equal arcs in the same circle, or in equal circles. § 194
6. If they are inscribed in the same segment of a circle, or in equal seg- ments of the same circle, or of equal circles. § 226
7. Two angles are unequal,*
a. If they are central angles subtended by unequal arcs in the same circle, or in equal circles. § 195
8. An angle is measured,
a. If it is a central angle, by the intercepted arc. § 224
h. If it is an inscribed angle, by one half the intercepted arc. § 225
c. If it is between a tangent and a chord, by one half the intercepted arc. § 231
d. If it is a right angle, by one half a semicircumference. § 232
e. If it is between- two intersecting chords, by one half the sum of the intercepted arcs. § 230
/. If it is between a tangent and a secant, by one half the difference of the intercepted arcs. § 233
g. If it is between two secants intersecting without the circle, by one half the difference of the intercepted arcs. § 234
9. Two arcs are equal,
a. If they are arcs of the same circle, or of equal circles and their ex- tremities can be made to coincide. § 193
b. If they subtend equal central angles in the same circle, or in equal circles. ' § 194
c. If they are subtended by equal chords In the same circle, or in equal circles. § 196
d. If they are intercepted on a circumference by parallel lines. § 206
10. Two arcs are unequal,
a. If they subtend unequal central angles in the same circle, or in equal circles. § 195
6. If they are subtended by unequal chords in the same circle, or in equal circles. § 197
11. An arc is bisected,
a. By the radius perpendicular to the chord that subtends the arc. § 198 6. By a line through the center perpendicular to the chord. § 200
12. Two circles are equal,
a. If their radii or diameters are equal. Ax. 16
h. If they circumscribe equal triangles. § 208
13. A line is tangent to a circle,
a. If it is perpendicular to a radius at its extremity. § 205
124 PLANE GEOMETRY. — BOOK II.
SUPPLEMENTARY EXERCISES
Ex. 236. ABC is an inscribed isosceles triangle ; the vertical angle G is three times each base angle. How many degrees are there in each of the &TCSAB, ^(7, and 50?
Ex. 237. If a hexagon is circumscribed about a circle, the sums of its alternate sides are equal.
Ex. 238. Two radii of a circle at right angles to each other are inter- sected, when produced, by a line tangent to the circle. Prove that the tangents drawn to the circle from the points of intersection are parallel to each other.
Ex. 239. Two circles are tangent to each other externally and each is tangent to a third circle internally. Show that the perimeter of the triangle formed by joining the three centers is equal to the diameter of the exterior circle.
Ex. 240. From two points, A and B, in a diameter of a circle, each the same distance from the center, two parallel lines AE and BF are drawn toward the same semicircumference, meeting it in E and F. Show that FF is perpendicular to AE and BF.
Ex. 241. Two circles are tangent externally at A J5C is a tangent to the two circles at B and C. Prove that the circumference of the circle described on 50 as a diameter passes through A.
Ex. 242. OA is a radius of a circle whose center is 0 ; -B is a point on a radius perpendicular to OA ; through B the chord ^O is drawn ; at O a tangent is drawn meeting OB produced in D. Prove that CBD is an isosceles triangle.
Suggestion. Draw a tangent at A.
Ex. 243. Through a given point P without a circle whose center is O two lines FAB and PCD are drawn, making equal angles with OP and inter- secting the circumference in A and B, C and Z>, respectively. Prove that AB equals OD, and that AP equals CP.
Suggestion. Draw OM and ON perpendicular to AB and CD, respect- ively.
Ex. 244. If the angles at the base of a circumscribed trapezoid are equal, each non-parallel side is equal to half the sum of the parallel sides.
Ex. 245. If a circle is inscribed in a right triangle, the sum of the diameter and the hypotenuse is equal to the sum of the other two sides of the triangle.
Ex. 246. Any parallelogram which can be circumscribed about a circle is equilateral,
Ex. 247. AB and CD are perpendicular diameters of a circle ; E is any point on the arc ACB. Then, D is equidistant from AE and BE.
PLANE GEOMETRY. — BOOK II. 125
Ex. 248. If two equal chords of a circle intersect, their corresponding segments are equal.
Ex. 249. If the arc cut off by the base of an inscribed triangle is bisected and from the point of bisection a radius is drawn and also a line to the oppo- site vertex, the angle between these lines is equal to half the difference of the angles at the base of the triangle.
Ex. 250. The two lines which join the opposite extremities of two par- allel chords intersect at a point in that diameter which is perpendicular to the chords.
Ex. 251. If a tangent is drawn to a circle at the extremity of a chord, the middle point of the subtended arc is equidistant from the chord and the tangent.
Ex. 252. A line is drawn touching two tangent circles. Prove that the chords, that join the points of contact with the points in which the line through the centers meets the circumferences, are parallel in pairs.
Ex. 253. "Two circles intersect at the points A and B ; through A a secant is drawn intersecting one circumference in G and the other in D ; through B a secant is drawn intersecting the circumference CAB in E and the other circumference in F. Prove that the chords CE and I^F are parallel.
Suggestion. Refer to Ex, 198 and 201.
Ex. 254. The length of the straight line joining the middle points of the non-parallel sides of a circumscribed trapezoid is equal to one fourth the perimeter of the trapezoid.
Ex. 255. A quadrilateral is inscribed in a circle, and two opposite angles are bisected by lines meeting the circumference in A and B. Prove that AB is a diameter.
Ex. 256. The centers of the four circles circumscribed about the four triangles formed by the sides and diagonals of a quadrilateral lie on the vertices of a parallelogram.
Ex. 257. If an equilateral triangle is inscribed in a circle, the distance of each side from the center is equal to half the radius of the circle.
Ex. 258. The vertical angle of an oblique triangle inscribed in a circle is greater or less than a right angle by the angle contained by the base and the diameter drawn from the extremity of the base.
Ex. 259. If from the extremities of any diameter of a given circle per- pendiculars to any secant that is not parallel to this diameter are drawn, the less perpendicular is equal to that segment of the greater which is con- tained between the circumference and the secant.
Ex. 260. Two circles are tangent internally at J., and a chord BC of the larger circle is tangent to the smaller at D. Prove that AD bisects the angle CAB.
Suggestion. Draw AT, the common tangent of the circles.
126 PLANE GEOMETRY.^BOOK II
Ex. 261. The tangents at the four vertices of an inscribed rectangle form a rhombus.
Ex. 262. If a line is drawn through the point of contact of two circles which are tangent internally, intersecting the circle whose center is A at (7, and the circle whose center is 5 at Z>, AC and BD are parallel.
Ex. 263. If lines are drawn from the center of a circle to the vertices of any circumscribed quadrilateral, each angle at the center is the supplement of the central angle that is not adjacent to it.
Ex. 264. Three circles are tangent to each other externally at the j)oints Af J5, and C. From A lines are drawn through B and C meeting the cir- cumference which passes through B and C at the points D and E. Prove that BE is a diameter.
Ex. 265. If an angle between a diagonal and one side of a quadrilateral is equal to the angle between the other diagonal and the opposite side, the same will be true of the three other pairs of angles corresponding to the same description, and the four vertices of the quadrilateral lie on a circum- ference.
Ex. 266. Let the diameter AB of a circle be produced to O, making BC equal to the radius ; through B draw a tangent, and from C draw a second tangent to the circle at D, intersecting the first at E ; AD and BE produced meet at F. Prove that DEF is an equilateral triangle.
Suggestion. Draw a line from the center to E.
Ex. 267. If from any point without a circle tangents are drawn, the angle contained by the tangents is double the angle contained by the line joining the points of contact and the diameter drawn through one of them.
Ex. 268. The lines, which bisect the vertical angles of all triangles on the same base and on the same side of it, and having equal vertical angles, meet at the same point.
Ex. 269. AB and AC are tangents at B and C respectively, to a circle whose center is 0; from D, any point on the circumference, a tangent is drawn, meeting AB in E and ^O in i^. Prove that angle EOF is equal to one half angle BOC.
Ex. 270. A circle whose center is 0 is tangent to the sides of an angle ABC at A and C ; through any point in the arc AC, as Z>, a tangent is drawn, meeting AB in E, and CB in F. Prove (1) that the perimeter of the triangle BEF is constant for all positions of Bin AC; (2) that the angle EOF is constant.
Ex. 271. If AE and BD are drawn perpendicular to the sides BC and AC, respectively, of the triangle ABC, and DE is drawn, the angles AED and ABD are equal.
Suggestion. Describe a circle passing through A, D, and B.
PLANE GEOMETRY. — BOOK 11. 127
Ex. 272. The perimeter of an inscribed equilateral triangle is equal to one half the perimeter of the circumscribed equilateral triangle.
Ex. 273. In the circumscribed quadrilateral ABCD, the angles A, J5, and Care 110°, 95°, and 80°, respectively, and the sides AB, BC, CB, and DA touch the circumference at the points E, F, G, and H respectively. Find the number of degrees in each angle of the quadrilateral EFGH.
Ex. 274. If an inscribed isosceles triangle has each of its base angles double the vertical angle, and its vertices the points of contact of three tan- gents, these tangents form an isosceles triangle each of vv^hose base angles is one third its vertical angle.
Ex. 275. ABQ is a triangle inscribed in a circle ; the bisectors of the angles A, B, and C meet in D ; AD produced meets the circumference in E. Prove that DE equals CE.
Suggestion. Produce CD to meet the circumference at F, and draw AF.
Ex. 276. If the diagonals of a quadrilateral inscribed in a circle inter- sect at right angles, the perpendicular from their intersection upon any side, if produced, bisects the opposite side.
Ex. 277. If the opposite angles of a quadrilateral are supplementary, the quadrilateral may be inscribed in a circle.
Suggestion. § 207. A circumference may be passed through A, B, and Z>, and if it does not pass through O, it will cut DC, or DC produced, as at jE'. Draw ^ji5. j),,- >.,^
Then, from data and the supposition that the circum- /j -^-.^^X
ference passes through E, it may be shown by raeas- / / \ \
urement of inscribed angles, that ZDCB = /.DEB. \ / \ j
But, § 11 5, Z DEB = Z CBE ■\-Z.DCB', a{ ■ fB
ZDCB=zZCBE+ZDCB, V., ,-'"'
which is absurd, and the supposition that the cir- cumference passes through E and not through C is untenable.
Hence, the circumference through A, B, and D passes through C.
Ex. 278. The four lines bisecting the angles of any quadrilateral form a quadrilateral which may be inscribed in a circle.
Suggestion. Kefer to Ex. 179 and 277.
Ex. 279. The line, drawn from the center of the square described upon the hypotenuse of a right triangle to the vertex of the right angle, bisects the right angle.
Ex. 280. AB and CD are two chords of a circle intersecting at E ; through A a line is drawn to meet a line tangent at C so that the angle AFC equals the angle BEC. Then, EF is parallel to BC.
Ex. 281. ABC is a triangle ; AD and BE are the perpendiculars from A and B upon BC and J. C respectively ; DF and EG are the perpendiculars from D and E upon ^C and BC respectively. Then, FG is parallel to AB.
128 PLANE GEOMETRY. — BOOK II,
PROBLEMS
255. Problems are valuable for developing the ingenuity of the student in discovering the auxiliary lines necessary for solution or demonstration, and for fixing clearly in mind the knowledge previously acquired. They do not, however, involve any new fundamental fact or principle of geometry, and may therefore be omitted without impairing the logical development of the science.
No definite rules can be given for solving problems, but close attention to the following suggestions, and a thorough study of the Summary will be of great assistance in developing a proper and logical method of procedure.
I. Study carefully the data of the problem to discover every fact that is giveuj and 7iotice also what is required.
II. From the facts given deduce all possible conclusions, and try to relate them to what is required.
III. If the 2^rinciples upon which the solution is based are not readily discovered from the data, try to get some clew to the solution by studying the Summary and by drawing lines perpendicular or parallel to other lines; by forming triangles; by joining given points; by describing circles; etc.
IV. The solution is often readily discovered by drawing a figure representing the problem solved, and then from a study of the relor tions of the known and unknown parts of the figure discover the facts which bear upon the solution.
Ex. 282. Construct the complement of a given angle.
Ex. 283. Construct the supplement of a given angle.
Ex. 284. At a given point in a given straight line, to construct an angle of 45°.
Ex. 285. Divide an angle into four equal parts.
Ex. 286. At a given point in a given straight line, to construct an ang^e of 60°.
Ex.^ 287. Trisect a right angle.
Ex. 288. Construct a square, having given one side.
Ex. 289. Construct an isosceles triangle, having given the base and the perpendicular from the vertex to the base.
Ex. 290. Construct an equilateral triangle, having given the perimeter.
PLANE GEOMETRY. — BOOK II. 129
Ex. 291. Construct an isosceles triangle, having given the perimeter and base.
Ex. 292. Construct a rectangle, having given two adjacent sides.
Ex. 293. Construct a rectangle, having given the shorter side and the difference of two sides.
Ex. 294. Construct a rectangle, having given the longer side and the difference of two sides.
Ex. 295. Construct a rectangle, having given the sum and difference of two adjacent sides.
Ex. 296. Construct a rhombus, having given one of its angles and the length of its side.
Ex. 297. Construct an isosceles triangle, having given the base and one of the two equal angles at the base.
Ex. 298. Construct an isosceles right triangle, having given its hypot- enuse.
Ex. 299. Construct a rhomboid, having given the perimeter, one side, and one angle.
Ex. 300. Construct a right triangle, having given the hypotenuse and one side.
Ex. 301. Construct a right triangle, having given the hypotenuse and one acute angle.
Ex. 302. Construct an isosceles trapezoid, having given two sides and the included angle.
Ex. 303. Construct a trapezoid, having given two adjacent sides, the included angle, and the angle at the other extremity of th^ given parallel
Ex. 304. From a given point without a given line, to draw a line making a given angle with the given line.
Ex. 305. Construct a square, having given a diagonal.
Ex. 306. Construct a rectangle, having given one side and the angle included between it and a diagonal.
Ex. 307. Construct a rectangle, having given a diagonal and an angle between it and a side.
Ex. 303. Construct a rhombus, having given its perimeter and one diagonal.
Ex, 309. Construct a rhomboid, having given two diagonals and an angle between them.
Ex. 310. Construct a rectangle, having given one diagonal and the angle included between the two diagonals.
Ex. 311, Construct a rectangle, having given the perimeter and one side. milne's geom. — 9
130 PLANE GEOMETRY. — BOOK II.
Ex. 312. Construct a trapezoid, having given two sides, tlie included angle, and the difference between the two parallel sides.
Ex. 313. Construct a trapezium, having given three consecutive sides and the two included angles.
Ex. 314. Construct a trapezium, having given two adjacent sides and the three angles adjacent to these sides.
Ex. 315. Construct an isosceles triangle, having given the base and the vertical angle.
Ex. 316. Construct an equilateral triangle, having given its altitude.
Ex. 317. Construct a triangle, having given two sides and the angle opposite one.
Ex. 318. Construct a triangle, having given its base, the median to the base, and the angle included between them.
Ex. 319. Construct a right triangle, having given its hypotenuse and having one of its acute angles double the other,
Ex. 320. Construct a trapezoid, having given the sum and difference of the parallel sides, and the sum and difference of the angles at the base.
Ex. 321. Construct a rhombus, having given its diagonals.
Ex. 322. Construct a rhomboid, having given two adjacent sides and an angle not included by them.
Ex. 323. Construct a rhomboid, having given one side and the angles included between it and the diagonals.
Ex. 324. Construct an isosceles trapezoid, having given the bases and the altitude.
Ex. 325. Construct an isosceles trapezoid, having given the altitude and the sum and difference of the parallel sides.
Ex. 326. Construct a triangle, having given two angles and a side oppo- site one.
Ex. 327. Draw a line which shall pass through a given point and make equal angles with two given intersecting lines.
Ex. 328. Construct a right triangle, having given one side and the angle opposite.
Ex. 329. Construct an isosceles trapezoid, having given the bases and a diagonal.
Ex. 330. Construct an isosceles trapezoid, having given the bases and one angle.
Ex. 331. From two given points, draw two equal straight lines which shall meet in the same point of a line given in position.
Ex. 332. ABC is an isosceles triangle. Draw a straight line parallel to the base AB and meeting the equal sides in E and F, so that BF^ EF, and EA are all equal.
PLANE GEOMETRY. — BOOK II. 131
Ex. 333. Given two straight lines which cannot be produced to their intersection, to draw a third line which would pass through their intersec- tion and bisect their contained angle.
' Ex. 334. Constrnct a triangle, having given the altitude and the angles at the base.
Ex. 335. Given the middlQ point of a chord in a given circle, to draw the chord.
Ex. 336. Construct a circle to pass through two given points and have its center on a given straight line. When is this impossible ?
Ex. 337. Draw a tangent to a circle parallel to a given straight line.
Ex. 338. Draw a tangent to a circle perpendicular to a given straight line.
Ex. 339. Draw a straight line tangent to a given circle and making with a given line a given angle.
Ex. 340. Construct a circle of given radius to pass through two given points. When is this impossible ?
Ex. 341. Construct a circle tangent to two intersecting lines with its center at a given distance from their intersection. How many such circles can be drawn ?
Ex. 342. From a given point as a center, to describe a circle tangent to a given circle. How many solutions may there be ?
Ex. 343. Construct a circle of given radius tangent to a given circle at a given point. How many solutions may there be ?
Ex. 344. Draw a common tangent to two given circles. How many solutions may there be ?
Ex. 345. In a given circle, to inscribe a triangle equiangular to a given triangle.
Ex. 346. About a given circle, to circumscribe a triangle equiangular to a given triangle.
Ex. 347. Construct a triangle, having given the vertical angle, one of the sides containing it, and the altitude.
Ex. 348. Construct a triangle, having given the base, the vertical angle, and one other side.
Suggestion. On the given base construct a segment that will contain an angle equal to the given angle.
Ex. 349. Construct a triangle, having given the base, the vertical angle, and the foot of the perpendicular from the vertex to the base.
Ex. 350. Construct a triangle whose vertex is on a given straight line, and having given its base and vertical angle.
Ex. 351. Construct a triangle, having given the base, the vertical angle, and the altitude.
Ex. 352. J^escribe a circle with a given center to intersect a given circle at the extremities of a diameter. Is this ever impossible ?
132 PLANE GEOMETRY, --BOOK IL
Ex. 353. Construct a circle to pass through a given point and be tangent to a given circle at a given poiut. When is this impossible ?
Ex. 354. Construct a circle to pass through a given point and touch l» given straight line at a given point.
Ex. 355. Construct a circle to touch three given straight lines.
Ex. 356. Within an equilateral triangle, to describe three circles each tangent to the other two and to two sides of the triangle.
Ex. 357. Construct a circle of given radius to touch two given straight lines.
Ex. 358. Construct a circle of given radius, having its center on a given straight line and touching ano£her given straight line. How many solutions may there be ?
Ex. 359. Construct a right triangle, having given the radius of the inscribed circle and one of the sides containing the right angle.
Ex. 360. Construct a triangle, having given the base, the vertical angle, and the length of the median to the base.
Ex. 361. Construct a triangle, having given the three middle points of its sides.
Ex. 362. Construct a circle of given radius to pass through a given point and touch a given straight line.
Ex. 363. From the vertices of a triangle as centers, to describe three circles which shall be tangent to each other.
Ex. 364. Construct a triangle, having given the base, the altitude, afld the radius of the circumscribed circle.
Ex. 365. Three given straight lines meet at a point ; draw another straight line so that the two portions of it intercepted between the given lines are equal. How many solutions may there be ?
Suggestion. Form a parallelogram.
Ex. 366. Through a given point, between two intersecting straight lines, to draw a line terminated by the given lines and bisected at the given point.
Ex. 367. Construct a circle to intercept equal chords of given length on three given straight lines.
Ex. 368. Construct a triangle, having given one angle, the opposite side, and the sum of the other two sides.
The Locus of a Point
256. When a point equidistant from the extremities of a straight line is to be found, the middle point of the line meets the conditions. But there are other points which also fulfill the required conditions, for every point in the perpendicular to the given line at its middle point is equidistant from the extremities of the line.
PLANE GEOMETRY. — BOOK 11. 133
Such a perpendicular is called the locals of the points which are equidistant from the extremities of the line.
The *line, or system of lines, containing every point which satisfies certain given conditions, and no other points, is called the Locus of those points.
A locus may also be described as a line, or the lines, traced by a point which moves in accordance with given conditions.
To prove that a certain line, or system of lines, is the required locus, it must be shown :
1. That every point in the lines satisfies the given conditions.
2. That any point not in the lines cannot satisfy the given con- ditions.
Ex. 369. Find the locus of a point which is equidistant from two inter- secting straight lines.
Data : Any two straight lines, as AB and CD, ^
intersecting at the point K. /|\ !
Required to find the locus of a point equidistant / • -"j^j
from AB and CD. ^^^^ / • A\ ^^
Solution. A point equidistant from two inter- • -^^^"""<^^!i>^^^ xr secting straight lines suggests a point in the bisec- 3^'lS^'"^^
tor of an angle. a^^'^^'^ i ' ^"^^D
Draw EF bisecting A CKB and AKD., and also i
GH bisecting A AKC and BKD.
Then, § 134, every point in EF is equidistant from AB and CD, and every point in GH is equidistant from AB and CD.
If all other points are unequally distant from AB and CD., then EF and GH is the required locus.
From P any point without the bisectors draw TM ± CD,^ and BE L AB., intersecting EF in J. From J draw JL ± CD, and also draw BL.
Then, § 61, PL> PM, and, § 125, P./ ^JL>PL\ PJ^JL> PM.
But .JLz=JB\ Why?
PJ+JB> PM, or PB > PM.
That is, the point P is unequally distant from AB and CD. Hence EF and GH is the required locus.
Ex. 370. Find the locus of "a point at a given distance from a given point.
Ex. 371. Find the locus of a point equidistant from two parallel straight lines.
* In this and the next four paragraphs the lines mentioned may be either straight or curved.
134 PLANE GEOMETRY. — BOOK II,
Ex. 372. Find the locus of a point at a given distance from a given straight line.
Ex. 373. Find a point vv^hich is equidistant from three given points not in the same straight line.
Ex. 374. Find the locus of a point equidistant from the circumferences of two concentric circles.
Ex. 375. Find a point in a given straight line wliich is equidistant from two given points.
Ex. 376. Find the locus of the center of a circle tangent to each of two parallel lines.
Ex. 377. Find the locus of the center of a circle which touches a given line at a given point.
Ex. 378. Find the locus of the center of a circle of given radius that passes through a given point.
Ex. 379. Find the locus of the center of a circle which is tangent to a given circle at a given point.
Ex. 380. Find the locus of the center of a circle of given radius and tangent to a given circle.
Ex. 381. Find the locus of the center of a circle passing through two given points.
Ex. 382. Find the locus of the center of a circle of given radius and tangent to a given line.
Ex. 383. Find the locus of the center of a circle tangent to each of two intersecting lines.
Ex. 384. Find the locus of the middle points of a system of parallel chords drawn in a circle.
Ex. 385. Find the locus of the middle points of equal chords of a given circle.
Ex. 386. Find the locus of the extremities of tangents of fixed length drawn to a given circle.
Ex. 387. Find the locus of the middle points of straight lines drawn from a given point to meet a given straight line.
Ex. 388. Find the locus of the vertex of a right triangle on a given base as hypotenuse.
Ex. 389. Find the locus of the middle points of all chords of a circle drawn from a fixed point in the circumference.
Ex. 390. Find the locus of the middle point of a straight line moving between the sides of a right angle.
Ex. 391. Find the locus of the points of contact of tangents from a fixed point to a system of concentric circles.
Ex. 392. Find the locus of the middle points of secants drawn from a given point to a given circle.
BOOK III
RATIO AND PROPORTION
257. 1. How is a magnitude measured ?
2. What is the numerical measure of a magnitude ?
3. What is the common measure of two or more magnitudes ?
4. What is meant by the ratio of two magnitudes ?
5. How may the ratio of two magnitudes be determined ?
6. Since the ratio of two magnitudes is the ratio of their numerical measures, what is the relation of two magnitudes whose numerical measures are 8 and 16 respectively? 5 and 10? 12 and 36? 15 and 45?
7. How does 8 compare with 2 ? What is the relation of 3 to 9? Of 12 to 4? Of 18 to 3? Of 20 to 40? Of 25 to 75? Of 35 to 70 ?
8. What is the ratio of 1 ft. to 1 yd. ? 3 in. to 1 ft. ? 2*=«» to jdm 9 5dm to 2^" ? 2 sq. ft. to 2 sq. yd. ? 3 cu. ft. to 1 cu. yd. ?
258. The quantities compared are called the Terms of the ratio. A ratio is denoted by a colon placed between the terms.
The ratio of 2 to 5 is expressed 2 : 5.
259. The first term of a ratio is called the Antecedent of the ratio. The second term of a ratio is called the Consequent of the ratio.
260. The antecedent and consequent together form a Couplet.
261. Since the ratio of two quantities may be expressed by a
fraction, as -, it follows that: b
The changes which may be made upon the terms of a fraction
ivithout altering its value may be made upon the terms of a ratio
without altering the ratio.
136 PLANE GEOMETRY. — BOOK III.
262. 1. What two numbers have the same relation to each other as 3 has to 6 ? 2 to 8 ? 5 to 15 ? 8 to 4?
2. What numbers have the same relation to each other that 4 in. has to 2 f t. ? 5 ft. to 2 yd. ? 5^"^ to 1™ ? S'^" to 8^'" ?
3. What number has the same relation to 6 that 2 has to 4 ?
4. What number has the same relation to 12 that 3 has to 9 ?
5. What number has the same ratio to 8 that 5 has to 15 ?
263. An equality of ratios is called a Proportion.
The sign of equality is written between the equal ratios.
a:h = c:d is di> proportion^ and is read : the ratio of a to 6 is equal to the ratio of c to d, or a is to 6 as c is to d.
The double colon, : :, is frequently used instead of the sign of equality.
264. The antecedents of the ratios which form a proportion are called the Antecedents of the proportion, and the consequents of those ratios are called the Consequents of the proportion.
In the proportion a:h — c-.d, a and c are the antecedents^ and h and d are the consequents of the proportion.
265. The first and fourth terms of a proportion are called the Extremes and the second and third terms are called the Means of the proportion.
In the proportion a : 6 = c : d, a and d are the extremes^ and 6 and c are the means.
266. A quantity which serves as both means of a proportion IS called a Mean Proportional.
In the proportion a : 6 = 6 : c, 6 is a mean proportional. *
267. Since a proportion is an equality of ratios, and the ratio of one quantity to another is found by dividing the antecedent by the consequent, it follows that :
A proportion may he expressed as an equation in which both mem- bers are fractions.
The proportion a:b=.C'.d may be written - = -.
Such an expression is to be read as the ordinary form of a proportion is read.
PLANE GEOMETRY.-rBOOK III, 137
268. Since a proportion may be regarded as an equation in whicli both members are fractions, it follows that:
1. The changes which may be made upon the members of an equor Hon without destroying the equality may be made upon the couplets of a proportion without destroying the equality of the ratios.
2. Tlie changes which may be made upon the terms of a fraction without altering the value of the fraction may be made upon the terms of each ratio of a proportion without destroying the propor- tion.
Proposition I
269. 1. Form several proportions, as 3 : 5 = 9 : 15, and discover how the product of the extremes compares with the product of the means in each.
2. If the means in any proportion are the same, how may the means be found from the product of the extremes?
3. Form a proportion whose consequents are equal. How do the antecedents compare?
4. Form a proportion in which either antecedent is equal to its conse- quent. How does the other antecedent compare with its consequent?
Theorem, In any proportion, the product of the extremes is equal to the product of the means.
Data : a : d = c : cL
To prove cm? = 6c
Proof. From data, § 267, % = %;
0 a
Multiplying each member of this equation by 6d,
Therefore, etc q.b.d.
270. Cor. I. A mean proportional between two quantities is equal to the square root of their product.
Lf Jib = b:c, find the value of b.
271. Cor. II. If in any proportion any antecedent is equal U* its consequent, the other antecedent is equal to its consequent.
138 PLANE GEOMETRY. — BOOK III.
272. Cor. III. If the consequents of any proportion are equaly the antecedents are equal, and conversely.
For, if a:h = c:hf
a c
1=1'
and multiplying by 6, a = c.
Proposition II
273. 1. If the product of the extremes of a proportion is 48, -what may the extremes be? If 72? If 30? If 36? If 6a2? If 12 aft? If ahcl
2. If the product of the means is 48, what may the means be ? If 96? If 108? If Qhcdl If ahcdl If a%2? If ahc'i
3. Form a proportion the product of whose, extremes or means is 60 ; 72; 84; 80; 64; 144; x^y'^ , xyz; xyzv.
Theorem, If the product of two quantities is equal to the -product of two others, eithsr two -may he D%ade the extremes of a proportion of which the other two are the means.
Data : ad — be.
To prove a:h = c:d.
Proof. Data, ad = be.
Dividing each member of this equation by bd,
|
a c b~d' |
|
|
at is, |
a:b = c:d. |
|
Therefore, etc. |
Q.E.D.
Ex. 393. If the vertical angle of an isosceles triangle is 30°, what is its ratio to each of the base angles ?
Ex. 394. If the exterior angle at the base of an isosceles triangle is 100°, what is its ratio to each angle of the triangle ?
Ex. 395. If one of the acute angles of a right triangle is 40°, what is its ratio to the other acute angle ? To the right angle ?
Ex. 396. The interior angles on the same side of a transversal cutting two parallel lines are to each other as 8 to 2. How many degrees are there in each angle ?
Ex. 397. The vertical angle of an isosceles triangle has the same ratio to a right angle that an angle of 40*^ has to an angle of an equilateral triangle. How many degrees are there in each angle of the isosceles triangle ?
PLANE GEOMETRY. — BOOR til. 139
Proposition III
274. 1. Form a proportion and transpose the means. How do the resulting ratios compare?
2. Transpose the extremes. How do the resulting ratios compare?
3. Transform similarly and investigate other proportions.
Theorem, In any proportion, the first term is to the third as the second is to the fourth; that is, tl%e terms are in pro- portion by alternation.
Data : a'.h = c:d.
To prove a : c = 6 : c?.
Proof. From data, § 267, ? = -•
0 a
Multiplying each member of this equation by -,
c
that is, a : c = 6 : d.
Therefore, etc. q.e.d.
Proposition IV
275. 1. Form a proportion. If the antecedent of each ratio becomes the consequent, and the consequent the antecedent, how do the resulting ratios compare ?
2. Transform similarly and investigate other proportions.
Theorem, In any proportion, the ratio of the second term to the first is equal to the ratio of the fourth term to the third; that is, the terms are in proportion hy inversion.
Data : a-.h — cd.
To prove 6 : a = c? : c.
Proof. From data, § 269, 6c = ad.
Dividing each member of this equation by o/a^
b^d,
a c ' that is, b:a = d:c.
' Therefore, etc. q.e.d.
140 PLANE GEOMETRY. — BOOK III.
Proposition V
276. 1. Forni a proportion. How does the ratio of the sum of the first two terms to either term compare with the ratio of the sum of the last two terms to the corresponding term ?
2. Transform similarly and investigate other proportions.
Theorem. In any -proportion, the ratio of tlw sum of the first two terms to either term is equal to the ratio of the sum of the last two terms to the eorresponding term; that is, the terms are in proportion hy composition.
Data : a\h — c.d.
or
|
To prove a + bi |
6 = c-hd: |
d, and a + b |
|
Proof. § 267, |
a_ b^ |
_c ~~d |
|
Adding 1 to each member of this equation. |
||
|
i^'- |
=1+1' |
|
|
a-\-b_ b |
c + d, - d ' |
that is, a + b:b — c-\-d:d.
In like manner it may be shown that a -\-b:a = c + d:c. Therefore, etc. q.e.d.
Proposition VI
277. 1. Form a proportion. How does the ratio of the difference of the first two terms to either term compare with ratio of the difference of the last two terms to the corresponding term?
2. Transform similarly and investigate other proportions.
Theoretn, In any proportion, the ratio of tJie difference between the first two terms to either terin is equal to tJie ratio of the difference between the last two terms to the cor- responding term; that is, the term^s are in proportion hy division.
Data : a:b — c'.d.
To prove a — 6:6 = c — d:c?, and a — b:a = c — d'.c.
PLANE GEOMETRY. — BOOK III. 143
Proof. §267, ? = ^.
h d
Subtracting 1 from each member of this equation,
h ^^~d ^' a — h c — d
that is, a — b:b = c — d:d.
In like manner it may be shown that a — b:a=c — d:c. Therefore, etc. q.e.d.
Proposition VII
278. 1. Form a proportion. How does the ratio of the sum of the first two terms to their difference compare with the ratio of the sum of the last two terms to their difference?
Theorem, In amj proportion, the ratio of the sum of the first two terirvs to their difference is equal to the ratio of the sum of ths last two terms to their difference; that is, the terms are in proportion by composition and division*
Data : a : ft = c : c?.
To prove a -{- b i a — b == c -\- d : c — d.
Proof. §§276,267, <^l±A=,^±A, (1)
b d
and, §§ 277, 267, ^^^ = ^-^. (2)
b d
Dividing (1) by (2),
a-{-b _c-{-d^ a — b G — d^
that is, a-\-b:a — b — c-\-d'.Q — d.
Therefore, etc. q.e.d.
142 PLANE GEOMETRY,^ BOOK IIL
Proposition VIII
279. 1. Form a series of equal ratios, as 2 : 3 = 4 : 6 = 8 : 12 = 10 : 15. How does the ratio of the sum of the antecedents to the sum of the con- sequents compare with the ratio of any antecedent to its consequent?
2. Transform similarly and investigate other series.
Theorem. In a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any ante- cedent is to its consequent.
Data : Any series of equal ratios, as a:b = C'.d = e :/= g : h. To prove a-f c ^ e -^g-.h-^-d +/+ A = a : 6, or c : d, etc.
Proof. Denoting each ratio by r, " = - = - = ^ = r. (1)
b d f h
From (1), a^hr, c^dr, e^ftf and g^hr. (2)
Adding equations (2),
a-hc-he4-g' = (&4-d 4-/-I- h)r. (3)
Dividing (3) by(b^d +/+ h),
g-f c4-e4-9^y b-^d 4-/+ h Substituting the value of r in (1),
b + d+f+h-b' ""' d' ^' that is, a-^c-^e-^g:b + d +/+ fe = a : 6, or c : d, etc. Therefore, etc. q.e.to.
• Ex. 398. It a:b = c:d, prove that a:a + 6 = C:c + d.
Ex. 399. liaib = b ic, prove that a -\- b : b + c = a :b.
Ex. 400. AD bisects angle A at the base of the isosceles triangle ABC^ and meets the side BO in D. If angle 0 is 68°, what is its ratio to angle ADB?
Ex. 401. The sura of the angles of a polygon expressed in right angles is to the number of its sides as 4 is to 3. How many sides has the polygon ?
Ex. 402. If the angle formed by two secants intersecting without a circle is 30° and the smaller of the intercepted arcs is 20°, what is the ratio of the smaller arc to the larger ?
Ex. 403. If the angle formed by two intersecting chords of a circle is 40° and one of the intercepted arcs is 30°, what is the ratio of that arc to the opposite intercepted arc ?
PLANE GEOMETRY. — BOOK III. 143
Proposition IX
280. Form two or more proportions in which the corresponding consequents are equal, as 2:3 = 6:9 and 5 : 3 = 15 : 9. How does the ratio of the sum of the antecedents of the first couplets to their common consequent compare with the ratio of the sum of the antecedents of the second couplets to their common consequent?
Theorem, When two or more proportions have the same quantity as the consequents of the first couplets and an- other quantity as the consequents of the second couplets, the sum of the antecedents of the first couplets is to their common consequent as the sum of the antecedents of the second couplets is to their common consequent.
Data : a:b = c:d, (1)
e:b=f:d, (2)
and g:b = h:d. (3)
To prove a-\-e-{-g:b = c -|-/+ h : d,
(4)
. (6) (6)
|
Proof. |
From (1) |
a c I'd' |
|
|
from (2) |
h d' |
||
|
from (3) |
9. = ^., b d' |
||
|
...(4) + (5) + (6), « |
+ e + ^ c-hf+h, b ~ d ' |
||
|
that is, |
a + ' |
e+9 |
■.b = c+f+h:d. |
|
Therefore, etc. |
Q.E.D.
281. Cor. When two or more proportions have the same quan- tity as the antecedents of the first couplets, and another quantity as the antecedents of the second couplets, the common antecedent of the first couplets is to the sum of their consequents as the common ante- cedent of the second couplets is to the sum of their consequents.
}lx. 404. It a:b = c:d, prove that 2 a + h : b =2 c + d: d.
Ex. 405. It a:b = c :d, prove that a :S a + b = c :3 c + d,
Ex. 406. If a :b = b :c, prove that 2 a — b : a = 2b — c :b,
Ex. 407. Ua:b = b:c, prove that a + 36:d = 6 + 3c:c
144 PLANE GEOMETRY. — BOOK III,
Proposition X
282. 1. Form a proportion ; multiply or divide the terms of either ratio by any number. How do the resulting ratios compare ? 2. Transform similarly and investigate other proportions.
Theorem. If in a proportion the terms of either coup- let are inultiplied hy any quantity, the resulting ratios forin a proportion.
Data : a:h =C'.d.
|
To prove |
ma |
:7nb = |
= c:d. |
|
|
Proof. |
a |
c d |
||
|
Multiplying both terms of the first fraction |
by |
|||
|
ma mb |
c ^d' |
|||
|
that is, |
ma |
:mb = |
= c:d. |
|
|
Therefore, etc. |
Q.E.D.
283. Cor. If in a proportion the terms of either couplet are divided by any quantity, the resulting ratios form a proportion.
Proposition XI
284. 1. Form a proportion; multiply or divide the antecedents or the consequents by any number. How do the resulting ratios compare?
Theorem, If in any proportion the antecedents or the consequents are multiplied hy the same quantity, the re- sulting ratios are in proportion.
Data : a\b = c:d.
To prove ma:b = mc: d,
and a:nb = c: nd.
M- (1)
Multiplying (1) by m, ^ = ^ ; that is, ma:b = mc:d.
PLANE GEOMETRY. — BOOK IIL 145
a c
Dividing (1) by 71, ^ = ^;
that is, a:nh = c'. nd.
Therefore, etc. q.e.d.
285. Cor. If in any proportion the antecedents or the conae" quents are divided by the same' quantity, the resulting ratios are in proportion.
Proposition XII
286. 1. Form several proportions. Multiply together their correspond- ing terms, and discover whether the resulting quantities form a pro- portion.
2. If there is an equal antecedent and consequent in the same couplet, or in corresponding couplets, cancel them from the products of the corre- sponding terms. Do the resulting quantities form a proportion ?
Theorem. The products of the corresponding terms of any number of proportions are in proportion.
Data : a:b = c:d, e :f= g : h, and k: l = m:o.
To prove aek : bji = cgm : dho.
^ ^ a c e q ^ h m
Proof. T = :5' >=?? ^^d y = —
b d f h I o
Multiplying these equations together,
aek _ cgm
'bfl~~dho'^ that is, aek : bjl = cgm : dho.
Therefore, etc. q.e.d.
287. Cor. In finding the proportion formed by the products of the corresponding terms of any number of proportions, an equal antecedent and consequent in the same couplet, or in corresponding couplets, may be dropped,
Por, if a:b = ci d,
and b:e—f:c,
§ 286, ab:be = cf: do.
Dividing the terms of the first couplet by b and the terms of the second by c, § 283, a : e =/; d,
milne's geom. — to
146 PLANE GEOMETRY, — BOOK III,
Proposition XIII
288. 1. Form a proportion; raise the terms of both ratios to the same power. How do the resulting ratios compare ?
2. Extract the same root of the terms of both ratios in a proportion, as 4 : 9 = 16 : 36. How do the resulting ratios compare ?
3. Transform similarly and investigate other proportions.
Theorem. In any proportion, like powers or like roots of the terms are in proportion.
Data : a : 6 = c d
Li II To prove i" : &" s=. c" • d", and a** : &" =s c" : d*.
Proof. J = |. (1)
Raising both fractions in (1) to the wth power,
that is, a''\b^=:c''i <t.
Extracting the nth root of both fractions io (1),
n — i»
that 13, or" ; ft'' » (f* : d*.
Therefore, etc. Q.I5.D.
Ex. 408. Make the changes that may be made upon the following propor- tion without destroying the equality of the ratios : 16 : 36 :a 4 : 9. Ex. 409. If a : 6 = c : df, prove that ma inh = 71(10'. nd. Ex. 410. If a:6 = c:(f, prove that a -h^b :b = c + id:d, Ex. 411. If a : 6 = 6 : c, prove that a^ -{• ab : b^ + be = a iC Ex. 412. If a : 6 = 6 : c, prove that a : c = (a + 6)2 : (6 + c)*. Ex. 413. If a: b = m:n, and b:c = n:Of prove that aic^mti^ Ex. 414. If a : 6 = c : ^, prove that
ma — nb : ma •{■ nb = mc -~ nd :mc ^ nd Ex. 415. Ua:b = c:d, prove that
So + 4(>:4o-6& = 8c + 4d:4c-6<i
BOOK IV
PROPORTIONAL LINES AND SIMILAR FIGURES
Proposition I
289. 1. Draw a line parallel to the base of a triangle through the middle point of one side and cutting the other side. How do the seg- ments of the other side compare in length ?
2. Draw a line parallel to the base one fourth, one sixth, or any part of the distance from the extremity of the base to the vertex. How do the segments of the other side compare ?
3. How does the ratio of one of these sides to either of its segments compare with the ratio of the other to its corresponding segment ?
Theorem, A line which is parallel to one side of a tri- angle and meets the other two sides divides those sides proportionally.
c Data: Any triangle, as ABC, and any /\
line parallel to AB, as de, meeting AC /......X^^
and BC in D and E, respectively. / ]7Z ^v
To prove CD \DA = CE :EB. ^Z '— —^5 -M
Proof. Case I. When CD and DA are commensurable. Suppose that Jif is a unit of measure common to CD and DA, and that M is contained in CD 3 times and in D^ 2 times.
Then, hyp., CD \DA = Z:2.
Divide CD and DA into parts each equal to the common measure if, and from each point of division draw lines parallel to AB. § 157, these lines divide CE into 3 and EB into 2 equal parts ;
.-. CE:EB = ^.2,
and, Ax. 1, CD : DA = CE : EB.
147
148 PLANE GEOMETRY. — BOOK IV.
Case II. When CD and DA are incommensurable.
Since CD and DA are incommensurable,
c suppose that CD and DF are commensur- A.
able and that FA < M. / ^\
Draw FG II DE. ^/ ~ \. ^
Case I, CD :DF = CE : EG. ^ ~~ ^-^
If M is indefinitely diminished, the ratios CD : DF and CE : EG
remain equal, and indefinitely approach their limiting ratios
CD : DA and CE : EB, respectively. Hence, § 222, CD : DA = CE : EB.
Therefore, etc. q.e.d.
290. Cor. ' A line which is parallel to one side of a triangle and meets the other two sides divides them so that one side is to either of its segments as the other side is to its corresponding segment.
Proposition II
291. Draw a line dividing each of two sides of a triangle into halves, or into other proportional parts. What is the direction of this line with reference to the third side ?
Theorem. A lUie which divides two sides of a triangle proportionally is parallel to the third side. (Converse of Prop. I.)
Q
Data: Any triangle, as ABC, and /\ the line DE dividing AC and ^C so / ^vs^ that CA : CD = CB : CE. r/- ^^
To prove DE II AB. / \.
A^ ^J5
Proof. If DE is not parallel to AB, some other line drawn through D will be parallel to AB.
Suppose that DF is that line.
Then, § 290, CA\CD = CB -. CF\
but, data, CA :CD = CB:GE',
CB : CF = CB : CE j hence, § 272, OF = CE.
PLANE GEOMETRY.— BOOK IV. 149
But this is impossible unless F coincides with E ; that is, Ax. 11, unless DF coincides with DE.
Therefore, the hypothesis, that some line other than DE drawn through D is parallel to AB, is untenable.
Hence, DE II AB.
Therefore, etc. ^ q.e.d.
Proposition III
292. Draw a triangle whose sides are 6", 5", and 3", or any other dimensions ; bisect any one of its angles and produce the bisector to meet the opposite side.
How does the ratio of the segments of this side made by the bisector compare with the ratio of the sides of the triangle adjacent to these segments ?
Theorem. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.
Data: Any triangle, as ABC, and CD
the bisector of one of its angles, ACB.
To prove AD : DB = AC : CB.
D
Proof. From B draw a line parallel to CD and meeting AC pro- duced in E.
|
Then, § 289, |
AD |
: DB = AC : CE, |
|
|
also. |
Zr = Zs, |
Why? |
|
|
and |
Zt = Zv', |
Why? |
|
|
but, data, |
Zr = Zt; |
||
|
.-. |
Z s = Zv, |
Why? |
|
|
and, § 118, |
CB = CE. |
||
|
Substituting CB |
in the proportion for its equal CE, |
||
|
AD |
: DB :=AC: CB. |
||
|
Therefore, etc. |
Q.E.D. |
150
PLANE GEOMETRY,-^ BOOK IV.
Proposition IV
293. Draw a triangle whose sides are 6", 5", and 3", or any other
dimensions; bisect an exterior angle at any vertex and produce the bisector to meet the opposite side produced. How does the ratio of the distances from the point of meeting to each extremity of the opposite side compare with the ratio of the other sides of the triangle ?
Theorem, The bisector of an exterior angle of a triangle meets the opposite side produced at a point the distances of which from the extremMies of this side are proportional to the other two sides.
Data: A triangle, as ABO\ an exterior angle, as BC7i> ; and its bisector CE meeting AB pro- duced in E.
To prove AE : BE=AC . BO.
Proof. Draw BF II CE.
Then, § 290, also, and but, lata,
and, § 118,
AE:BE = AO: FO,
Z 5 as Z Vy
FC = BQ.
Substituting BQ in the proportion for its equal FC^
AE '. BE = AC : BC.
Therefore, etc.
Why? Why?
Why?
Q.B.D.
294. Sch. This proposition is not true, if the triangle is equi- lateral. • Why ?
Ex. 416. The base of a triangle is 10 ft. and the other sides 8 ft and 12 ft. Find the segments of the base made by the bisector of the vertical angle.
Ex. 417. The sides ^C and 5(7 of the triangle ABC dire 5 ft. and 8 ft. respectively. If a line drawn parallel to the base divides AC into segments of 2 ft. and 3 ft., what are the segments into which it divides BC2
PLANE GEOMETRY. — BOOK IV. 151
295. If a straight line is divided at a point between its extrem- ities, it is said to be divided internally. The line is equal to the sum of the internal segments.
If a straight line is produced and divided at a point on the part produced, it is said to be divided externally. The line is equal to the difference between the external segments.
c
Fig.l.
In Fig. 1, AB is divided internally at C, and AB = AG+ GB.
Fig. 2.
In Fig. 2, AB is produced and divided externally at E^ and AB = BE- AE.
296. If a line is divided at a given point so that one segment is a mean proportional between the whole line and the other seg- ment, it is said to be divided in extreme and mean ratio.
In Fig. 1, if AB'.AC^AC: CB,
4.J5 is divided internally in extreme and mean ratio at the point C.
In Fig. 2, if AB : AE = AE : BE,
AB is divided externally in extreme and mean ratio at the point E.
297. If a line is divided internally and externally into seg- ments which have the same ratio, it is said to be divided har- monically.
-E
c B Fig. 8.
InFig. 3, if ^C:C5 = 6:3,
and if AE:BE = Q'.^,
AC : CB = AE : BE,
and AB is divided harmonically at the points C and E.
Ex. 418. A line drawn parallel to the base of triangle ^BC divides AC into segments of 3^'" and 8<^"> respectively, and the segment of J5C, corre- sponding to S*!™, is 6^"". What is the length ot BC?
152 PLANE GEOMETRY. — BOOK IV.
Proposition V
298. Draw a triangle whose sides are 6", 5", and 3", or any other dimensions ; bisect an interior and an exterior angle at one vertex and produce the bisectors to meet the opposite side and the opposite side produced respectively. How do the ratios of the internal and external segments of the opposite side compare ?
Theorem, The bisectors of an interior and of an exterior angle at one vertex of a triangle divide the opposite side harmonically.
Data: Any triangle, as ABC, ^
and the bisectors CE and CF of the interior and an exterior angle at C, respectively.
To prove AE : EB == AF : BF.
Proof. § 292, AE : EB =z AG \ BC,
§293, AF:BF=:AC:BC',
hence, AE : EB = AF : BF. Why ?
Therefore, etc. q.e.d.
299. Polygons whose homologous angles are equal, and whose homologous sides are proportional, are called Similar Polygons.
In similar polygons, points, lines, and angles that are similarly situated are called homologous points, lines, and angles.
Equal polygons are similar, but similar polygons are not necessarily equal.
Thus, all equilateral triangles are similar, but not all equilateral triangles are equal.
Ex. 419. The sides AB, BC, and AC of the triangle ABC are respec- tively 8 in., 5 in., and 10 in. The bisector of the exterior angle at C meets AB produced in E. What is the length of BE ?
Ex. 420. In triangle ABC, AC is 16"" and J?C is 5>n. The bisector of the exterior angle at C meets AB produced in E. If AE is 21'", what is the length of the side AB ?
Ex. 421. The bisectors of an interior and exterior angle at C of the triangle ABC meet the opposite side and the opposite side produced in E and F, respectively. If AB is 14 in. and EB is 4 in., what are the interna' and external segments of AB ?
PLANE GEOMETRY. — BOOK IV, 153
Proposition VI
300. Draw a triangle ; draw another whose angles are equal, each to each, to the angles of the first and one of whose sides is double, or any other number of times, the homologous side of the first. How do the ratios of any two pairs of homologous sides jcompare? What kind of triangles are they? Why?
Theorem, Two triangles are similar, if the angles of one are equal to the angles of the other, each to each.
Data: Any two trian- gles, as ABO and BEF, in which angle A — angle D, angle B = angle E, and angle c = angle F.
To prove A ABC and ^'
DEF similar.
Proof. In- the greater triangle, ABC, measure CG equal to DF, CH equal to EF, and draw GH.
Then, § 100, A GHC = A DEF,
and /.CGH = Z.D = ZA', Why?
GH W AB. Why?
Hence, § 290, AC:GC = BC: HC,
or, substituting DF for its equal GC, and EF for its equal HC, AC:BF = BC:EF.
In like manner, AB : DE = BC : EF.
Since, in the A ABC and DEF the homologous sides are pro- portional, and, from data, the homologous angles are equal, § 299, A ABC and DEF are similar.
Therefore, etc. q.e.d.
301. Cor. I. Two triangles are similar , if two angles of one are equal to ttuo angles of the other, each to each.
302. Cor. II. Two right triangles are similar, if an acute angle of one is equal to an acute angle of the other.
Ex. 422. The sides of a triangle are 5, 7, and 9. If the side of a similar triangle homologous to 7 is 8, what are the other sides of the triangle ?
154
PLANE GEOMETRY. — BOOK IV,
Proposition VII
303. Draw any two triangles such that the sides of one are propor- tional to the sides of the other ; for example, draw one triangle whose sides are 3", 4", and 5", and another whose sides are 6", 8", and 10". How do the homologous angles compare in size ? What kind of triangles are they? Why?
TTieorem, Two triangles are similar, if the sides of one are proportional to the sides of the other, ea^ch to eaeh.
Data: Any two tri- angles, as ABC and DEF^ such that
AC :DF = BC:EF — AB : DE.
To prove A ABC and. ^ ^EF similar.
Proof. In the greater triangle, ABC, measure CG equal to DF^ GH equal to EF, and draw GH.
Data,
Then, § 291, hence, and, § 301, .-. § 299, that is,
But, dat^,
whence, § 272, and, § 107,
But hence.
AC iBF = BC lEFj AC: GC = BC\HC. GH II AB ; Z.A=/. CGH, /.B =iZ. CHG, A ABC and GHC are similar; AC: GC = AB: GH\ AC : DF = AB : GH. AC:DF = AB:DE; AB : GH = AB : BE ; GH = BE, A GHC = A DEF. A ABC and GHC are similar ; A ABC and BEF are similar.
Why?
Q.E.D.
304. Sch. In § 299 the characteristics of similar polygons were defined as :
1. Their homologous angles are equal.
2. Their homologous sides are proportional.
From § 300 and ^ 303 it is seen that in the case of triancjles,
PLANE GEOMETRY. — BOOK IV.
166
either condition involves the other; that is, if the homologous angles of two triangles are equal, the homologous sides are pro- portional, and conversely; hence, triangles are similar, if theii homologous angles are equal or if their homologous sides are proportional. In the case of polygons of more than three sides either condition may exist without involving the other.
Thus, a square and a rhombus may Rave their sides all equal and, conse- quently, proportional, but the angles of the square are right angles, and 'those of the rhombus are oblique ; therefore, the figures are not similar. Also a square and a rectangle have their angles all equal, but their sides may not be proportional ; consequently, the figures are not similar.
Proposition VIII
305. Draw two similar triangles whose sides are 3", 4", and 5", and 6", 8", and 10" respectively, or any two similar triangles; draw lines representing their altitudes. How does the ratio of their altitudes com- pare with the ratio of any two homologous sides ?
Theorem. The altitudes of similar triangles are to each other as any two homologous sides.
Data: Any two similar triangles, as ABC and DEF, and their altitudes, as CQ and FH, respectively.
To prove CG : FH = AC : DF = BC:FF = AB : DE,
Proof. Data, .-. § 299, §94, .-. § 302, and, § 299,
A ABC Sind DEF are similar; ZA = ZDy
Aagc and DHF are rt. A\ rt. Aagc and DHF are similar, CG : FH = AC : DF.
In like manner it may be shown that
CG:FH = BC:EF. But, § 299, BC'.EF = AB\DE\
hence,
CG : FH = AC : DF = BC I EF == AB : DE.
Therefore, etc.
Why?
Q.E.D.
166 PLANE GEOMETRY.^ BOOK IV.
Proposition IX
306. Draw two triangles such that an angle of one is equal to an angle of the other, and the including sides in the first triangle are 3" and 5", and in the second 6" and 10". How do the homologous angles compare? How do the ratios of any two pairs of homologous sides compare? What name is given to triangles that have such relations to 3ach other?
Theorem. Two triangles are similar, if an angle of one is equal to an angle of the other, and the sides about these angles are in proportUm.
Data: Any two trian- gles, as ABC and VEF, in which angle C = angle F, and AO:BO = DF: EF,
To prove A ABC and DEF similar.
Proof. In the greater triangle, ABC, measure CQ equal to BFf CH equal to EF, and draw QH.
Then, since, da.ta, Z.C = Af,
§100, AOHC^^ADEF.
Data, AC:BC = J)F:EF;
AC:BC=GC:HC. Why?
Hence, § 291, GH II AB,
ZA^Z CQH, and Z B = Z CHO ; Why ?
hence, § 301, A ABC and GHC are similar;
that is, A ABC and DEF are similar.
Therefore, etc. q.e.d.
Ex. 423. The sides of a triangle are 8^™, 10<*™, and 12^™ in length respec- tively. If a line 9'^"* long, parallel to the longest side, terminates in tlie other two, what are the segments into which it divides them ?
Ex. 424. If 'the bisector of an interior angle of a triangle divides the side opposite the angle into two segments which are 6 ft. and 8 ft. respectively, and if the side of the triangle adjacent to the 8 ft. segment is 20 ft., what is the length of the other side of the triangle f
PLANE GEOMETRY. — BOOK IV.
157
Proposition X
307. Draw two triangles whose sides are parallel, each to each, or per- pendicular, each to each. What may be inferred regarding the relative size of the homologous angles? Then, what kind of triangles are they?
Theorem. Two triangles are similar, if their sides are parallel, each to eojch, or are perpendicular, each to each.
Data : Any two tri- angles, as ABC and DBF, in which AB, AC, and BC are par- allel or perpendicular ^z i ^ ^^ j. ^^
respectively to DE, DFy and EF.
To prove A ABC and DEF similar.
Proof. By §§ 81, 83, angles which have their sides either paraL lei or perpendicular are either equal or supplementary.
1. Suppose that each of the angles of one triangle is supple- mentary to the corresponding angles of the other ; that is, suppose /.A-\-/:d=2 rt. Zs; Zj5 +Z^ = 2 rt. Zs; Zc + Zi^ = 2 rt. a
Then, the sum of the interior angles of the two triangles is equal to 6 rt. A, which is impossible.
2. Suppose that one angle of one triangle is equal to the cor- responding angle of the other, and that the other two angles of the triangles are supplementary, each to each; that is, suppose /La = /.D; Z^ + Z^=:2rt. Z; Z(7 + Zi^ = 2rt. Z.
Then, the sum of the angles of the two triangles exceeds 4 rt. Z, which is impossible.
3. Suppose that two angles of one triangle are equal to the corresponding angles of the other, each to each ; then the third angles must be equal ;
that is, suppose Z. A = Z D\ /.B = Z.E,
then, /.C = /.F;
that is, A ABC and DEF are mutually equiangular.
Hence, § 300, A ABC and DEF are similar.
Therefore, etc. Q.E.D.
158 PLANE GEOMETRY. — BOOK IV,
Proposition XI
308. 1. Draw three or more lines which meet in a point and two parallel lines cutting them. Discover whether any pairs of triangles thus formed are similar. Are the pairs of bases proportional?
2. If three non-parallel lines intersect two parallel lines, making the intercepted segments 4" and 6" on one side of the middle line and 8" and 12" on the other side, will the non-parallel lines meet in the same point if produced ?
Theorem, Lines which meet in a point intercept propor- tional segments upon two parallel lines; conversely, non- parallel lines which intercept proportional segments upon two parallel lines meet in a point.
Data: Any lines, as AH^ BH, and CH, which meet at a point, as H, and intercept the segments AB, BC, DE, and EF upon two parallel lines, AC and DF.
To prove ABiDE = BC: EF.
Proof. Zr = Zs,Zt = Zv,Ziv = Zx, and Z y
: § 301, Aabh and BEH are similar,
and , A BCH and EFH are similar.
Then, § 299, AB :DE =BH: EH,
*».nd BC: EF = BH:EHj
AB : BE = BC : EF. *' Q.E.D.
Conversely : Data : Non-parallel lines, as AB, BE, and CF, inter- secting parallel lines AC and DF, so that AB : BE = BC : EF.
To prove that AD, BE, and CF, if produced, meet in a point.
Proof. Produce AD and BE to meet in H and draw CH.
Suppose that J is the point in which DF intersects CH.
Then, AB: DE = BC : EJ, Why ?
but, since, data, ab : de = BC : EF,
this is impossible, unless EJ = EF, and J and F coincide.
Then, CF passes through H.
Consequently, AD, BE, and CF meet in a point.
Therefore, etc. q.b.d.
PLANE GEOMETRY. — BOOK IV. 159
Proposition XII
t
309. Draw two polygons such that they may be divided into the same number of triangles, similar, each to each, and similarly situated. How do the homologous angles of these polygons compare in size ? How do the ratios of any two pairs of homologous sides compare ? What kind ot polygons are they ? Why ?
Theorem, Two polygons are similar, if each is composed of the same number of triangles which are simAlar, each to eoAih, and similarly placed.
I^ata: Any two polygons, as ABODE and FGHJK, composed of triangles ABC, ACD, and ABE ; and FGH, FHJ, and FJK, respec- tively, which are similar, each to each, and are similarly placed.
To prove ABODE and FGHJK similar.
Proof. /Lb =Zg. Why ?
Also, ' Z.r = /.Sy
and At = /.v,
/.BOD = Z.GHJ. Why?
In like manner it may be shown that Z ODE = Z HJK, etc.
Hence, the homologous angles of the polygons are equal.
Again, § 299, AB : FG = BO : GH = AC : FH = OD : HJ = etc., or AB : FG = BC : GH = OD : HJ = etc. ;
that is, the homologous sides of the polygons are proportional.
Hence, § 299, ABODE and FGHJK are similar.
Therefore, etc. q.b.d.
Ex. 425. If a stick .3 ft. long, in a vertical position, casts a shadow 1 ft. 7^ in. long, how high is a church steeple which at the same time cast* a shadow 78 ft. in length ?
160
PLANE GEOMETRY. — BOOK IV.
Proposition XIII
• 310. Draw two similar polygons and from two homologous vertices draw diagonals dividing the polygons into triangles. How many tri- angles are there in each polygon ? How do the homologous'^ angles of the corresponding triangles compare in size? How do the ratios of their sides compare ? Then, what kind of triangles are they ?
Theorem, If two polygons are similar, they may he divided by diagonals into the same numher of triangles which are similar, each to each, and similarly placed. (Converse of Prop. XII.)
Data: Any two similar polygons, as ABCDE and FGHJK. ' To prove that the polygons ABODE and FGHJK may be divided
by diagonals into the same number of triangles which are similar,
each, to each, and are similarly placed.
Proof. From any two homologous vertices, as A and F, draw
the diagonals AC, AD, FH, and FJ.
. In A ABC and FGH, Z. B = Z. G,
and AB :FG =BC:GH',
.'. § 306, A ABC and FGH are similar,
Z r = Zs]
Zbcd = Z ghj'j
Zt = Zv',
BC: GH= ACiFH, BC: GH= CDiHJj AC'.FH = CD :HJ, and, § 306, A ACD and FHJ are similar.
In like manner, A ADE and FJK are similar.
Therefore, etc.
and but
and
Why?
Why? Why? Why? Why? Why? Why?
Q.E.D.
PLANE GEOMETRY. — BOOK IV. 161
Proposition XIV .
311. Draw two similar polygons; measure the sides of each. How does the ratio of the perimeters, or the sums of the homologous sides, compare with the ratio of any two homologous sides?
Theorem, The perimeters of similar -polygons are to each other as any two homologous sides.
Data: Any two similar polygons, as ABODE and FGHJK, Denote their perimeters by P and Q respectively.
To prove P : Q = AB : FG = etc.
Proof. AB : FG = BC : GH = CD : HJ = etc. ; Why ?
.-. § 279, AB -i-BC-^- etc. : FG + GH-{- etc. = AB:FG = etc. ; that is, P : Q = AB : FG = etc.
Therefore, etc. q.e.d.
Proposition XV
312. 1 . Draw a right triangle whose sides are 3", 4^', and 5", or any other right triangle ; from the vertex of the right angle draw a perpen- dicular to the hypotenuse. How do the angles of each of the triangles thus formed compare in size with the homologous angles of the original triangle"? How does the ratio of the longer segment of the hypotenuse to the perpendicular compare with the ratio of the perpendicular to the shorter segment ?
2. How does the ratio of the hypotenuse to either side about the right angle compare with the ratio of the same side to the segment of the hypotenuse adjacent to it ?
3. Draw a circle and its diameter ; from any point in the circumfer- ence draw a perpendicular to the diameter. How does the ratio of the longer segment of the diameter to the perpendicular compare with the ratio of the perpendicular to the shorter segment ?
MILNfi's OEOM. — 11
162 PLANE GEOMETRY. — BOOK IV.
Theorem, If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the perpendicular is a mean proportional between the seg- ments of the hypotenuse.
Data: Any right triangle, as ABC, and the perpendicular CD, from the vertex of the right angle C, upon AB.
To prove AD : CD = CD : BD.
Proof. In the rt. A ABC and ACDy
Za is common ;
.*. § 302, A ABC and ACD are similar.
In like manner it may be shown that A ABC and CBD are similar ; hence, A ^ CD and C^i) are similar. Why?
Now, AC, AD, and CD are the sides of A ACD homologous respectively to BC, CD, and BD of A CBD ;
hence, § 299, AD : CD =z CD :BD.
Therefore, etc. q.e.d.
313. Cor. I. Each side about the right angle is a mean propor- iional between the hypotenuse and the adjacent segment.
314. Cor. II. The perpendicular to a diameter from any point in the circumference of a circle is a mean proportional between the segments of the diameter.
Proposition XVI
315. From a point without a circumference draw a tangent and a secant; from the point of tangency draw chords to the points at which the secant intersects the circumference. What angles of the figure are equal? What two triangles are similar? Then, how does the ratio of the secant to the tangent compare with the ratio of the tangent to the external segment of the secant ?
Theorem. If from a point without a circle a secant and a tangent are drawn, the tangent is a mean proportional between the whole secant and the external segment.
PLANE GEOMETRY. — BOOK IV,
Data: Any circle, as BCD', any point without, as A ; any secant from A, as ADB ; and the tangent from ^, as ^C.
To prove
AB : AC = AC: AD,
Proof. Draw BC and DC.
In A ABC and ADC, Z A is common, § 225, Z ^ is measured by | arc DC,
and, § 231, Z ACD is measured by ^ arc DC;
.'. ■ Zb ==Zacd.
Hence, § 301, A ABC and ADC are similar, and, §299, AB :AC = AC:AD.
Therefore, etc.
Proposition XVII
Why?
Q.E.D.
316. JProhlem, To divide a straight line into parts pro- portional to any number of given lines.
n , H G r.
D^'
^-0
Data : Any straight line, as JIS ; also the lines I, m, and n.
Required to divide AB into parts proportional to I, m, and n.
Solution. From one extremity of AB, as A, draw a line, as AC, making with AB any convenient angle.
On ^C measure AD, DE, and EF equal to I, m, and n respec- tively. Draw FB.
Through D and E draw lines parallel to FB, meeting AB in H and G respectively.
Then, AH, HG, and GB are the parts required. q.e.f.
Proof. By the student. Suggestion. Refer to § 289.
164 PLANE GEOMETRY. — BOOK IV.
Proposition XVIII
317. Problem. To find a fourth* proportional to three £iven lines.
A-^--
Y
F'
"0
Data : Any three lines, as I, m, and n. Required a fourth, proportional to I, m, and n. Solution. Draw any two lines, as AB and ACj forming any convenient angle at ^. On AB take AD = m. On AC take AE = I, and EF = n. Draw ED.
From F draw a line parallel to ED meeting AB in G. Then, DG is the proportional required. q.e.f.
Proof. By the student. Suggestion. Refer to § 289.
Proposition XIX
318. Problem, To find a third "t proportional to two given lines.
I . m D G „
Data : Any two lines, as I and m. Required a third proportional to I and m.
Solution. Draw any two lines, as AB and AC, forming any convenient angle at ^. On ^j5 take AD = m. On ^C take AE = l, and EF = m.
* When a : 6 = c ; d, d is termed the fourth proportional to a, ft, and c. t When a : 6 = 6 : c, c is termed the third proportional to a and 6.
PLANE GEOMETRY. — BOOK IV. 165
Draw ED.
From F draw a line parallel to ED meeting AB in G.
Then, DG i^ the proportional required. q.e.p.
Proof. By the student. Suggestion. Refer to § 289.
Proposition XX
319, Problem, To find a mean proportional between two given lines. .
m [^ I
A C D B
Data : Any two lines, as I and m. Required a mean proportional between I and m. Solution. Draw any line, as AB. On AB take AG = 1, and GD = m. On AD as a diameter describe a semicircumference. At G erect a perpendicular to AD meeting the semicircumfer- ence in E.
Then, GE is the required proportional. q.e.f.
Proof. By the student. Suggestion. Refer to § 314.
Ex. 426. Three lines are 10^% 12c'n, and 16°™. Construct their fourth proportional.
Ex. 427. Two lines are 11cm and 9*5™. Construct their third proportional.
Ex. 428. Two lines are 6^™ and 2°™ Construct their mean propor- tional.
Ex. 429. Tangents are drawn through a point 6™ from the circumference of a circle whose radius is 9">. Find the length of the tangents.
Ex. 430. If one side of a polygon is 2 ft. 6 in. long, what is the length of the corresponding side of a similar polygon, if their perimeters are respectively 15 ft. and 25 ft. ?
Ex. 431. The shortest distance from a given point to the circumference of a given circle is 2 ft. The length of a tangent from the same point to the circumference is 3 ft. Find the diameter of the circle.
Ex. 432. Five straight lines passing through the same point intercept segments on one of two parallel lines, of 12^"^^ 20dm, 28^™, and 36<*m, ^he segment of the other parallel corresponding to the 20'*™ segment is 15"*™. Find the other segments.
166
PLANE GEOMETRY. — BOOK IV.
320. Problem,
ratio.
Proposition XXI To divide a line in extreme and mean
- .c
•-/P •
A F B O
Datum: Any line, as ^5.
Required to divide AB in extreme and mean ratio.
Solution. At one extremity of AB, as A, draw AC perpendicular to AB and equal to \AB.
With C as a center and ^ C as a radius describe a circumference.
Through C draw BE cutting the circumference in D and meeting it in E. On AB take BF = BB and on AB produced take BG=BE.
Then, AB : BF = BF : AF,
and AGiBG = BG : AB',
that is, § 296, AB is divided at F internally, and at G externally^ in extreme and mean ratio. q.e.f.
Proof. § 315, BE:AB=AB:BD',
.-. § 277, BE ~ AB : AB = AB — BD : BD, (1)
and, § 276, AB -f BE : BE = BD + AB : AB. (2)
Const., DE = 2 AC = AB',
hence, BE — AB = BE — be = BD = BF.
Substituting in (1) for BE — AB its equal BF, for AB — BD its equal AF, and for BD its equal BF,
BF: AB =AF: BF, or, § 275, AB:BF = BF: AF.
Const., AB + BE = AG, and BD-\-AB= BE.
Substituting in (2) for AB -^BE its equal AG, and for BD -\- AB its equal BE, AG : BE = BE : AB.
Since, const., BE = BG,
AQ : BG == BQ : AB,
PLANE GEOMETRY. — BOOK IV. 167
Proposition XXII
321. Broblem, Upon a given line to construct a polygon similar to a given polygon.
>b:
Data: Any polygon, as ABODE, and any line, as FG.
Required to construct on FG a polygon similar to ABODE.
Solution. Draw AO and AD.
At F and G construct Z t and Z v equal respectively to Z r and Z s.
Produce the sides from F and G until they meet at H.
In like manner on FH construct A FHJ, and on FJ, A FJK, similar respectively to A AOD and ADE.
Then, FGHJK is the required polygon. q.e.f.
Proof. By the student. Suggestion. Refer to §§ 301, 309.
SUMMARY 322. Truths established in Book IV.
1. Two lines are parallel,
a. If one divides two sides of a triangle proportionally and the other is the third side. § 291
2. Lines are in proportion,
a. If they are segments of two sides of a triangle made by a line parallel to the third side. ' § 289
b. If they are two sides of a triangle and their corresponding segments made by a line parallel to the third side. § 290
c. If they are two sides of a triangle and the segments of the third side made by the bisector of the angle opposite that side. § 292
d. If they are two sides of a triangle and the external segments of the third side made by the bisector of the exterior angle at the vertex opposite that side. §293
168 PLANE GEOMETRY. — BOOK IV.
e. If they are the internal and external segments of a side of a triangle made by the bisectors of an interior and exterior angle at the vertex opposite that* side. § 298
/. If they are the altitudes and homologous sides of similar triangles.
§305
g. If they are segments of parallel lines made by lines which meet in a point. § 308
h. If they are homologous sides of similar polygons. § 299
i. If they are perimeters of similar polygons and any two homologous sides. §311
3. A line is a mean proportional between two other lines,
a. If it is the perpendicular to the hypotenuse of a right triangle from the vertex of the right angle, and the other lines are the segments of the hypotenuse. § 312
b. If it is either side about the right angle of a right triangle, and the other lines are the hypotenuse and the segment of it adjacent to that side made by the perpendicular from the vertex of the right angle. § 318
c. If it is the perpendicular to the diameter of a circle from any point in the circumference, and the other lines are the segments of the diameter. § 314
d. If it is a tangent to a circle from any point without, and the other lines are a secant from the same point and its external segment. § 315
4. Lines pass through the same point,
a. If they are non-parallel lines that intercept proportional segments upon two parallel lines. § 308
6. Two angles are equal,
a. If they are homologous angles of similar polygons. § 299
6. Two triangles are similar,
a. If the angles of one are respectively equal to the angles of the other. § 300
b. If two angles of the one are respectively equal to two angles of the other. § 301
c. If they are right triangles and an acute angle of one is equal to an acute angle of the other. § 302
d. If the sides of one are proportional respectively to the sides of the other. § 303
e. If an angle of one is equal to an angle of the other and the including sides are in proportion. § 30(»
/. If their sides are parallel, each to each. § 307
PLANE GEOMETRY.— BOOK IV. 169
g. If their sides are perpendicular, each to each. § 307
h. If they are the corresponding triangles of similar polygons divided by
homologous diagonals. § 310
7. Two polygons are similar,
a. If they have their homologous angles equal and their homologous sides proportional. § 299
b. If each is composed of the same numfcer of triangles similar each to each and similarly placed. § 309
SUPPLEMENTARY EXERCISES
Ex. 433. Construct a triangle whose sides are 6, 8, and 10 ; then con- struct a similar triangle whose side homologous to 8 is 5.
Ex. 434. Divide a line lO'^™ long internally in extreme and mean ratio.
Ex. 435. The median from the vertex of a triangle bisects every line drawn parallel to the base and terminated by the sides, or the sides produced.
Ex. 436. Two circles intersect at A and JB, and at A tangents are drawn, one to each circle, to meet the circumference of the other in C and D respec- tively ; BC, BB, and AB are drawn. Prove that BD is a third proportional to ^C and AB.
Ex. 437. The diameter AB of a circle whose center is 0 is divided at any point O, and CD is drawn perpendicular to AB, meeting the circum- ference in D ; OD is drawn, and CE perpendicular to OD. Prove that DE is a third proportional to ^0 and DC.
Ex. 438. In the triangle ABC, AD is the median to BC; the angles ADC and ADB are bisected by DE and DE, meeting AC and AB in E and F respectively. Then, FE is parallel to BC.
Ex. 439. A secant from a given point without a circle is 1 ft. 6 in. long, and its external segment is 8 in. long. Find the length of a tangent to the circle from the same point.
Ex. 440. The radius of a circle is 6 in. What is the length of the tan- gents drawn from a point 12 in. from the center ?
Ex. 441. If the tangent to a circle from a given point is 2^^ and the radius of the circle is 16^"^, find the distance from the point to the circum- ference.
Ex. 442. If from the vertex D of the parallelogram ABCD a straight line is drawn cutting AB at E and CB produced at F, prove that OF is a fourth proportional to AE, AD, and AB.
170 PLANE GEOMETRY. — BOOK IV,
Ex. 443. If the segments of the hypotenuse of a right triangle made by the perpendicular from the vertex of the right angle are 6 in. and 4 ft. , find the length of the perpendicular and the length of each of the sides about the right angle.
Ex. 444. Eind the length of the longest and of the shortest chord that can be drawn through a point 7^ in. from the center of a circle whose radius is 19^ in.
Ex. 445. If the greater segment of a line divided internally in extreme and mean ratio is 36 in., what is the length of the line ?
Ex. 446. The shorter segment of a line divided externally in extreme and mean ratio is 240^™, Find the length of the greater segment in meters.
Ex. 447. Find the shorter segment of a line 12^™ long when it is divided internally in extreme and mean ratio. When it is divided externally in extreme and mean ratio.
Ex. 448. The tangents to two intersecting circles drawn from any point in their common chord produced are equal.
Ex. 449. If the common chord of two intersecting circles is produced, it will bisect their common tangents.
Ex. 450. ABC is a straight line, ABD and BCE are triangles on the same side of it, having angle ABD equal to angle CBE and AB:BC = BE : BD. If AE and CD intersect in i^, triangle AFC is isosceles.
Ex. 451. If in the triangle ABC^ CE and BD are drawn perpendicular to the sides AB and AC respectively, these sides are reciprocally propor- tional to the perpendiculars upon them ; that is, AB : AC = BD : CE.
Ex. 452. ABCD is a parallelogram. If through O, any point in the diagonal AC, EF and GH are drawn, terminating in AB and DC, and in AD and BC respectively, iE'//is parallel to GF.
Ex. 453. Lines are drawn from a point P to the vertices of the triangle ABC; through i>, any point in PA, a line is drawn parallel to AB, meeting PB at E, and through E a line parallel to BC, meeting PC at F. If FD is drawn, triangle DEF is similar to triangle ABC.
Ex. 454, If two lines are tangent to a circle at the extremities of a diameter, and from the points of contact secants are drawn terminated respectively by the opposite tangent and intereecting the circumference at the same point, the diameter is a mean proportional between the tangents.
Ex. 455. AB and AC are secants of a circle from the common point A, cutting the circumference in D and E respectively. Then, the secants are reciprocally proportional to their external segments ; that is, AB : AC = AE : AD.
Suggestion. Draw CD and BE, and refer to § 322, 6, b.
PLANE GEOMETRY.--- BOOK IV, 171
Ex. 456. AB and CD are two chords of a circle intersecting at E. Prove that AE : DE = CE : BE.
Ex. 457. Two secants intersect without a circle. The segments of one are 4 ft. and 20 ft., and the external segment of the second is 16 ft. Find the length of the second secant.
Ex. 458. From a point without a circle two secants are drawn, whose external segments are respectively 7<^™ and Q*^™, the internal segment of the latter being \S^^\ What is the length of the •first secant ?
Ex. 459. The segments of a chord intersected by another chord are 7 in. and 9 in., and one segment of the latter is 3 in. What is the other segment ?
Ex. 460. Two secants from the same point without a circle are 24'^'" and 32dm long. If the external segment of the less is 5<*™, what is the external segment of the greater ?
Ex. 461. Through a point ?•" from the circumference of a circle a secant 28'n long is drawn. If the internal segment of this secant is XT'", what is the radius of the circle ?
Ex. 462. If from any point in the <iiameter of a circle produced a tan- gent is drawn and a perpendicular from the point of contact is let fall on the diameter, the distances from the point without the circle to the foot of the perpendicular, the center of the circle, and the extremities of the diameter are in proportion.
Suggestion. Draw the radius to the point of contact.
Ex. 463. If the sides of a triangle are respectively 1.6^^^ .12Hm^ and 10™ long, what are the segments into which each side is divided by the bisector of the opposite angle ?
Ex. 464. If an angle of one triangle is equal to an angle of another, and the perpendiculars from the vertices of the remaining angles to the sides opposite are proportional, the triangles are similar.
Suggestion. Refer to § 322, 6, c and e.
Ex. 465. If two circles are respectively 6 in. and 3 in. in diameter and their centers are 10 in. apart, find the distance from the center of the smaller one to the point of intersection of their common exterior tangent with their line of centers produced.
Ex. 466. Two intersecting chords of a circle are 38 ft. and 34 ft. respec- tively ; the segments of the first are 8 ft. and 30 ft. Find the segments of the second.
Ex. 467. What is the length of a chord joining the points of contact of the tangents drawn from a point 13 in. from the center of a circle whose radius is 5 in. ?
Ex. 468. Chords AB and CD of a circle are produced in the direction of B and D respectively to meet in the point Ej and through E the line EF is drawn parallel to AD to meet GB produced in F. Prove that EF is a mean proportional between FB and FC
172 PLANS QnOMBTRY,^BOOK IP,
Ex. 469. AB is a diameter of a circle, and through A any straight line is drawn to cut the circumference in O and the tangent at ^ in Z>. Prove that ^C is a third proportional to AD and AB.
Ex. 470. From any point in the base of a triangle straight lines are drawn parallel to the sides. Prove that the intersection of the diagonals of every parallelogram so formed lies in a line parallel to the base of the triangle.
EXo 471. If E is the middle point of one of the parallel sides DC of the trapezoid ABGD^ and AE and BE produced meet BG and AD produced in F and G respectively, prove that QF is parallel to AB.
Ex. 472. If a line tangent to two circles cuts their line of centers, the segments of the latter are to each other as the diameters of the circles.
Ex. 473. The bisector of the vertical angle C of the inscribed triangle ABG cuts the base at D and meets the circumference in E. Prove that AG'.GD=GE'.BG.
Ex. 474. Through any point A of the circumference of a circle a tangent Is drawn, and from A two chords, AB and AG\ the chord FG parallel to the tangent cuts AB and ^ C in 2> and E respectively. Prove AB ; AE= A G : AD.
Ex. 475. The greatest distance of a chord 8 ft. in length from its arc is 4 in. Find the diameter of the circle.
Ex. 476. If two circles are tangent externally, their common exterior tangent is a mean proportional between the diameters of the circles.
Suggestion. Draw radii to the points of contact, draw the common interior tangent to intersect the common exterior tangent, and connect the point of intersection with the centers.
Ex. 477. The perpendicular from any point of a circumference upon a chord is a mean proportional between the perpendiculars from the same point upon the tangents drawn at the extremities of the chord.
Suggestion. Draw lines from the given point to the extremities of the chord, and refer to § 322, 6, c.
Ex. 478. From a point A tangents AB and ^C are drawn to a circle whose center is 0, and BD is drawn perpendicular to GO produced. Prove that BD is a fourth proportional to AO^ GD^ and GO.
Suggestion. Draw AO and BG.
Ex. 479. From a point E in the common base of two triangles ABG and ABD^ straight lines are drawn parallel to ^C and AD^ meeting BG and BD at F and G respectively. Prove that FG is parallel to GD.
Ex. 480. If tangents to a circle are drawn at the extremities of a diam- eter, the radius is a mean proportional between the segra«»nts of any third tangent intercepted between them and divided at its point of tangency.
Suggestion. Draw lines to form a right triangle, havmg the third tangen* for its hypotenuse and a vertex at the center.
BOOK V
AREA AND EQUIVALENCE
M
D
323. The amount of surface in a plane figure is called its Area.
A surface is measured by finding how many times it contains some given square which is taken as a unit of measure.
The ordinary units of measure for surfaces are the square inch, the square foot, the square centimeter, the square decimeter, etc.
Suppose that the square M is the unit of measure, and that ABCD is the rectangle to be meas- ured.
By applying M to ABCD it is evident that the rectangle may be divided into as many rows of squares, each equal to if, as the side of M is contained times in
the altitude of ABCB ; that in each row there are as many squares as the side of M is contained times in the base of ABCD-, and therefore, that the product of the numerical measures of the base and altitude of ABCD is equal to the number of times that M is contained m. ABCD.
In this case the side of M is contained 4 times \n AD and 6 times in AB ; consequently, M is contained 24 times in ABCD-, that is, the rectangle contains 24 square units.
Therefore, if the side of a square is a common measure of the base and altitude of a rectangle, the product of the numerical measures of the base and altitude expresses the number of times that the rectangle contains the square, and is the numerical measure of the surface, or the area of the rectangle.
324. For the sake of brevity, the product of the base and altitude is used instead of the product of the numerical measures of the base and altitude.
178
174
PLANE GEOMETRY. — BOOK V.
The product of two lines is, strictly speaking, an absurdity, but since the expression is used to denote the area of a rectangle it follows, that the geometrical concept of the prodiict of two lines is the rectangle formed by them.
Thus, AB X CD implies a product, which is a numerical result, but it must be interpreted geometrically to mean rect. AB • CD.
For similar reasons, if AB represents a line, AB^ must be inter- preted to mean geometrically the square described upon the line AB, and conversely, the square described upon a line may be indicated by the square of the line.
325. It has been stated in § 36 that equal figures may be made to coincide, consequently such figures have equal areas.
Figures, however, which cannot be made to coincide may have equal areas, and they are called equivalent figures.
All equal figures are equivalent, but not all equivalent figures are equal.
If a square and a triangle each contains one square foot of surface, they are equivalent ; but since they cannot be made to coincide, they are not equal.
The symbol of equivalence is =c=.
326. Since equivalent means equal in area, or in volume as will be shown, then, § 222 may be extended to apply to equivalent magnitudes; consequently, if, while approaching their respective limits, two variables are always equivalent, their limits are equivalent.
Proposition I
327. 1 . If a rectangle is 3" long and another of the same altitude is 6" long, how do they compare in area ? How, then, do rectangles having equal altitudes compare in area?
2. How do rectangles that have equal bases, but different altitudes, compare in area?
Theorem. Rectangles ivhich have equal altitudes are to ea/}h other as their bases.
D c H
|
1 |
E
PLANE GEOMETRY.— BOOK V.
175
Data : Any two rectangles, as ABCD and EFGH, whose altitudes, AD and EH, are equal.
To prove ABCD : EFGH = AB : EF.
Proof. Case I. When AB and EF are commensurable.
Suppose that iJ/ is a common unit of measure for AB and EF.
Apply M to each base, and suppose that it is contained in AB 7 times and in EF 4 times.
Then, AB :EF = 1 iL
Divide AB into 7 equal parts and EF into 4 equal parts, and at each point of division erect a perpendicular.
ABCD is thus divided into 7 rectangles, and EFGH into 4 rec- tangles.
Since, § 156, these rectangles are all equal, ABCD : EFGH = 7 : 4:. ABCD : EFGH = AB : EF. Why ?
Case II." When AB and EF are incommensurable.
KO
H
M
JB
Since AB and EF are incommensurable, suppose that iHf is a common unit of measure for AJ and EF, and that JB is less than M. Draw JK W AD.
Then, Case I, AJKD .: EFGH =AJ:EF; and JBCK is less than any one of the rectangles whose base is equal to M.
Now, if M is indefinitely diminished, the ratios AJKD : EFGH and AJ : EF remain equal and indefinitely approach the limiting ratios ABCD : EFGH and AB : EF respectively.
Hence, § 222, ABCD : EFGH = AB : EF.
Therefore, etc. q.e.d.
328. Cor. Rectangles which have equal bases are to each other as their altitudes.
176
PLANE GEOMETRY. — BOOK V,
Proposition II
329. Draw two rectangles whose bases are respectively 5'' and 3" and altitudes 2" and 4", or any other dimensions ; divide them into squares having a side of 1". How many square inches are there in the first rec- tangle? In the second? How does the ratio of the areas of the two rectangles compare with the ratio of the products of their bases by their altitudes ?
Theorem, Rectangles are to each other as the products of their bases by their altitudes.
Data : Any two rectangles, as A and B, of which d and m are the bases, and e and n the altitudes, respectively.
To prove A:B = dxe'.mxn.
Proof. Construct a rectangle c, having the base m and the altitude e.
^ : (7 = c? : m, C:5 = e: n; A:B =zd xeimxn,
Q.E.D.
Proposition III
Then, § 327, and, § 328, hence, § 287,
Therefore, etc.
330. How many square inches of surface are there in a rectangle that is 5" long and 1" wide? 5" long and 2" wide? 5" long and C" wide? 8'' long and 1" wide ? How may the amount of surface, or the area of any rectangle, be found?
Theorem, The area of a rectangle is equal to the prod- uct of its base by its altitude.
Data: Any rectangle, as A, whose base is d and altitude 6.
To prove area oi A = d xe.
PLANE GEOMETRY. — BOOK V. 177
Proof. Assume that the unit of measure is a square M, whose side is the linear unit.
§329, A:M = dxe:lxl,or ^ = ^^ = dxe.
M 1x1
But, § 323, the surface of A is measured by the number of times it contains the unit of measure M;
A
— = areaof^.
M
But — = c2 X e.
M
Hence, area oi A — dxe,
Therefore, etc. q.e.d.
Proposition IV
331. 1. Draw an oblique parallelogram, and on the same base a rec- tangle having an equal altitude. How do the triangles thus formed at the ends of this figure compare ? How does the area of the parallelo- gram compare with the area of the rectangle ?
2. What ratio do two rectangles have to each other ? (§329) What, then, is the ratio of two parallelograms to each other?
Theorem, A parallelogram is equivalent to the rectangle which has the same base and altitude.
Data: Any parallelogram, as ABCD, ^_d e c
whose base is AB and altitude BE. \ / /
To prove ABCB equivalent to the rec- \/ /
tangle whose base is AB and altitude BE. I v
Proof. Draw AF II BE and meeting OD produced in F.
Const., ABEF is a rectangle which has the same base and alti- tude as ABCD.
In rt. Abce and ADF, BE = AF, Why ?
and BC = AD', Why?
Abce = A adf. Why ?
Hence, Abce-\- abed ^Aadf + abed ;
that is, ABCD =o= ABEF.
Therefore, etc. q.e.d.
milne's geom. — 12
178 PLANE GEOMETRY. — BOOK V.
332. Cor. I. The area of a parallelogram is equal to the product of its base by its altitude.
333. Cor. II. Parallelograms are to each other as the products of their bases by their altitudes; conseqnentlj, parallelograms which have equal altitudes are to each other as their bases, imrallelograms which have equal bases are to each other as their altitudes, and parallelograms which have equal bases and equal altitudes are equivalent.
Proposition V
334. 1. Draw any triangle, and through two of its vertices draw lines parallel to the opposite sides, producing them until they meet. What part of the parallelogram thus formed is the original triangle? How does this parallelogram compare with a rectangle having the same base and altitude? What part of such a rectangle is the triangle?
2. What ratio do two rectangles have to each other? (§329) What, then, is the ratio of two triangles to each other?
Theorem, A triangle is equivalent to one half the rec- tangle which has the same base and altitude.
E O
Data: Any triangle, as ABC, whose base / ^' is AB and altitude CD. ! ^^ To prove A ABC =o= \ rect. AB • CB. l^^^ /
A B D
Proof. Draw AE W BC and CE il BA.
Then, ABCE is a parallelogram, AC \^ its diagonal, and, § 152, Aabc = A AEC,
or A ABC ^^ ABCE.
But, § 331, ABCE o= rect. AB • CD.
Hence, AABC -o^\ rect. AB • CD. q.e.d.
335. Cor. I. The area of a triangle is equal to one half the product of its base by its altitude.
336. Cor. II. Triangles are to each other as the produxits of their bases by their altitudes; consequently, triangles which have equal altitudes are to each other as their bases, triangles ivhich have equal bases are to each other as their altitudes, and triangles which have equal bases and equal altitudes are equivalent.
PLANE GEOMETRY. — BOOK V, 179
Proposition VI
337. Draw a trapezoid and one of its diagonals. How does the area of the trapezoid compare with the combined areas of the triangles thus formed? Since both triangles have the same altitude, how does the area of the trapezoid compare with the area of the rectangle which has the same altitude and a base equal to the sum of the parallel sides of the trapezoid ?
Theorem, A trapezoid is equivalent to one half the rec- tangle which has the same altitude and a base equal to the sum of its parallel sides.
D C
Data: Any trapezoid, as ABCD, whose altitude is CE and whose parallel sides are AB and CD.
To prove ABCD=o=^ rect. CE • {AB + en), a 'e~b
Proof. Draw the diagonal AC.
Then, ABCD <^ /\ ABC -\- J^ ADC.
§ 334, /^ABC ^\ rect. CE • ^5,
and A ^D C =0= i rect. CE - CD \
hence, t.ABC-\-I^ADC^\ rect. CE - AB ■\-\ rect. CE * CD -^
that is, ABCD ^\ rect. CE • {AB -f CD).
Therefore, etc. q.e.d.
338. Cor. Tlie area of a trapezoid is equal to one half the product of its altitude by the sum of its parallel sides.
339. Sch. It will be observed that the corollaries § 332, § 335, and § 338 are arithmetical rules for computing areas.
Such rules are readily formed from the theorems to which they are attached by employing the terms product and equal instead of rectangle and equivalent.
Ex. 481. Triangles on the same base and having their vertices in the same line which is parallel to the base are equivalent.
Ex. 482. The parallel sides of a trapezoid are 12d«n and S^"^, and their distance apart is 5<i"'. What is the area of the trapezoid ?
Ex. 483. The area of a trapezoid is 52 sq. in., and the sum of the two parallel sides is 13 in. What is the distance between the parallel sides ?
Ex. 484. The area of a triangle is 36 sq. ft. If its base is 9 ft, what is its altitude ?
180 PLANE GEOMETRY. — BOOK V.
Proposition VII
340. Draw a triangle one of whose sides is 5'', base 6", and altitude 3", or other dimensions ; draw another triangle having an equal side and altitude but any base whatever, as 10", and the angle between the base and given side equal to the corresponding angle of the first triangle. How does the ratio of the areas of these triangles compare with the ratio of the products of the sides that include their equal angles ?
Theorem. Two triangles having an angle of one equal to an angle of the other are to each other as the products of the sides including the equal angles.
Data: Any two triangles, as ABC and
DEC, having the common angle C.
To prove
AABC:ADEC = ACXBG:DCXEC,
A
Proof. Draw AE.
Since BC and EC may be regarded as the bases of the A ABC and AEC, their bases are in the same straight line ; and since they have their common vertex at A, they have the same altitude.
.-.§336, AaBC:AAEC = BC:EC.
In like manner, A AEC : A DEC = AC : DC.
Hence, § 287, A ABC : Adec = AC x BC : DC X EC. Therefore, etc. q.e.d.
341. Cor. If the products of the sides including the equal angles are equal, the triangles ore equivalent.
Ex. 485. Two triangles have an angle in each equal, the including sides (n one being 8 ft. and 12 ft., and in the other 6 ft. and 20 ft. The area of the smaller triangle is 27 sq. ft. Find the area of the larger triangle.
Proposition VIII
342. Find the area of a triangle whose base is 14", another side 8", and the altitude 6", or other dimensions; also the area of a similar tri- angle whose base is 7", or any other convenient length. How does tlie ratio of the areas of the two triangles compare with the ratio of the areas of the squares described upon their bases ? With the ratios of the squares upon other homologous sides or lines ?
PLANE GEOMETRY. — BOOK V. 181
Theorem. Similar triangles are to each other as the squares upon their homologous sides.
Data : Any two similar triangles, as ABC and DEF.
To prove AabcA DEF = Iff : ¥i? = etc.
Proof. §299, jiA=/!:D',
.-.§340, AABC:ADEF = ABXA0:DEXDF, (1)
But, § 299, AC '. DF = AB : DE. (2)
Multiplying the antecedents by AB and the consequents by DEy §284, AB X ACiDE X DF = Aff:DE\
Substituting in (1),
A ABC: A DEF = Iff : De\
In like manner the same may be proved for any homologous sides.
Therefore, etc. q.e.d.
343. Cor. Similar triangles are to each other as the squares upon any of their homologous lines.
Ex. 486. The homologous sides of two similar triangular fields are in the ratio of 5 : 3. How many times the area of the second field is the area of the first ?
Proposition IX
344. Divide two similar polygons into triangles by diagonals drawn from a pair of homologous vertices. Since the homologous triangles are similar, to what is the ratio of their areas equal ? (§ 342) Write all the ratios of the areas of each pair of triangles. Discover from the ratios whether every pair can be shown to have the same ratio. How, then, does the ratio of the sum of the triangles of one polygon — that is, its area — to the sum of the triangles of the other compare with the ratio of any two corresponding triangles ? (§ 279) Then, how does the ratio of the polygons compare with the ratio of the squares described upon their homologous sides ?
182 PLANE GEOMETRY. — BOOK V,
Theoretn. Similar polygons are to each other as the •squares upon their homologous sides,
D
J
Data : Any two similar polygons, as ABODE and FGHJK. To prove ABODE : FGHJK = AB' : F(? = etc.
Proof. Draw the homologous diagonals AO, AD, FH, and FJ.
Then, § 310, the corresponding triangles thus formed are similar,
and, § 299, AB : FG = BO : GH= OD: HJ=DE: Jir= etc.; .-. § 288, AB^ : FGT = BO' : GH' = cF : H? = DE^ : JK^ = etc., also, § 342, A ABO : A FGH = Iff : F(? = etc., A AOD : A FHJ = off : U? = etc., and A ABE : A FJK = D^ : J? = etc.
Since the ratios of these A are all equal, by § 279, A ABO+A AOD -{-A ADE : A FGH-\- A FHJ-\- A FJK=: A ABC : A FQH', that is, ABODE : FGHJK = A ABO : A FGH.
But A ABO: A FGH = Iff : pff = etc.
Hence, abode : FGHJK = Iff : i^ = etc.
Therefore, etc. q.e.d.
345. Cor. I. Similar polygons are to each other as the sqvxires upon any of their homologous lines.
346. Cor. II. The homologous sides of any similar polygons are to each other as the square roots of the area^ of those polygons.
Ex. 487. In two similar polygons two homologous sides are 15 ft. and 25 ft. The area of the smaller polygon is 450 sq. ft. Find the area of the larger one.
PLANE GEOMETRY. — BOOK V. 183
Proposition X
347. Draw two lines respectively 3" and 5" long, or any other lengths; construct a square on each and a square on their sum; also construct the rectangle of these lines. How does the area of the square on their sum compare with the combined areas of the other squares and double the area of the rectangle ?
Theorem. The square upon the sum of two lines is equivalent to the sum of the squares upon the lines plus twice the rectangle formed hy the lines*
Data: Any two lines, as AB and BC, and f
their sum AC.
To prove AC^ =o= JLB^ + BC^ + 2 rect. AB • BO.
.^D
r
^c
B
Proof. On reconstruct the square ACDE; draw BG I! CD; tak^ DK equal to BC; and draw FK II AC, cutting BG in H.
Since the sides of HKDG are respectively parallel to the sides of the square ACDE, its angles are rt. A.
§151, GD = HK=BCy
and GH=DK',
but, const, DK—BC)
GD = HK= GH= DK— BC, Why ?
and, § 143, HKDG is a square whose side equals BQ\ that is, HKDG = BC^.
Similarly, ABHF = Iff.
Since the sides of BCHK are respectively parallel to the sides of the square ACDE, its angles are rt. A; and since, HB = AB,
BCKH= rect. AB . BC.
Similarly, FHGE = rect. AB - BC.
But A CDE =o ABHF + HKD G -\- B CKH -\- FHQB ;
that is, AC^ =0= Iff H- BC^ •{■ 2 rect. AB - BC.
Therefore, etc. q.e.i>.
184
PLANE GEOMETRY. — BOOK V.
Proposition XI
348. Draw two lines respectively 3" and 5" long, or any other lengths; construct a square on each, and a square on their difference; also con- struct the rectangle of these lines. How does the area of the square on their difference compare with the combined areas of the other squares minus double the area of the rectangle ?
Theorem, The square upon the difference of two lines is equivalent to the sum of the squares upon the lines minus twice the rectangle formed hy the lines.
Data: Any two lines, as AG and BC, and ^' ^
their difference AB.
To prove A? =o= AC^ + BC^ — 2 rect. AC - BC.
Proof. On AC construct the square ACDE, on BC the square BGJC, and on AB the square ABHF. Produce FH to meet CD in K.
Z GBC is a rt. Z; Z.ABG is a rt. Z; Z ABH is a rt. Z ; GBH is a straight line. JCK is a straight line. A G and J are rt. Ay A GHK and HKJ are vt A) HGJK is a rectangle. HB = AB, and BG = BC'y EG = AC, GJ = BC', HGJK = rect. AC 'BC, FKDE = rect. AC'BC. Const., ABHF — Iff, ACDE = A^, and BGJC = B^. But ABHF o ACDE + BGJC — {HGJK -\- FKDE) ;
that is, Iff =0= ic^ + BC^ — 2 rect. AC • BC.
Therefore, etc.
Then,
but
Similarly, Now, also
hence, also
Similarly,
Why? Why?
Why?
Why? Why?
Why? Why?
Q.E.D.
PLANE GEOMETRY. — BOOK V.
185
Proposition XII
349. Draw a right triangle whose sides are 3", 4", and 5", or any- other right triangle ; construct a square on each side and find the area of each square. How does the square on the hypotenuse compare in area with the sum of the squares on the other sides ?
Theorem, The square upon the hypotenuse of a right triangle is equivalent to the sum of the squares upon the other two sides.
First Method
Data: Any right triangle, as ABC\ the square on the hypote- nuse, as ABDE ; and the squares on the other two sides, as BCGF and ACHJ respectively.
To prove ABDE ^ BCGF + ACHJ.
Proof. From C draw CKWbd, cutting AB in L and meeting ED in K, and draw CE and BJ.
A ACB, ACH, and BCG are rt. A ; § 58, ACG and BCH are straight lines.
In A AEC and AJB, AE = ab, ac= A J,
|
and |
Zeac=Zjab^, |
|
|
.-. §100, |
A AEC = A AJB ; |
|
|
but, § 334, |
AEKL ^ 2 A AEC, |
|
|
and |
ACHJ ^2 A AJB; |
|
|
.-. |
AEKL =0= ACHJ. |
|
|
In like manner, |
BDKL =o BCGF. |
|
|
But |
ABDE ^ AEKL -f BDKL. |
|
|
Hence, |
ABDE =c= BCGF + ACHJ, |
|
|
Therefore, |
etc. |
|
|
Prove that BDKL |
=o BCQF, |
Why? Why?
Q.S.D.
186
PLANE GEOMETRY. — BOOK V.
Second Method
Proof. Draw the perpendicular CD.
Then, § 313, AB:AC=AC: AD,
:2
or
and
or
Ax. 2, or
Therefore, etc.
AC =ABX AD, AB : BC = BC : DB, BC? = ABX DB.
AC ■\-BC r= AB {AD + DB) = AB X AB = Ab\
AB^ = AC^ + BG^
Q.E.D.
350. Cor. I. Either side of a right triangle is equal to the square root of the difference between the squares of the hypotenuse and the other side.
The following is an easy method of determining integral numbers which are measures of the sides of right triangles :
Write in a column the squares of the numbers of the scale as far as de- sired ; subtract each square from all of the others following it. When the remainder is a perfect square its square root is the measure of one side of a right triangle, the square root of the minuend of this subtraction is the meas- ure of the hypotenuse, and the square root of the subtrahend is the measure of the other side. By taking equimultiples of these numbers the measures of the sides of similar right triangles may be found.
The following sets of numbers are some of the integral measures of the sides of right triangles :
3 4 5
6 12 13
7 24 25
8 16 17
9 40 41
351. Cor. II. The ratio of the diagonal of a square to a side
is V2.
For, in the square ABCD, AG^ — AB^ -f- B(f ;
but BC = AB
hence, AC =2ab.
Dividing by ab\ ^^^ = 2 ; whence, ^ = V2. AB ^B
PLANE GEOMETRY. — BOOK V. 187
In the figure on page 185 :
Ex. 488. Prove CE perpendicular to JB.
Ex 489. If the lines AH and BG are drawn, prove that they are parallel.
Et:. 490. Prove that the sum of the perpendiculars from F and J to AB produced is equal to AB.
Ex. 491. Prove that J, (7, and F are in the same straight line.
Ex. 492. If the lines EJ and DF are drawn, prove that the sum of the angles AEJ^ AJE, BDF, and BFD is equal to one right angle.
Ex. 493. If EM and DK are drawn perpendicular respectively to J A and FB produced, prove that the triangles AEM and BBN are each equal to triangle ABC.
Ex. 494. If the lines JH, KC, and FG are produced, prove that they meet in a common point.
Ex. 495. If B is the middle point of the side BC of the right triangle ABC, and BE is drawn perpendicular to the hypotenuse AB, prove that
AtP-AE'^ -BE^.
Ex. 496. The area of a rectangle is 26.4081 ^m and its altitude is 4.8*™. Fird the length of its diagonal.
^x. 497. The perpendicular distance between two parallel lines is 20 in. aiAd a line is drawn across them at an angle of 45°. What is the length 0^ the part intercepted between the parallel lines ?
Ex. 498. Find the area of a right isosceles triangle, if the hypotenuse is 1^0 rd. in length.
Ex. 499. The diameter of a circle is 12<^™ and a chord of the circle is jQcm^ What is the length of a perpendicular from the center to this chord ?
Ex. 500. Two parallel chords in a circle are each 8 ft. in length, and the distance between them is 6 ft. Find the radius of the circle.
Ex. 501. Two sides of a triangle are IS^m and 15<i'" and the altitude on the third side is 12dm. pjnd the third side and also the area of the triangle.
Ex. 502. Two parallel lines are 12 ft. apart, and from a point on one of them two lines, one 20 ft. and the other 13 ft. long, are drawn to the other parallel. What is the area of the triangle thus formed ?
352. When from the extremities of a given straight line per- pendiculars are let fall upon an indefinite straight line, the portion of the indefinite line between the perpendiculars is called the projection of the given line.
MN is the projection of the C|-^,„^^
line CB upon the line AB. If | ^~^^^£)
the point B is in the line AB, \ ^ ^ ^
then ilfl> is the projection of Ol). M N M D
188
PLANE GEOMETRY. — BOOK V.
Proposition XIII
353. Draw a triangle whose sides are 2", 3", and 4", or any other oblique triangle ; construct a square on the side opposite an acute angle ; construct squares on the other two sides and also the rectangle of one of those sides and the projection of the other upon that side ; find the area of each figure constructed. How does the area of the first square com- pare with the combined area of the otner squares less twice the area of the rectangle?
Theorem. In any oblique triangle the square upon the side opposite an acute angle is equivalent to the sum of tlw squares upon the other two sides minus twice the rectangle formed by one of those sides and the projection of the other upon that side,
c
Data: Any oblique triangle, as ABC, in which A is an acute angle and AD the projection of AO on AB, or AB produced.
To prove
BC
AB -{-AC — 2 rect. AB . AD.
BD =o=AB -\-AD — 2 rect. AB - AD.
Proof. When CD lies within A ABC, BD = AB — AD\
when CD lies without A ABC,
BD = AD-AB;
and in either case,
§ 348,
Adding CD to both members of this equation,
BD^ -{-cff^ IS' 4- aS +Gff — 2 rect. AB But, § 349, bS -\-gS^ b^, .
and Iff ^'ci?^ Iff,
Hence, substituting sff and Iff for their equivalents, Bff =0= Iff + Iff — 2 rect. AB - AD.
AD.
Q.E.I>.
PLANE GEOMETRY. — BOOK V. 189
Proposition XIV
354. Draw a triangle whose sides are 2", 3'', and 4", or any other obtuse triangle ; construct a square on the side opposite the obtuse angle; construct squares on the other two sides and also the rectangle of one of those sides and the projection of the other upon that side produced ; find the area of each figure constructed. How does the area of the first square compare with the combined area of the other squares and twice the area of the rectangle ?
Theoretn, In any obtuse triangle the square upon the side opposite the obtuse angle is equivalent to the sum of the squares upon the other two sides plus twice the rec- tangle formed by one of those sides and the projection of the other upon that side.
Data : Any obtuse triangle, as ABC, in which B is the obtuse angle and BD the ^^
projection of BC upon AB produced. ^^ /
To prove ^^ /
AC^ ^ Iff + BC^ + 2 rect. AB • BD. /— /
Proof. . AD = AB-\-BD'^
then, § 347, Iff ^ Iff + sff + 2 rect. AB • BD.
Adding CD to both members of this equation,
Iff -^cff=o= Iff + sff +cff -\-2 rect. AB - BD. But, § 349, Iff -\- off ^ Iff,
and sff -\-c3^^ Bff.
Hence, substituting Aff and Bff for their equivalents,
Ic^ ^ Iff 4- BO^ + 2 rect. AB • BD. Therefore, etc. q.e.d.
Ex. 503. The diagonals of a rhombus are 30 in. and 16 in. What is the length of the sides ?
Ex. 504. A square lawn with the walk around it contains ^ of an acre. If the walk contains |f of the eniire area, what is the width of the walk ?
Ex. 505. Find the area of a field in the form of a trapezoid whose bases are 45 rd. and 27 rd. , and each of whose non-parallel sides is 15 rd.
190
PLANE GEOMETRY. — BOOK V.
Proposition XV
355. 1. Draw a triangle whose sides are 2", 3", and 4", or any other oblique trianglo,; construct the squares on any two sides; construct a square on one half of the third side and also a square on the median to that side; find the area of each of these squares. How does the com- bined area of the first two compare with double the combined area of the other two?
2. Construct and find the area of the rectangle of the third side and the projection of the median upon that side. How does the difference in the area of the first two squares compare with double the area of this rectangle ?
Theorem, In any oblique triangle the sum of the squares upon any two sides is equivalent to twice the square upon one half the third side, plus twice the square upon the median to that side.
E .
Data : Any oblique triangle, as ABC, and the median CD, making with AB the obtuse angle ADC and the acute angle BDC.
To prove AC^ -\-BC^ =o= 2 AD^ -\-2cff.
Proof. Draw CEJlab, or AB produced.
Then, in the A ADC and DBC respectively, § 354, AC^ ^Iff -{-CD^-{-2 rect. AD • DE, (1)
and, § 353, BC^ =^Dff +CD^ — 2 rect. DB ■ DE. (2)
But, data, AD = DB.
Substituting for DB in the second equation its equal AD and adding (1) and (2), ic' + 5c' =g= 2 Id' + 2 off.
Therefore, etc.
Q.E.D.
356. Cor. In any oblique triangle 'the difference of the squares upon any two sides is equivalent to twice the rectangle formed by the third side and the projection of the median upon that side.
PLANE GEOMETRY. — BOOK V, 191
Proposition XVI
357. Draw a circle and two intersecting chords ; draw two chords, which do not meet, to connect the extremities of the given chords, thus forming two triangles. What angles of the figure are equal? Are the triangles equal, equivalent, or similar ? How does the ratio of the longer segments of the given chords (sides of the similar triangles) compare with the ratio of their shorter segments? How does the rectangle formed by the segments of one chord compare with the rectangle formed by the segments of the other ?
Theorem. If two chords of a circle intersect, the rectangle formed by the segments of one chord is equivalent to the rectangle forrvied hy the segments of the other.
Data: Any two chords of a circle, as AB a/ and CDj intersecting, as at J&. /
To prove rect. ae-be^ rect. DE • CE. \
Proof. Draw AC and bd.
Then, in A AEC and DEB, § 225, Z. A= Z.D, each being measured by ^ arc CB, and /.G = Z. B, each being measured by -J- arc AD\
,', § 301, A AEG and DEB are similar.
Hence, AE\DE— GE : BE, Why ?
2tnd, § 269, AE X BE = DE X GE;
that is, § 324, rect. AE- BE^ rect. DE - GE.
Therefore, etc. q.e.d.
358. Cor. If a chord passes through a fixed point, the area of the rectangle formed hy its segments is constant in whatever direction the chord is drawn.
Ex. 506. A ladder 25"* long, with its foot in the street, will reach on one side to a window 20"» high, and on the other to a window 16™ high. What is the distance between the windows ?
192
PLANE GEOMETRY. — BOOK V.
Proposition XVII
359. From a point without a circle draw two secants ; draw two inter- secting chords to connect the points of intersection of tlie secants and the circumference ; select two triangles each of which has a secant for one of its sides. What angles of these triangles are equal? Are the triangles equal, equivalent, or similar ? How does the ratio of the secants (sides of the similar triangles) compare with the ratio of their external segments? How does the rectangle formed by one secant and its ex- ternal segment compare with the rectangle formed by the other and its external segments?
Theorein, If from a point without a circle two secants are drawn, the rectangle formed hy one secant and its ex- ternal segment is equivalent to the rectangle formed by the other secant and its external segment.
Data : Any point without a circle, as A, and any two secants from A, as AB and AC, cutting the circumference in E and D re- spectively.
To prove rect. AB • AE =0= rect. AC • AD.
Proof. Draw BD and CE.
Then, in A ABD and ACE,
Z^ is common, and Z.B = Z.C\
A ABD and ACE are similar.
Hence, ab\AC= ad\ ae,
and rect. AB- AE^ rect. AC- AD.
Why? Why? Why?
Therefore, etc.
Q.E.D.
Ex. 507. Find the area of a triangle each of whose sides is 12 ft.
Ex. 508. The side of a rhombus is 29<='" and one of its diagonals is 40"°. What is the length of the other diagonal ?
Ex. 509. The area of a rhombus is 1176 sq. in, and one of its diagonals is 42 in. What are its sides and the other diagonal ?
Ex. 510. The radius of a circle is S^m and a tangent to the circle is IS**™. What is the length of a secant drawn from the same point as the tangent, if the secant is 6*™ from the center ?
PLANE GEOMETRY, — BOOK V. 193
Proposition XVIII
360. Draw a triangle and its circumscribing circle ; bisect the vertical angle and produce the bisector to meet the circumference ; connect this point of meeting with the point of intersection of the base and shortest side. What angles of the figure are equal? What triangles are similar? From the ratios of the sides of similar triangles and of the segments of intersecting chords discover how the rectangle formed by the sides of the given triangle compares with the rectangle formed by the segments of the base plus the square upon the bisector of the vertical angle?
Theorem, If the bisector of the vertical angle of a tri- angle intersects the base, the rectangle formed by the two sides is equivalent to the rectangle formed by the seg- ments of the base plus the square upon the bisector.
Data: Any triangle, as ABC, and the bi- /''^-^"^^ / \,\
sector of its vertical angle, as CD, intersect- ^[^^ 4 — -t^^b
ing the base in D. \ j / \
To prove \ / /'' /
rect. AC . -BC7=c= rect. AD - BD -^ cff. \^ // y
"'-- — ic-'-''
E
Proof. Circumscribe a circle about A ABC, produce CD to meet the circumference in E, and draw EB.
Then, in A ADC and EBC, data, Zacd=z/. ecb,
and /.A = ZE; Why?
A ADC and EBC are similar. Why ?
Hence, ' AC : EC = CD : BC,
and rect. AC - BC ^ rect. EC • CD,
or rect. AC • BC ^ rect. (DE + CD) • CD ^ rect. DE - CD -\- cff.
But, § 357, rect. DE - CD ^ rect. AD • BD.
Hence, rect. AC > BC =o= rect. AD * BD -{- cff.
Therefore, etc. q.e.d.
Ex. 511. A pole standing on level ground was broken 75 ft. from the top and fell so that the end struck 60 ft. from the foot. Find the length of the pole.
milne's geom- — 33
J 94
PLANE GEOMETRY. — BOOK V,
Proposition XIX
361. Draw a triangle, a line representing its altitude, the circumscrib ing circle, and a diameter from the vertex ; connect the other extremity of this diameter with the point of intersection of the base and shortest side. What right triangles are similar? How does the ratio of their longest sides compare with the ratio of their shortest" sides ? How does the rectangle formed by the sides of the given triangle compare with the rectangle formed by its altitude and the diameter of the circumscribing circle ?
Theorem. The rectangle formed hy any two sides of a triangle is equivalent to the rectangle formed hy the alti- tude upon the third side and the diameter of the circum- scribing circle.
Data : Any triangle, as ABC; a diameter of tbe circumscribing circle, as CD ; and the altitude upon AB, as CE.
To prove rect. AC - BC ^ rect. CD • CE.
Proof. Draw DB.
Data, § 94, Z AEC is a rt; Z,
§ 227, Z DBC is a rt. Z.
Then, in rt. A AEC and DBC,
Za = ZD; .'. § 302, A AEC and DBC are similar.
Why?
Hence, and
Therefore, etc.
AC: CD = CE: BC, rect. AC ' BC^ rect. CD • CE.
Q.E.D.
Ex. 512. Upon the diagonal of a rectangle 28™ by 21™ a triangle equiva- lent to the rectangle is constructed. What is the altitude of the triangle ^
Ex, 513. The base and altitude of a right triangle are 6 ft. and 8 ft. respectively. What is the length of the perpendicular drawn from the vertex of the right angle to the hypotenuse ?
Ex. 514. The parallel sides of a trapezoid are 12 in. and 16 in. and the non-parallel sides are 10 in. What is the area of the triangle formed by joining the middle point of the shorter base with the extremities of the longer ?
PLANE GEOMETRY.— BOOK V,
195
Proposition XX
362. Problem. To construct a square equivalent to the sum of two given squares.
|
E |
1 i |
c |
" 1 1 |
||
|
A |
1 I |
||||
|
B |
'\ |
1 |
F D
Data : Any two squares, as A and B. Required to construct a square equivalent to A-^B. Solution. Draw FB equal to a side of A. At one extremity, as Al F, draw FF _L FB and equal to a side of B. Draw ED. . Construct a square C, having each of its sides equal to EB. Then, C is the required square. q.e.f.
^roof . By the student. Suggestion. Refer to § 349.
Proposition XXI
363. JProblem. To construct a square equivalent to the difference of two given squares.
Data : Any two squares,, as A and B. Required to construct a square equivalent to A — B. Solution. Draw an indefinite line, as GB. At G, erect a perpendicular to GB, as GE, equal to a side of B. With J5^ as a center and a radius equal to a side of A, describe an arc intersecting GB at F. Draw EF.
Construct a square C, having each of its sides equal to GF. Then, C is the required square. q.e.f.
Proof. By the student. Suggestion. Refer to § 349o
196
PLANE GEOMETRY. — BOOK V.
Propositioa XXII
364. Problem, To construct a square equivalent to tha sum of any number of given squares.
^ " re
-Fs
H
D E
Data : Any squares, as A, B, and C.
Required to construct a square equivalent to A + B -\- c.
Solution. Draw DE equal to a side of A.
At D erect a perpendicular to DE, as DF, equal to a side of B.
Draw FE.
At F erect a perpendicular to FE, as FG, equal to a side of C.
Draw GE.
Construct a square H, having its sides each equal to GE.
Then, H is the required square. q.e.d.
Proof. By the student. Suggestion. Eefer to § 349.
Ex. 515. Divide a triangle into two equivalent triangles by a line drawn through any vertex.
Ex. 516. Construct a triangle equivalent to a given triangle and having the same base.
Ex. 517. Construct an isosceles triangle equivalent to a given triangle and having the same base.
Ex. 518. Construct a right triangle equivalent to a given triangle.
Ex. 519. Construct a triangle equivalent to a given triangle, having the same base and an angle at the base equal to a given angle.
Ex. 520. Construct a triangle similar to a given triangle and four times the given triangle.
Ex. 521. Divide a parallelogram into two equivalent parts by a line through any point in its perimeter.
Ex. 522. Divide a rectangle into four equivalent parts by lines through any vertex.
Ex. 523. Construct a square equivalent to a triangle whose base is 18cm and altitude 4"".
Ex. 524. Construct a square equivalent to a rectangle whose dimensions are 16<^'" and 4cm.
Ex. 525. Construct a square equivalent to the difference between two squares whose areas are 25'<i «'" and IQn cm.
PLANE GEOMETRY. — BOOK V,
197
Proposition XXIII
365. JProblem, To construct a polygon similar to two given similar polygons and equivalent to their sum.
< c
Data : Any two similar polygons, as A and B.
Required to construct a polygon similar to A and B and equiva- lent to ^ + ^•
Solution. Draw a line, as FD, equal to m, a side of A.
At one extremity, as F, erect a perpendicular FF, equal to n, the homologous side of B. Draw FD.
Taking I, a line equal to FD, as homologous to m and n, construct a polygon C, similar to A and B.
Then, C is the required polygon. q.e.f.
~" Why?
|
Proof. |
FD^ -{- FF^ = FD^ ; |
|
.*. |
m^ + n^ = E |
|
Now, § 344, |
A: C = m^:^, |
|
and |
B: C=n^:f'j |
|
.', § 280, |
A + B: C = m^-{-n^:P. |
|
But |
m' + n' = E |
|
Hence, § 271, |
A + B^C. |
Ex. 526. Construct a right triangle equivalent to a given square.
Ex. 527. Construct a right triangle equivalent to a given rectangle.
Ex. 528. Construct a right triangle equivalent to a given parallelogram.
Ex. 529. Construct an isosceles triangle equivalent to a given square.
Ex. 530. Construct a square equivalent to the sum of two squares whose sides are 5 in. and 10 in.
Ex. 531. Construct a square equivalent to the difference of two squares whose sides are IS''™ and 17<=™.
Ex. 532. Construct a polygon similar to two given similar polygons, and equivalent to their difference.
198
PLANE GEOMETRY. — BOOK V.
Proposition XXIV
366. Problem, To construct a square having a given ratio to a given square.
K
/F
/H
Data : Any square, as A, and any ratio, as m : n.
Required to construct a square B, such that B -. A = m : n.
Solution. Draw EF equal to a side of A, and draw ED making any acute angle with EF.
On ED take EG equal to w, and GJ equal to m.
Draw GF\ also draw JH II GF, meeting EF produced at H.
On EH describe a semicircumference, and at F erect FK per- pendicular to EH and meeting the semicircumference in K.
On a line equal to FK as a side, construct the square B.
Then, B : A = m -. n. , q.e.f.
Proof. § 314, hence, FK^ = EF x FK Why ?
Now,
EF:FK=FK:FH', FK^ = EF X FK FK^ : eP = FK^ : eP ; FK^ :EF^ = EF X FH: EF , or, dividing each term of the second ratio by EF,
FK : EF :
FH.EF
FK^ : EF^ ■
FH'.EF. GJ:EG-- m:n:
m: n
B: A = m
§283, Const.,
that is,
Ex. 533. Ex. 534.
gram.
Ex. 535. Construct a square which shall be equivalent to a right isos- celes triangle, having given the perpendicular from the vertex of the right angle upon the hypotenuse.
Construct an isosceles triangle equivalent to a given rectangle. Construct an isosceles triangle equivalent to a given parallelo-
PLANE GROMETRY. — BOOK V. 199
Proposition XXV
367. Problem, To construct a triangle equivalent to a given polygon. ^ j)
Datum : Any polygon, as ABCDEF. - // 1 y^^/
Required to construct a triangle / ^\/ [/
equivalent to ABCDEF. /y^^ i ,\''1
J HA B G
Solution. Draw DB, and from C tc AB produced draw CG il DB ; also draw DG.
Draw FA, and from F to BA produced draw FH II EA ; also draw EH.
Draw DH, and from E to BA produced draw EJ II BH-, also draw DJ.
Then, JGD is the required triangle. q.e.f.
Proof. In the polygons AGDEF and ABCDEF^ ABDEF is common, and, § 336, A DBG ^ A BBC ;
AGDEF ^ ABCDEF. In the polygons HGDE and AGDEF,
AGDE is common, and Aj^^^IT^A^^J?'; Why?
HGDE =^ AGDEF. In the polygons JGD and ir(?Z)^,
ZTGD is common, and Ahdj<^AHDE. Why?
Hence, A JGD =0 ITG^D^ ^ AGDEF ^ ABCDEF.
Ex. 536. Construct a parallelogram equivalent to a given parallelogram and having an angle equal to a given angle.
Ex. 537. Bisect a given parallelogram (1) by a line passing through a given point within ; (2) by a line perpendicular to a side ; (3) by a line parallel to a side.
200
PLANE GEOMETRY. — BOOK V,
Proposition XXVI
368. Problem* To construct a square equivalent to a given parallelogram.
J
E G
H
A B F a K
Datum : Any parallelogram, as ABCD. Required to construct a square equivalent to ABCD. Solution. Draw the altitude BE-^ also draw FG equal to BE. Produce FG to K, making GK equal to AB. On FK as a diameter describe a semicircumf erence, draw GJ meeting it in J and X FK. On a line equal to GJ as a side, construct the square H. Then, H is the required square. q.e.f.
Proof. By the student. Suggestion. Refer to § 314.
' 369. Sch. A square may be constructed equivalent to a given triangle by taking for its side a mean proportional between the base and one half the altitude of the triangle.
To construct a square equivalent to any given polygon, first re- duce the polygon to an equivalent triangle, and then construct a square equivalent to this triangle.
Proposition XXVII 370. Problem, To construct a rectangle equivalent to a given square, and having the sum of its base and altitude ^qual to a given line.
i
H
Data : Any square, as A, and the line BC.
Required to construct a rectangle equivalent to A, and having the sum of its base and altitude equal to BC.
PLANE GEOMETRY. — BOOK V.
201
Solution. Upon J5C as a diameter describe a semicircumference.
At one extremity of BC, as B, erect a perpendicular to BC, as DBj equal to a side of A. Draw DE II 5(7 meeting the semicir- cumference in E. Draw EF II Z)J5 meeting BC m. F.
With base GF and altitude -Bi^ construct rectangle H.
Th-en, H is the required rectangle. q.e.f.
Proof. DB = EfI Why?
nff = EF^ <> A.
But, § 314, BF:EF = EF:CF)
BF X OF = E?; that is, H^A.
Proposition XXVIII
371. Problem, To construct a rectangle equivalent to a given square, and having the difference of its base and altitude equal to a given line.
|
D N l\ 1 \ 1 \ |
|
|
A |
|
|
V v |
\c
H
Data : Any square, as A, and the line BC.
Required to construct a rectangle equivalent to A, and having the difference of its base and altitude equal to BC.
Solution. On BC as a diameter describe a circumference.
At one extremity of BC, as B, erect a perpendicular to BC, as BD, equal to a side of A.
Through 0, the center of the circle, draw DF intersecting the circumference in E and meeting it in F.
Then, FD — ED = EF, or BC.
With base FD and altitude ED construct rectangle H.
Then, H is the required rectangle. q.e.f.
Proof. By the student. Suggestion. Refer to §§ 316, 269,
i02
PLANE GEOMETRY. — BOOK V.
Proposition XXIX 372. Problem, To construct a polygon similar to a given Tiolygon and equivalent to any other given polygon.
H
I
Data : Any two polygons, as A and B.
Required to construct a polygon similar to A and equivalent to B.
Solution. Find c, the side of a square equivalent to A, and d, the side of a square equivalent to B, and let e be a side of A. Find a fourth proportional to c, d, and e, as /. Upon / homologous to e construct H similar to A. Then, H is the required polygon. q.e.f.
Proof. Const., Also
But, const..
But, § 344,
Hence, § 272,
H is similar to A.
c :d =e 'J;
(?:d^ = e^\f\ 4=o=c^, and £=od^;
A'.B = e^:f.
A'.H=e^'.p',
A:H=A:B. H^B.
Why
SUMMARY "373. Truths established in Book V. 1. A rectangle is equivalent,
a. If it is the rectangle formed by the segments of one of two intersecting chords, to the reetangle formed by the segments of the other. § 357
b. If it is formed by a secant and its external segment, to a rectangle formed by another secant from the same point and its external segment. § 359
c. If it is formed by the two sides of a triangle, to the rectangle formed by the segments of the base, made by the bisector of the vertical angle, plus the square upon the bisector. § 360
PLANE GEOMETRY. — BOOK V. 203
d. If it is formed by two sides of a triangle, to the rectangle formed by the altitude upon the third side and the diameter of the circumscribing circle. " § 361
2. Rectangles are in proportion,
a. If they have equal altitudes, to their bases. § 327
b. If they have equal bases, to their altitudes. § 328
c. To the products of their bases by their, altitudes. § 329
3. A parallelogram is equivalent,
a. To the rectangle which has the same base and altitude. § 331
b. To another parallelogram which has an equal base and an equal alti- tude. . § 333
4. Parallelograms are in proportion,
a. If they have equal altitudes, to their bases. § 333
b. If they have equal bases, to their altitudes. § 333
c. To the products of their bases by their altitudes. § 333
5. A triangle is equivalent,
a. To one half the rectangle which has the same base and altitude. § 334
b. To another triangle which has an equal base and an equal altitude. § 336
c. To another triangle which has an angle equal to an angle of the first, and the products of the sides, including the equal angles, equal. § 341
6. Triangles are in proportion,
a. If they have equal altitudes, to their bases. § 336
b. If they have equal bases, to their altitudes. § 336
c. To the products of their bases by their altitudes. § 336
d. If they have an angle of one equal to an angle of the other, to the prod- ucts of the sides including the equal angles. § 340
e. If they are similar triangles, to the squares upon their homologous sides. § 342
/. If they are similar triangles, to the squares upon any of their homolo- gous lines. § 343
7. A trapezoid is equivalent,
a. To one half the rectangle which has the same altitude and a base equal to the sum of the parallel sides. ~ § 337
8. The square upon a line is equivalent,
a. If the line is the sum of two lines, to the sum of the squares upon the lines plus twice the rectangle formed by them. § 347
b. If the line is the difference of two lines, to the sum of the squares upon the lines minus twice the rectangle formed by them. § 348
c. If the line is the hypotenuse of a right triangle, to the sum of the squares upon the other two sides. § 349
204 PLANE GEOMETRY. — BOOK V.
d. If the line is the side of an oblique triangle, opposite an acute angle, to the sum of the squares upon the other two sides minus twice the rectangle formed by one of those sides and the projection of the other upon that side.
§353
e. If the line is the side opposite an obtuse angle of a triangle, to the sum of the squares upon the other two sides plus twice the rectangle formed by one of those sides and the projection of the other upon that side. § 354
9. The sum of two squares is equivalent,
a. If they are the squares upon any two sides of an oblique triangle, to twice the square upon one half the third side plus twice the square upon the median to that side. § 356
10. The difference of two squares is equivalent,
a. If they are the squares upon any two sides of an oblique triangle, to twice the rectangle formed by the third side and the projection of the median upon that side. § 356
11. Similar polygons are in proportion,
a. To the squares upon their homologous sides. § 344
6. To the squares upon any of their homologous lines. § 345
12. The area of a figure is equal,
a. If it is a rectangle, to the product of its base by its altitude. § 330
6. If it is a parallelogram, to the product of its base by its altitude. § 332
c. If it is a triangle, to half the product of its base by its altitude. § 335
d. If it is a trapezoid, to half the product of its altitude by the sum of its parallel sides. § 338
SUPPLEMENTARY EXERCISES
Ex. 538. The straight line joining the middle points of the parallel sides of a trapezoid bisects the trapezoid.
Ex. 539. The lines joining the middle point of either diagonal of a quad- rilateral to the opposite vertices divide the quadrilateral into two eq-uivalent parts.
Ex. 540. Two triangles are equivalent, if they have two sides of one respectively equal to two sides of the other, and if the included angles are supplementary.
Ex. 541. 0 is any point on the diagonal ^C of the parallelogram ABCD. If the lines DO and BO are drawn, prove that the triangles AOB and AOD are equivalent.
Ex. 542. A rhombus and a rectangle have equal bases and equal areas. One side of the rhombus is 15™ and the altitude of the rectangle is 12™. What are their perimeters ?
PLANE GEOMETRY.^ BOOK V. 205
Ex. 543. The area of a rhombus is equal to one half the product of its diagonals.
Ex. 544. The diagonals of a rhombus are 64 rd. and 37 rd. What is the area of the rhombus ?
Ex. 545. The base of a triangle is 75™, and its altitude is 60™. Find the perimeter of an equivalent rhombus, if its altitude is 45™.
Ex. 546. Find the area of a rhombus, if the sum of its diagonals is 12 in. and their ratio is 3 : 5.
Ex. 547. A man travels 25 miles east from a certain town, and another man travels 36 miles north from the same town. How far apart are the men ?
Ex. 548. The shortest side of a triangle acute-angled at the base is 45 ft. long, and the segments of the base made by a perpendicular from the vertex are 27 ft. and 77 ft. How long is the other side ?
Ex. 549. The sides of a triangle are 25™ and 17™, and the lesser segment of the base made by a perpendicular from the vertex is 8™. What is the length of the base ?
Ex. 550. In a right triangle the base is 3<i™, and the difference between the hypotenuse and perpendicular is 1^™. What are the hypotenuse and perpendicular ?
Ex. 551. In a right triangle the hypotenuse is 5<i™, and the difference between the base and perpendicular is l^^™. Find the base and perpendicular.
Ex. 552. The sides of a right triangle are in the ratio of 3, 4, and 5, and the perpendicular upon the hypotenuse from the vertex of the right angle is 20 yd. What is the area of the triangle ?
Ex. 553. If in any triangle a perpendicular is drawn from the vertex to the base, the difference of the squares upon the sides is equivalent to the difference of the squares upon the segments of the base.
Ex. 554. In a right triangle the square on either side containing the right angle is equivalent to the rectangle contained by the sum and the difference of the other sides.
Ex. 555. If the diagonals of a quadrilateral intersect at right angles, prove that the sum of the squares upon one pair of opposite sides is equiva- lent to the sum of the squares upon the other pair.
Ex. 556. The altitude of an equilateral triangle is 60 in. How long are its sides ?
Ex. 557. Through J) and E, the middle points of the sides AC and BC of the triangle ^J5 Cany two parallel straight lines are drawn meeting ^5 or AB produced in the points F and G. Prove that the parallelogram DFQE is equivalent to half the triangle ABC.
Ex. 558. The four triangles into which a parallelogram is divided by its diagonals are equivalent.
206 PLANE GEOMETRY. — BOOK V.
Ex. 559. The diagonals of a trapezoid divide it into four triangles, two of which are si^nilar, while the other two are equivalent.
Ex. 560. .kU any trapezoid the triangle, having for its base one of the non- parallel sides and for its vertex the middle point of the opposite side, is equivalent to one half of the trapezoid.
Ex. 561. The triangle, formed by drawing a line from any vertex of a parallelogram to the middle point of one of the opposite sides, is equivalent to one fourth of the parallelogram.
Ex. 562. A triangle is equivalent to one half the rectangle of its perim- eter and the radius of the inscribed circle.
Suggestion. Draw radii to the points of contact and lines from the ver- tices of the triangle to the center of the circle.
Ex. 563. If the perimeter of a quadrilateral circumscribed about a circle is 400 ft. and the radius of the circle is 25 ft. , what is the area of the quadri- lateral ?
Ex. 564. The area of a triangle is 875 sq. yd. Find its base and altitude, if they are in the ratio of 14 to 5.
Ex. 565. The homologous sides of two similar fields are in the ratio of 3 to 5, and the sum of their areas is 416Ha. What is the area of each field ?
Ex. 566. A board 12 ft. long is 10 in. wide at one end and 6 in. at the other. What length must be cut from the narrower end to contain a square foot?
Ex. 567. The side of one equilateral triangle is equal to the altitude of another. What is the ratio of their areas ?
Ex. 568, The perimeter of an isosceles triangle whose base is its shortest side is lOO^^"" ; the difference between the base and an adjacent side is 23'^'". , What is the altitude of the triangle ? What is its area ?
Ex. 569. Two chords on opposite sides of the center of a circle are parallel ; one is 16 ft. long and the other is 12 ft. If the distance between them is 14 ft,, what is the diameter of the circle ?
Ex. 570. If from the vertex of an acute angle of a right triangle a straight line is drawn bisecting the opposite side, the square upon that line is less than the square upon the hypotenuse by three times the square upon half the line bisected.
Ex. 571. In the right triangle ABC, BG^ = 3 AC\ If CD is drawn from the vertex of the right angle to the middle point of AB, angle ACD equals 60°.
Ex. 572. If ACB and ADB are two angles inscribed in a semicircle, and AE and BF are drawn perpendicular to CD produced, prove that C^^ -j- CF^ = DE^ + DF^.
PLANE GEOMETRY. — BOOK V. 207
Ex. 573. If lines are drawn perpendicular to the diagonals of a square at their extremities, a second square is formed equivalent to twice the original square.
Ex. 574. The square upon the base of an isosceles triangle whose vertical angle is a right angle is equivalent to four times the triangle.
Ex. 575. The sum of the squares on the lines joining any point in the cir- cumference of a circle with the vertices of an inscribed square is equivalent to twice the square of the diameter of the circle.
Ex. 576. If AF and BE are the medians drawn from the extremities of the hypotenuse of the right A ABC, prove that 4 AF^ + 4 BE^ = 5 AB^.
Ex. 577. If perpendiculars PF, PD, and PE are drawn from any point P to the sides AB, BC, and ^C of a triangle, prove that
AF^ + BD^ + CE^ = AE^ + BF^ + CL^.
Ex. 578. If any point P within the rectangle ABCD is joined to the vertices, prove that PA^ + PG'^ = PB^ + PL^.
Ex. 579. If CD and AE are the perpendiculars from the vertices C and A of the acute triangle ABC to the opposite sides, prove that
AC'^ = BCxCE-\-ABx AD.
Suggestion. Refer to § 373, 8, d.
Ex. 580. If ^C and BG are the equal sides of an isosceles triangle, and AD is drawn perpendicular to BC, prove that AB^ = 2 BC x BD.
Ex. 581. The sum of the squares on the diagonals of a parallelogram is equivalent to the sum of the squares on its four sides.-
Ex. 582. Three times the sum of the squares on the sides of a triangle is equivalent to four times the sum of the squares on the medians of the triangle.
Ex. 583. Two sides of an oblique triangle are 137 and 111 respectively, and the difference of the segments of the third side made by a perpendicular from the opposite vertex is 52. What is the third side ?
Ex. 584. The chord of an arc is 80 in. ; the chord of half the arc is 41 in. What is the diameter of the circle ?
Ex. 585. Froni a point without a circle two tangents are drawn which with the chord of contact form an equilateral triangle whose side is 18 in. Find the diameter of the circle.
Ex. 586. If the center of each of two equal circles is on the circum- ference of the other, the square on the common chord is equivalent to three times the square on the radius.
Ex. 587. A tangent and a secant meet without a circle, forming an angle of 45° ; the tangent is 2 ft. long and the diameter of the circle is 4 ft Find the length of the secant.
208 PLANE GEOMETRY. — BOOK V.
PROBLEMS OF CONSTRUCTION
Ex. 588. Divide a given parallelogram into two equivalent parts by a line drawn parallel to a given line.
Ex. 589. Divide a given triangle into two parts, whose ratio is 3 : 4, by a line drawn from one vertex.
Ex. 590. Divide a given parallelogram into two parts, whose ratio is 2 : 3, by a line parallel to a side.
Ex. 591. Construct a parallelogram equivalent to the sum of two given parallelograms of equal altitude.
Ex. 592. Construct a parallelogram equnralent to the difference of two given parallelograms of equal bases.
Ex. 593. Construct a square equivalent to five times a given square.
Ex. 594. Transform a given trapezoid into an equivalent isosceles trape- zoid.
Ex. 595. Construct a rhombus equivalent to a given parallelogram, and having one side of the parallelogram for a diagonal.
Ex. 596. Construct a square equivalent to a given rectangle.
Ex. 597. Construct a square equivalent to four sevenths of a given square.
Ex. 598. Construct a rectangle equivalent to a given square, and having a given side.
Ex. 599. Construct a rectangle equivalent to a given rectangle, and hav- ing a given side.
Ex. 600. Construct a square equivalent to a given rhombus.
Ex. 601. Construct a parallelogram having a given altitude, and equiva- lent to a given parallelogram.
Ex. 602. Construct a rectangle having a given altitude and equivalent to a given parallelogram.
Ex. 603. Construct a rhombus having a given side, and equivalent to a given parallelogram.
Ex. 604. Construct a rhombus having a given altitude and equivalent to a given parallelogram.
Ex. 605. Transform a triangle into an equivalent parallelogram whose base shall be the base of the triangle and one of whose base angles shall be equal to a base an^le of the triangle.
Ex. 606. Construct a triangle having a given angle, and equivalent to a given parallelogram.
Ex. 607. Construct a triangle equivalent to a given trapezium.
Ex. 608. Construct a parallelogram equivalent to a given trapezium.
PLANE GEOMETRY. — BOOK V. 209
Ex. 609. Construct an isosceles triangle on a given base and equivalent to a given trapezium.
Ex. 610. Construct a right triangle equivalent to a given triangle, having given one of the sides about the right angle.
Ex. 611. Construct a right triangle equivalent to a given triangle, having given the hypotenuse.
Ex. 612. Construct a triangle equivalent- to a given triangle and having its base and altitude equal.
Ex. 613. Construct an equilateral triangle equivalent to a given triangle.
Ex. 614. Construct an equilateral triangle equivalent to a given square.
Ex. 615. Construct a rectangle having a given diagonal and equivalent to a given rectangle.
Ex. 616. Construct a rectangle having a given diagonal and equivalent to a given square.
Ex. 617. Construct a square equivalent to a given trapezoid.
Ex. 618. Construct a triangle equivalent to a given trapezoid.
Ex. 619. Construct a parallelogram equivalent to a given trapezoid and having for its base the longer base of the trapezoid.
Ex. 620. Construct a triangle equivalent to a given triangle and similar tto another given triangle.
Ex. 621. Construct a parallelogram equivalent to the sum of two given parallelograms.
Ex. 622. The area of a square is 16. Construct a square that shall be to it in the ratio of 5 to 3.
Ex. 623. Construct a hexagon similar to a given hexagon, having its ratio to the given hexagon as 5 is to 3.
Ex. 624. Construct a square equivalent to two thirds of a given hexagon.
Ex. 625. Construct a square equivalent to the sum of a given pentagon and a given parallelogram.
Ex. 626. Divide a given triangle into two equivalent parts by a line per- pendicular to one side.
Ex. 627. Divide a given triangle into two equivalent parts by a line paral- lel to one side.
Ex. 628. Bisect a given trapezoid by a line parallel to the bases.
Ex. 629. Bisect a given quadrilateral by a line drawn from one of the -vertices.
Ex. 630. Bisect a given quadrilateral by a line drawn from any point an its perimeter.
milne's geom. — 14
210
PLANE GEOMETRY. — BOOK V.
ALGEBRAIC SOLUTIONS
Ex. 631. Given the sides of any triangle, to compute the altitude.
Solution. In /lABC, suppose that
angle A is acute.
A c B
Then, § 353, a"^ = b^ + d^ -2c x AD;
AD = ^' + ''-^\ 2c
In A ADC, h^ = b^-A&.
Substituting for AD^ its value,
\ 2c I 4c2
_ (2 6c + b^+ c2- gg) (2 bc-b^- cH g^) ~ 4c2
_ {(5 + c)2 - g2}{g2 - (6 - c)^} 4c2
_ (6 + c + g) (6 + c - g)(g + b - c)(a -b + c) ~ 4c2
g + 6 + c = 2s,
6 + c — g = 2(s — g),
g + 6 — c = 2(s — c),
g - & + c = 2(s-6).
2s X 2(s-g)x2(s- 6)x2(s-c).
ADC
Let
then,
and Hence,
h^
4c2
;i = * Vs(s - g) (s - &) (s - c). c
Ex. 632. G^ivew «^e sides of any tri- angle, to compute its area.
Denote the area by ^.
a \h
Solution. Ex. 631, h=- V8(« - g) (s - b) (s - c), c
then,
c ..2
A = -x^ Vs(s - g)(s - 6)(« - c) ^ c
= V8(«-a)(s- 6)(«- c).
PLANE GEOMETRY. — BOOK V.
211
Ex. 633. Given the sides of a triangle, to compute the medians.
Solution. §355, a2 + 52 = 2m2 + 2(|y.
Hence, 4 m2 = 2(a2 + 62) - c2,
and m = ^V2(a2 + 6?)- c2.
Ex. 634. Given the sides of a tnangle, to compute the bisectors of the angles.
Solution. Circumscribe a circle about A ABC; produce CD to meet the circumference in E; and draw BE,
|
§360, |
ah = ADxBD + k^; |
|
.-. |
k^ = ah-ADy.BD, |
|
§292, |
AD:BD = b:a; |
|
.*. |
AD + BD : AD = a + b :bj |
|
and |
AD + BD:a + b = AD : 6 = BD : a ; |
|
that is, |
c AD BD a-\-b 6 a |
|
whence, |
AD= ^\, and BD = «\; a+b a+b |
|
hence, |
(a + 6)2 |
|
^abd ^' "i |
|
|
""T (a + 6)2J |
|
|
a6{(a + 6)2-c2} |
|
|
(a + 6)2 |
|
|
a6(a + 6 + c)(a + 6-c) |
|
|
• |
(a + 6)2 |
|
06 X 2 s X 2(s — c) |
|
|
(a + 6)2 0 |
|
|
Hence, |
k = -^y/abs(is-c). |
1/ /
E
a + b
212
PLANE GEOMETRY. — BOOK V,
Ex. 635. Given the sides of a triangle and the radius of the circumscribing circle, to compute the area of the triangle.
Solution. Denote the radius of the circle by r.
Then, § 361, ab = 2rh;
abc = 2 rch. But ch = 2A\
abc = 4:Ar. abc
Hence,
A =
4r
Ex. 636. Given the sides of a triangle, to compute the radius of the circumscribed circle.
Solution. § 361, But, Ex. 631,
whence,
ab = 2 rh.
|
h-- |
a)is- |
b){s- |
c); |
|
|
abz |
= 'i^Ks |
-«)(« |
-b){s |
-c), |
|
r = |
_ |
abc |
4 Vs(s — a) (s — b) (s — c)
Ex. 637. The three sides of a triangle are 58 ft., 51 ft., and 41 ft. in length. What is the area of the triangle ?
Ex. 638. Find the altitude on each of the sides of a triangle whose sides are respectively 7 in., 9 in., and 11 in.
Ex. 639. If the sides of a triangle are respectively 4™, 6™, and S^ long, what are its three medians ?
Ex. 640. What is the area of a triangle, if the radius of the circumscribing circle is 6.196™ and the sides of the triangle are respectively 9'", 6™, and 12" in length ?
Ex. 641. The sides of a triangle are respectively 12 in., 11 in., and 9 in. in length. Find the radius of the circumscribing circle.
Ex. 642. The sides of a triangle are respectively 30^"", 50^", and 70^". Find the lengths of the three angle bisectors.
Ex. 643. If two sides and one of the diagonals of a parallelogram are respectively 12 in., 15 in., and 18 in., what is length of the other diagonal ? What is the area of the parallelogram ?
BOOK VI
KEGULAR POLYGONS AND MEASUREMENT OF THE CIRCLE
374. A polygon which is equilateral and equiangular is called a Regular Polygon.
An equilateral triangle and a square are regular polygons.
Proposition I
375. Draw a circle and inscribe in it any equilateral polygon. How do the arcs subtended by the sides of the polygon compare ? How do the arcs intercepted by the sides of the angles of the polygon compare ? How do the angles themselves compare? What may any equilateral polygon that is inscribed in a circle be called ?
Theoretn. Any equilateral polygon inscribed in a circle is a regular polygon.
Data: Any equilateral polygon, as ^ ABODE, inscribed in a circle.
To prove ABODE a regular polygon.
Proof. § 196, arc AB = arc BO — arc OD = etc. ;
arc BODE = arc ODEA = arc DEAD — etc., and Za = Zb = Zo= etc. ; Why ?
that is, ABODE is equiangular.
But, data, abode is equilateral ;
hence, § 374, ABODE is a regular polygon.
Therefore, etc. q.e.d.
213
214 PLANE GEOMETRY. — BOOK VL
Proposition II
376. 1. Divide the circumference of a circle into any number of equal arcs; draw the chords of these arcs in succession. What kind of an inscribed polygon is thus formed?
2. Draw tangents to the circle at the extremities of the chords and produce them until they intersect. How do they compare in length? How do the angles formed by each pair of tangents compare in size? What kind of a circumscribed polygon has been formed?
Theorem. If the circumference of a circle is divided into any number of equal arcs,
1. The chords joining the extremities of the arcs in sue- cession form a regular inscribed polygon.
2. The tangents drawn at the extremities of the arcs form a regular circumscribed polygon.
Data : Any circumference divided into equal arcs at A, B, C, etc. ; the chords AB, BC, CD, etc. ; and the tangents GBH, HCJ, etc.
To prove
ABODE and FGHJK regular polygons.
Proof. 1. AB — BC=CD = etc.,
and Za = Zb = Zc = etc. ;
.*. § 374, ABODE is a regular polygon.
2. Z BAG = Z ABG = Z OBH=z etc. ; Why ?
Aabg, boh, ODJ, etc., are equal isosceles A. Why ?
Hence, Zg = Zh=Zj= etc.,
and GB = BH=HO= etc. ; Why ?
tangents GBH. HO J, etc., are equal.
Hence, fghjk is a regular polygon.
Therefore, etc. . q.e.d.
377. The radius of a circle inscribed in a regular polygon is called the apothem of the polygon.
PLANE GEOMETRY. — BOOK VL 215
378. The radius of a circle circumscribed about a regular poly- gon is called the radius of the polygon.
379. The common center* of the circles inscribed in and circum- scribed about a regular polygon is called the center of the polygon.
380. The angle between the radii drawn to the extremities of any side of a regular polygon is called the angle at the center of the polygon.
Proposition III
381. 1. Draw any regular polygon; pass a circumference through three of its vertices. Does it pass through the other vertices? Why?
2. With the same center and a radius equal to the apothem describe a circle. How many sides of the polygon does this circle touch? Why?
Theorem. A circle may he circumscribed about, and a circle may be inscribed in, any regular polygon.
D
Datum : Any regular polygon, as ABODE.
To prove 1. That a circle may be circum- iribed about ABODE. 2. That a circle may be inscribed in ABODE.
Proof. 1. Describe a circle passing through A, B, and (7, and from the center 0, draw OA, OB, 00, OD, and OE.
In the A BOO, OB =00; Why ?
Z OBO = Z OOB. Why ?
But, § 374, Z ABO = Z BOD ;
.-. Ax. 3, Z OB A = Z OOD.
Also AB = OD, and 0B = 00;
hence, § 100, Aabo = AODO,
and OA = OD.
Consequently, the circle passing through A, B, and O passes through D.
In like manner it may be shown that this circle passes through E.
Therefore, a circle described with the center 0 and a radius equal to OA will be circumscribed about the polygon.
216 PLANE GEOMETRY. — BOOK VI.
2. Since the sides AB, BC, CD, etc., are equal chords of the circumscribed circle, they are equally distant from the center and the perpendiculars drawn from the center to the chords are all equal.
Hence, a circle described with the center 0 and a radius equal to one of these perpendiculars, as OH, will be inscribed in the polygon, for each of the sides of the polygon will be perpendicular to a radius at its extremity and tangent to the circle. § 205
Therefore, etc. q.e.d.
382. Cor. I. The radius drawn to the vertex of any interior angle of a regular polygon bisects the angle,
383. Cor. II. Each angle at the center of a regular polygon is equal to four right angles divided by the number of sides of the polygon.
Proposition IV
384. Draw a circle and a regular inscribed polygon; at the middle points of the arcs subtended by its sides draw tangents and produce them until they intersect. How do they compare in length ? How do the angles formed by each pair of tangents compare in size? What kind of a circumscribed polygon has been formed?
Theorem, Tangents to a circle at the middle points of the arcs subtended hy the sides of a regular inscribed poly- gon form a regular circumscribed polygon.
Data : A circle whose center is 0 ; any regular inscribed polygon, as ABODE-, and the tangents FG, GH, HJ, etc., at the middle points L, M, P, etc., of the arcs AB, BO, CD, etc.
To prove FGHJK a regular circum- scribed polygon
Proof.
|
\^^^ ^Bl |
|
|
F |
L G |
|
arc AB = arc BC = arc CD = etc.. |
Why? |
|
arc AL = arc LB = arc BM = etc. ; |
Why? |
|
arc LM = arc MP = etc. |
PLANE GEOMETRY. — BOOK VL
217
But the sides of FGHJK are tangents at the extremities of these arcs.
Hence, § 376, FGHJK is a regular circumscribed polygon. Therefore, etc. q.e.d.
385. Cor. Regular inscribed and circumscribed polygons of the same number of sides may be so placed that their sides are parallel and their vertices will then lie upon the radii (prolonged) of the inscribed polygon.
Proposition V
386. Draw two regular polygons of the same number of sides. How do the homologous angles compare in size ? How do the ratios of any two pairs of homologous sides compare? What name is given to poly- gons that have such relations to each other?
Theorem, Regular polygons which have the same num- ber of sides are similar,
D
Data : Any two regular polygons, as ABODE and FGHJK, which have the same number of sides.
To prove ABCDE and FGHJK similar.
Proof. By § 166, the sum of the angles of each polygon is equal to twice as many right angles as the polygon has sides less two.
Since, § 374, each polygon is equiangular, and since each con- tains the same number of angles ;
all the angles of both polygons are equal.
§ 374, AB = BC = CD = etc., and FG = GH = HJ = etc. ; AB:FG= BC:GH=etG.
Hence, § 299, ABODE and FGHJK are similar.
Therefore, etc
Q.E.D.
218
PLANE GEOMETRY, — BOOK VL
Proposition VI
387. Draw two regular polygons of the same number of sides ; draw radii to the extremities of a pair of homologous sides. What kind of triangles are formed ? How does the ratio of their bases compare with the ratio of the radii? With the ratio of the apothems? Since the polygons are similar, how does the ratio of their perimeters compare with the ratio of any two homologous sides ? With the ratio of their radii ? Of their apothems ?
Theorem, The perimeters of regular polygons of the same number of sides are to eaxih other as their radii and also as their apothems.
D
Data : Any two regular polygons of the same number of sides, as ABODE and FGHJK, whose radii are OA and PF, and apothems OL and PM, respectively.
Denote their perimeters by Q and S respectively.
To prove Q:S = OA:PF= OL: PM.
Proof. Draw OB and PG.
In the A AOB and FPG, §383, Zo = Zp,
and OA:PF=OB:PG',
.-. § 306, A AOB and FPG are similar.
Hence, AB:FG= OA: PF,
and AB:FG= OL: PM.
But, § § 386, 311, Q:S = AB:FG.
Hence, Q:S= OA:PF= OL: PM.
Therefore, etc. q.e.d.
Ex. 644. -il'he apothem of a regular pentagon is 41<=™ and a side is G^m. Find the perimeter of a regular pentagon whose apothem is 82^".
Why?
Why?
PLANE GEOMETRY. — BOOK VI. 219
Proposition VII
388. Draw a regular polygon and draw the radii to the vertices of its angles. How does each triangle thus formed compare with the rectangle of its base and altitude ? How does the sum of the bases of the triangles compare with the perimeter of the polygon ? Since the triangles are of equal altitude, how does the polygon compare with the rectangle of its perimeter and apothem?
Theorem, A regular polygon is equivalent to one half the rectangle formed hy its perimeter and apothem.
Data: Any regular polygon, as ABODE,
and its apothem OF. ■^^-
Denote its perimeter by P. \
To prove ABODE ^ h rect. P-OF, ^
Proof. Draw the radii OA, OB, 00, OD, and OE.
These radii divide the polygon into triangles whose altitudes- are each equal to the apothem and the sum of whose bases i^ equal to the perimeter.
Now, § 334, A ABO =c= I rect. AB - OF,
A BOO =0= J rect. BO • OF, etc.
Hence, A ABO + A BOO + etc. =o= I rect. (AB -\-BC+ etc.) • OF ; that is, ABODE o ^ rect. P • OF.
Therefore, etc. q.e.d.
389. Cor. I. The area of a regular polygon is equal to one half the product of its perimeter by its apothem.
390. Cor. II. Regular polygons of the same number of sides are to each other as the squares upon their radii and also as the squares upon their apothems. §§ 386, 345
Ex. 645. The sides of a regular circumscribed polygon are bisected at the points of tangency.
Ex. 646. The angle at the center of a regular polygon is the supplement of the angle of the polygon.
220 PLANE GEOMETRY. — BOOK VI.
Proposition VIII
391. Draw a circle and circumscribe a polj^gon about it. How does the circumference of the circle compare in length with the perimeter of the polygon? How does the circumference compare with any enveloping line?
Theorem, The circumference of a circle is less than the perimeter of a circumscribed polygon or any enveloping line.
Data: Any circumference, as ABC, and any- enveloping line, as FGHJK.
To prove ABC < fghjk.
Proof. Of all the lines enveloping the area ABC there must be a least line.
Draw DE tangent to ABC, and cutting FGHJK in D and E.
Then, Ax. 10, ED < EHJD\
FGEDK < FGHJK; hence, FGHJK is not the least enveloping line.
Similarly, it may be shown that no other line than ABC can be the least line enveloping the area ABC.
Hence, ABC < FGHJK
Therefore, etc. q.e.d.
Ex. 647. Find the angle and the angle at the center of a regular dodecagon. '
Ex. 648. If the radius of a regular inscribed hexagon is r, prove that its rfde = r, and its apothem = ^ rVS.
Ex. 649. If the radius of an inscribed equilateral triangle is r, prove that its side = rVS, and its apothem = ^ r.
Ex. 650. If the radius of an inscribed square is r, prove that its side = rV2, and its apothem = I r V2.
Ex. 651. The radius of a circle being r, find the area of an inscribed equilateral triangle.
Ex. 652. The radius of a circle being r, find the area of an inscribed square.
Ex. 653. Find the area of a regular hexagon whose side is 10 ft.
PLANE GEOMETRY. — BOOK VL
221
Proposition IX
392. Draw a circle and inscribe in it a regular polygon ; circumscribe about it a similar polygon whose sides are parallel to the sides of the inscribed polygon. If the number of sides of each polygon is increased indefinitely, what line will their perimeters approach as a limit ? What will their areas approach as a limit ?
Theorem, If a regular polygon is circumscribed about a circle and a similar polygon is inscribed in the circle, and if the number of their sides is indefinitely increased,
1. Their perimeters approach the circumference as a common limit.
2. The polygons approach the circle as a common limit.
Data : Any two regular polygons of the same number of sides, as A and -B, respec- tively circumscribed about and inscribed in a circle whose center is O. "
Denote the circle by S, its circumference by C, and the perimeters of the polygons by P and Q respectively.
To prove that, if the number of sides of the polygons is indefinitely increased,
1. P and Q approach C as their common limit.
2. A and B approach 8 as their common limit.
Proof. 1. Place the polygons so that their sides are parallel ; draw a line from 0 to G^ the point of contact of the side BE of the circumscribed polygon and draw OD which by § 385 will pass through the extremity H of the side HJ of the inscribed polygon^
Since the polygons have the same number of sides,
|
§387, |
P\Q=OJ)\OG\ |
||
|
.-. § 277, |
P—Q:Q=OD~OG:OG, |
||
|
and, § 269, |
0GX(P-Q)=QX{0D- OG); |
||
|
whence, |
P-Q = —X(0D-0G). OG ^ ^ |
(1) |
|
|
Kow, Ax. |
10, |
OD<OG-{- DG, or OD — 0G< BG. |
(2) |
But, if the number of sides of each polygon is increased indefi-
222
PLANE GEOMETRY. — BOOK VL
nitely, the two polygons continuing to have the same number of
sides, the length of each side decreases
indefinitely and approaches the limit 0;
therefore DG^ which is half the side DE,
approaches the limit 0, and in (2), OD — OG
approaches the limit 0.
Hence, from (1), P —Q approaches the limit 0.
Since, § 391, P is always greater than C, and. Ax. 10, Q is always less than C,
the difference between G and either P or Q is less than P — Q] therefore, the difference approaches the limit 0.
Consequently, P and Q approach G as their common limit.
2. Since the polygons are regular and similar, § 390, A'.B=ob^:OC^',
A — B'.B=off—off'.0^ Now 0b^—^^ = D^;
A — B:B = D^ : 0^,
Why? Why?
wrhence.
£=^v5?.
OG
(3)
But DG, which is half of the side BE, approaches the limit 0.
Hence, from (3), A — B approaches the limit 0.
Since. Ax. 8, ^ is always greater than S, cLud B is always less than S,
the difference between S and either ^ or ^ is less than A — B', therefore, the difference approaches the limit 0.
Consequently, A and B approach S as their common limit.
Therefore, etc. q.e.d.
393. Sch. It is evident that this is a special case of the theorem, which may be proved, that the perimeters of polygons inscribed in and circumscribed about a closed convex curve,* when the number of their sides is indefinitely increased, approach the curve as a limit, and the polygons approach the figure bounded by the curve as a limit
*A convex curve is a curve which a straight line can cut in only two points.
PLANE GEOMETRY, — BOOK VI.
223
Proposition X
394. Draw two circles and inscribe in them regular polygons of the same number of sides. How does the ratio of their perimeters compare with the ratio of their radii ? How does the ratio of the circumferences compare with the ratio of their radii ? Of their diameters ?
Theoi^em. Circumferences are to each other as their radii. ^-r^^'^^r-^c
B E
Data : Any two circumferences, as ABC and DEF, whose radii are OB and P^, respectively.
To prove ABC : DEF — OB : PE.
Proof. In ABC and DEF inscribe regular polygons of the sanw number of sides, and denote their perimeters by Q and S, respeo tively.
Then, § 387, Q:S=OB: PE.
If the number of sides of the polygons is indefinitely in- creased, the polygons still remaining regular and similar, § 392, Q and s approach ABC and DEF, respectively, as their limits.
Hence, ABC : DEF = OB : PE. q.e.d.
395. Cor. The ratio of the circumference of a circle to its diame- ter is constant.
The ratio of the circumference of a circle to its diameter is represented by the Greek letter tt whose approximate value, as shown in Ex. 698, is 3.1416.
If the circumference of a circle is denoted by C, its diameter by
D, and its radius by i?, q
7r=—} D
.'. C=7rD, OT 2 7rR;
that is, the circumference of a circle is equal to ir times the diam .
eter or2ir times the radius.
224 PLANE GEOMETRY. — BOOR Vl.
396. Similar arcs, similar sectors, and similar segments are those which, correspond to equal central angles.
Proposition XI
397. Draw a circle and circumscribe about it any regular polygon. How does the apothem of the polygon compare with the radius of the circle ? If the number of sides of the polygon is indefinitely increased, how does the limit of its perimeter compare with the circumference of the circle? How, then, does the polygon at its limit compare with the circle ? Since the polygon is equivalent to one half the rectangle of its perimeter and apothem, how does the circle compare with the rectangle of its circumference and radius?
Theorem, A circle is equivalent to one half the rectangle formed hy Us circumference and radius.
Data: Any circle, as ABG, whose center is O. ^
Denote its circumference by C and its radius by R.
To prove ABG^\ rect. C • R.
Proof. Circumscribe about the circle any regular polygon, as DEF, and denote its perimeter by P.
Then, the apothem of the polygon is equal to R,
and, § 388, DEF =0= i rect. P - i?.
Now, if the number of sides of the polygon is indefinitely in- creased,
§ 392, P indefinitely approaches C as its limit,
and BEF indefinitely approaches ABQ as its limit.
But, however great the number of sides, DEF ^ ^ rect. P • R.
Hence, § 326, ABG 0= ^ rect. C - R.
Therefore, etc. q.e.d^..
Arithmetical Rule: To be framed by the student. § 339"
eLANE GEOMETRY. — BOOK VL 226
398. Cor. I. The area of a circle is equal to it times the square of its radius.
§395, C=2 7rR;
.'. Area = |(2 Tri? x R), or vl^.
399. Cor. II. The areas of circles are to each other as the squares of their radii.
400. Cor. III. The area of a sector is equal to one half the prod- uct of its arc hy its radius.
401. Cor. IV. Similar sectors are to each other as the squares of
their radii. _
Ex. 654. If the circumferences of two circles are 314.16cm and 157.08cm respectively, and the radius of the first is 60^^, what is the radius of the second ? What is the area of the first ?
Ex. 655. Find the circumference and area of a circle whose radius is 2.5^"'.
Ex. 656. What is the ratio of the radii of two circles, if the area of one circle is twice that of the other ?
Ex. 657. What is the area of a sector whose arc is | of the circumference, if the radius of the circle is 18^™ ?
Ex. 658. If the radius of a circle is 63 in., how long is the arc of a sector whose angle is 45° ?
Ex. 659. Calling the equatorial radius of the earth 3962.8 miles, what ia the length of a degree on the equator ?
Ex. 660. Find the radius of a circle whose area is 6'^ ™.
Ex. 661. Find the circumference of a circle whose area is 100 sq. in.
Ex. 662. What is the area of a circle whose circumference is 100 ft. ?
Ex. 663. What is the area of a circle circumscribed about a square whose side is a ?
Ex. 664. The diagonals drawn from a vertex of a regular pentagon to the opposite vertices trisect the angle at that vertex.
Ex. 665. If the chord of an arc is 72dm and the chord of half that arc is 86.9dm, what is the diameter of the circle ?
Ex. 666. The chord of an arc is 24 in. and its altitude is 9 in. What is the diameter of the circle ?
Ex. 667. The chord of half an arc is 12m and the radius of the circle is 18m. What is the altitude of the arc ?
Ex. 668. The altitude of an equilateral triangle is equal to one and a half times the radius of the circumscribed circle.
Ex. 669. Find the central angle subtended by an arc whose length is equal to the radius of the circle.
226
PLANE GEOMETRY.-^ BOOK VI.
Proposition XII
402. Draw two similar segments and their radii. What kind of sectors and what kind of triangles are thus formed? How does the difference in the area of each sector and the corresponding triangle com- pare with the area of the corresponding segment? Since the ratio of the sectors and also the ratio of the triangles equals the ratio of the squares of the radii, how does the ratio of the segments compare with the ratio of the same squares?
Theorem, Similar segments are to eaxih other as the squares upon their radii.
Data : Any two similar segments, as ^5Cand DEF, whose radii are AO and DP respectively.
To prove
seg. ABC ; seg. DBF
— 2 2
= A0: DP . Proof. Draw OC and PF. In the A ACQ and DFP, AO — /.P, and AO'.DP=CO\FP\
A ACQ and dfp are similar; hence, § 342, AACOiA DFP = AO^ : Dp. But, § 401, sect. ABCO : sect. DFFP = Iff : D?;
sect. ABCO : sect. DFFP = A ACQ : A DFP, and sect. ABCO : A ^CO=sect. DFFP : A DFP.
Hence, § 277, sect. ABCO —AAC0:AAC0 = sect. DFFP — A DFP : A DFP ; that is, seg. ABC : A ACO = seg. DEF : A DFP,
or seg. ABC : seg. DEF = A ACO: A DFP.
Why? Why? Why?
Why?
AAC0:ADFP = A0 :DP.
But
Hence, seg. ABC : seg. DEF
Therefore, etc.
AO .DP
Why? Why?
Q.E.D.
PLANE GEOMETRY. — BOOK VL 227
Proposition XIII 403. Problem, To inscribe a square in a circle.
Datum : Any circle. /
Required to inscribe a square in the circle. I
Solution. Draw any two diameters at right angles to each other, as AB and CD.
Draw AD, DB, BC, and CA.
Then, ADBC is the required square. q.e.f.
Proof. By the student.
Proposition XIV 404. Problem, To inscribe a regular hexagon in a circle^
Datum : Any circle. //' \\
Required to inscribe a regular hexagon ^C^ "K
in the circle. \\ //
Solution. From A, any point in the circumference, as a center, and with a radius equal to the radius of the circle, describe an arc intersecting the circumference as at B.
From ^ as a center with the same radius, describe another arc intersecting the circumference as at C
In like manner determine the points D, E, and F.
Draw chords connecting these points in succession.
Then, ABCDEF is the required hexagon. q.e.f
Proof. By the student.
Ex. 670. To circumscribe an equilateral triangle about a circle. Ex. 671. To circumscribe a square about a circle.
228 PLANE GEOMETRY. — BOOK VI.
Proposition XV 405. Problem, To inscribe a regular decagon in a circle.
Datum : Any circle. jf' \^
Required to inscribe a regular decagon ; ^p \
m the circle. A / \ ]d
B
Solution. Draw any radius, as OA, and divide it in extreme and mean ratio as at P ; that is, so that AG : PC = PO : AP.
From ^ as a center and with PO as a radius, describe an arc intersecting the circumference as at B. Draw AB.
From 5 as a center with the same radius, or AB, describe an arc intersecting the circumference as at C.
In like manner determine the points D, E, F, G, H, J, and K.
Draw chords connecting these points in succession.
Then, ABCD-K is the required decagon. q.e.f.
Proof. Draw BP and BO.
Const., AO : PO = PO : AP,
and AB = PO',
.-. in A ABO and ABP, A0:AB = AB: AP, and Zais common ;
A ABP and ABO are similar. Why ?
Now, A ABO is isosceles ;
A ABP is isosceles, and AB = BP = PO; Why?
hence, A BPO is isosceles.
But Z APB = Zo + Z PBO = 2 ZO. Why ?
Then, Za-^Z ABP + Z APB = 5 Z O,
and Z 0 = I of 2 rt. A, or j\ of 4 rt. A;
arc AB is ^-^ of the circumference,
and the chord AB, which subtends the arc AB, is a side of the regular inscribed decagon ABCD-K.
PLANE GEOMETRY, — BOOK VL 229
Proposition XVI
406. Problem, To inscribe a regular pentadecagon in a circle. k ^
Datum : Any circle. . / W
Required to inscribe a regular pentadeca- I m
gon in the circle. V /
c
Solution. Draw the chord AB equal to a side of the regular inscribed hexagon, and from A draw the chord AC equal to a side of the regular inscribed decagon. Draw CB.
From 5 as a center with CB as a radius, describe an arc inter* secting the circumference as at B.
In like manner determine the points E, F, G, H, etc.
Draw chords connecting these points in succession.
Then, CBBEF etc., is the required pentadecagon. q.e.f.
Proof. Arc ^^ = i of the circum.,
and arc AG = ^ oi the circum. ;
.*. arc BG = arc AB — arc ^C = i — J^, or -^^ of the circum.,
and the chord CB, which subtends the arc CB, is a side of the regular inscribed pentadecagon GBDEF etc.
407. Cor. I. By joining the alternate vertices of a,ny regular inscribed polygon of an even number of sides, a regular polygon oj half the number of sides is inscribed.
408. Cor. II. By joining to the vertices of any regular inscribed polygon the middle points of the arcs subtended by its sides, a regular polygon of double the number of sides is inscribed.
Ex. 672. To inscribe a regular octagon in a circle.
Ex. 673. To inscribe a regular dodecagon in a circle.
Ex. 674. To circumscribe a regular hexagon about a circle.
Ex. 675. To circumscribe a regular octagon about a circle.
Ex. 676. To inscribe a regular hexagon in an equilateral triangle.
230
PLANE GEOMETRY. — BOOK VL
Ex. 677. To divide an angle of an equilateral triangle into five equal parts.
Ex. 678. The segment of a circle is equal to f of a similar segment. What is the ratio of their radii?
Ex. 679. . How many degrees are there in an arc 18 in. long on a circum- ference whose radius is 5 ft. ?
Ex. 680. The radii of two similar segments are as 3:5. What is the ratio of their areas ?
Ex. 681. In a circle 3 ft. in diameter an equilateral triangle is inscribed. What is the area of a segment without the triangle ?
Ex. 682. Two chords drawn from the same point in a circumference to the extremities of a diameter of a circle are 6 in. and 8 in. respectively. What is the area of the circle ?
MAXIMA AND MINIMA
409. Of any number of magnitudes of the same kind tlie greatest is called, the Maximum, and the least is called the Minimum.
Of all chords of any circle the diameter is the maximum ; and of all lines from any point to a given line the perpendicular is the minimum.
410. Figures which have equal perimeters are called Isoperi- metric.
Proposition XVII
411. Theorem, Of all triangles having two given that in which these sides are perpendicular to each is the Tnaxijnujn.
Data: Any two triangles, as ABC and ABD, such that AC = AD, AB is common, and AC and AB are perpen- dicular to each other.
To prove A ABC the maximum.
Proof. Draw EDl^AB.
Area of A ABC = -^ ^c x AB, and area of A ABD = ^ED x AB.
sides, other
But
and Hence, Therefore, etc.
AD > ED, and AC= AD)
AC > ED, \ACX AB>^ED X AB. A ABC is the maximum.
Why? Why?
Why?
Q.E.D.
PLANE GEOMETRY. — BOOK VI. 231
Proposition XVIII
412. Theorem. Of all isoperimetric triangles which have the same hase, the isosceles triangle is the maximum.
Data: Any two isoperimetric triangles c.-'' _ i^ upon the same base AB, as ABC and ABD^ ^^'^^^^^ \ of which ABC is isosceles. - ^<^^^ j. ^^-^
To prove A ABC the maximum. ^^^--^^T^^'i-^
Proof. Produce ^C to ^, making CE equal to AC. Draw EB. Since C is equidistant from A, B, and E, Z ABE may be inscribed in a semicircumf erence ;
Z ^5^ is a right angle.
Draw DF equal to i>5 meeting EB produced in i^ ; CG and DH parallel to AB ; CJ and DiT perpendicular to AB ; and draw ^J^. Then, AE = AC + BC = AD -[- BD = AD -{- FD. Why ?
But AD-\- FD> AF] Why?
hence, § 133, EB > FB.
But G-B = 1^5, and HB = ^i^5; Why ?
GB>HB.
Also, (?s = CJ, and flS = D-ff, the altitudes of A ^5C and ABDj /espectively ;
CJ>DK.
Now, area of A iBC = \AB x CJ,
and area of A^5Z) = ^^5 X D^; Why?
area of A ABC > area of A ABD ; that is, A ABC is the maximuin.
Therefore, etc. q.e.d.
413. Cor. Of all isoperimetric tHangles, the equilateral triangle is the maximum.
Ex. 683. Of all equivalent parallelograms having equal bases, the rec- tangle has the least perimeter.
232
PLANE GEOMETRY. — BOOK VI.
Proposition XIX
414. Theorem, Of isoperimetric polygons which have the same nUmber of sides, the Tnaximum is equilateral.
Data : The maximum of isoperimetric poly- -
gons of a given mimber of sides, as ABCDEF.
To prove
ABCDEF equilateral.
Proof. Draw AE. b
Then, A AEF must be the maximum of all the A that can be formed upon AE with a perimeter equal to that of A AEF, for if not, a greater A, as AEG, could be substituted for A ^^i^ without changing the perimeter of ABCDEF.
But it would be impossible to enlarge ABCDEF, for, data, it is the maximum.
Hence, § 412, A AEF is isosceles,
and AF = EF.
Similarly any two consecutive sides may be shown equal.
Hence, ABCDEF is equilateral. q.e.d.
Proposition XX
415. Theorem, Of isoperimetric regular polygons, that which has the greatest number of sides is the m^ajcimum.
Data : Any two isoperimetric regular polygons, as ABC and D, of which D has one side more than ABC.
To prove D the maximum.
Proof. To E, any point in AB, draw CE and construct the A CEF ec^ual to the A ACE.
PLANE GEOMETRY. — BOOK VL
233
Q.E.D.
Then, EBCF ^ ABC,
and EBCF and D are isoperimetric.
But, §414, D>EBCF',
D>ABG\ that is, D is the maximum.
Therefore, etc.
416. Cor. The area of a circle is greater than the area of any ^
isoperimetric polygon.
Proposition XXI
417. Theorem. Of regular polygons which have equal areas, that which has the greatest number of sides has the least perimeter.
Data : Any regular polygons which have equal areas, as A and Bj of which A has a greater number of sides than B.
To prove the perimeter of A less than the perimeter of B.
Proof. Construct the regular polygon C, having its perimeter equal to that of A and having the same number of sides as B.
Then, § 415, A>G. _
But, data, A^^B-,
B>C, the perimeter of B is greater than that of C. the perimeter of C is equal to that of A ; the perimeter of B is greater than that of A ; the perimeter of A is less than the perimeter of B. Therefore, etc. q.e.d.
418. Cor. The circumference of a circle is less than the perimeter of any polygon which has an equal area.
Ex. 684. Of all rectangles of a given area, the square has the least perimeter.
and, § 346, But
that is,
234
PLANE GEOMETRY. — BOOK VI.
M-
SYMMETRY
419. If a point bisects the straight line joining two other points, the two points are said to be symmetrical with respect to a point, and this point is called the center of symmetry.
M and N are symmetrical with respect to the center A, if A bisects the straight line MN.
420. If a straight line is the perpen- dicular bisector of the straight line joining two points, the points are said to be symr metrical with respect to a straight line, and this line is called the axis of symmetry.
M and N are symmetrical with respect to the axis XX', if XX is the perpendicular bisector of the straight line MN.
421. If every point of one figure has a corresponding symmetrical point in another, the two figures are said to be symmetrical with respect to a center or an axis.
'V'o
If every point in the figure ABC has a sym- metrical point in A'B' C with respect to O as a center, then, the figures ABC and A'B'C are symmetrical with respect to the center O.
If every point in the figure DEF has a sym- metrical point in D'E'F' with respect to XX as an axis, then, the figures DEF and D'E'F' are symmetrical with respect to the axis XX'.
Two plane figures that are symmetrical with respect to an axis can be applied one to the other by revolving either one about the axis ; consequently they are equal, and if two figures can be made to coincide by revolving one of them about an axis through 180®, they are symmetrical with respect to the axis.
422. If a point bisects every straight line drawn through it and terminated in the boundary of a figure, the figure is said to be symmetrical with respect to a point.
PLANE GEOMETRY,-~BOOK VI.
235
If 0 bisects every straight line drawn through it and terminated by the boun- dary of ABCDEF, then, ABCDEF is symmetrical with respect to the point O.
423. If a straight line divides a plane figure into two parts which are symmetrical with respect to the line, the figure is said to be symmetrical with respect to a straight line.
If the parts ABCD and AFED are symmetrical with respect to XX', then, the figure ABCDEF is symmetrical with respect t(* the straight line XX'.
Proposition XXII
424. Theorem, A quadrilateral which has two adja/^ent sides equal and the other two sides equal, is symmetrical with respect to the diagonal joining the vertices of the angles formed hy the equal sides.
Data: A quadrilateral, as ABCD, having AB = AD, CB = CD, and the diagonal AG,
To prove ABCD symmetrical with respect to ^C.
Proof. In the A ABC and ADC, data, AB = AD, CB = CDy
and AC is common ;
A ABC = A ADC, Why ?
Z BAC = Z DAC, and Z BCA = Z DC A. Why ?
Hence, if ADC is turned on AC as an axis, it may be made to coincide with ABC.
.'. § 421, ADC and ABC are symmetrical with respect to AC-, that is, § 423, ABCD is symmetrical with respect to AC.
Therefore, etc. q-E-d.
236
PLANE GEOMETRY. — BOOK VL
|
G |
Y F |
||
|
y_ |
^- \9 |
||
|
H |
^'' 1 |
E |
|
|
A |
M; X"" |
0 |
D |
|
B |
C Y' |
Proposition XXIII
425. Theorem. If a figure is syjnmetrical with respect to two ojces perpendicular to each other, it is syminetricaZ with respect to their intersection as a center.
Data: A figure, as ABCD-Hy symmetrical with respect to the two perpendicular axes, XX' and yy\ which intersect at o.
To prove ABCD-H symmet- rical with respect to O as a center.
Proof. From any point in the perimeter, as P, draw PMP' _L xx* and PNQ A. YY^.
Draw MN, P'O, and OQ.
Now, §420, PM=P'M,
and ^M=ON', Why?
P'M= ON, and, §71, P'M W 0N\
consequently, § 150, MP' ON is a parallelogram ;
P'O is equal and parallel to MN.
Similarly, OQ is equal and parallel to MN.
Hence, points P', O, Q are in the same straight line P'OQ, whict* is bisected at 0. Why 5
But since P is any point in the perimeter, P'OQ is any straigh* line drawn through 0.
Hence, § 422, ABCD-H is symmetrical with respect to o as a center.
Therefore, etc. q.e.d.
Ex. 685. A segment of a circle is symmetrical with respect to the per^ pendicular bisector of its chord as an axis.
Ex. 686. A circle is symmetrical with respect to its center or with re- spect to any diameter as an axis.
Ex. 687. A parallelogram is symmetrical with respect to the point oi Intersection of its diagonals as a center.
PLANE GEOMETRY. — BOOK VL 237
SUMMARY 426. Truths established in Book VL
1. Two lines are equal,
a. If they are sides of a regular polygon. § 374
2. Lines are in proportion,
a. If they are the perimeters of regular polygons of the same number of sides, and their radii. § 387
b. If they are the perimeters of regular polygons of the same number of sides, and their apothems. § 387
c. If they are circumferences and their radii. § 394
3. Two angles are equal,
a. If they are angles of a regular polygon. § 374
4. An angle is bisected,
a. If it is an interior angle of a regular polygon, by the radius drawn to its vertex. § 382
5. A polygon is regular,
a. If it is equilateral and equiangular. § 374
b. If it is equilateral and inscribed in a circle. § 375
c. If it is formed by chords joining the extremities of arcs which are equal divisions of the circumference of a circle. § 376
d. If it is formed by tangents drawn at the extremities of arcs which are equal divisions of the circumference of a circle. § 376
e. If it is formed by tangents to a circle at the middle points of the arcs subtended by the sides of a regular inscribed polygon^ § 384
6. Polygons are similar,
a. If they are regular and have the same number of sides. § 386
7. A regular polygon is equivalent,
a. To half the rectangle formed by its perimeter and apothem. § 388
8. A circle is equivalent,
a. To half the rectangle formed by its circumference and radius. § 397
9. A circumference is the limit,
a. Of the perimeter of a regular inscribed polygon when the number of its sides is indefinitely increased. § 392
b. Of the perimeter of a regular circumscribed polygon when the num- ber of its sides is indefinitely increased. § 892
238 PLANE GEOMETRY. — BOOK VI.
10. A circle is the limit,
a. Of a regular inscribed polygon when the number of its sides is indefi- nitely increased. § 392
b. Of a regular circumscribed polygon when the number of its sides is indefinitely increased. § 392
11. Figures are in proportion,
a. If they are regular polygons of the same number of sides, to the squares upon their radii. § 390
b. If they are regular polygons of the same number of sides, to the squares upon their apothems. § 390
c. If they are circles, to the squares of their radii. § 399
d. If they are similar sectors, to the squares of their radii. § 401
e. If they are similar segments, to the squares of their radii. § 402
12. The area of a figure is equal,
a. If it is a regular polygon, to one half the product of its perimeter by its apothem. § 389
b. If it is a circle, to one half the product of its circumference by its radius. § 397
c. If it is a circle, to w times the square of its radius. § 398
d. If it is a sector, to one half the product of its arc by its radius. § 400
SUPPLEMENTARY EXERCISES
Ex. 688. If the perimeter of each of the figures, equilateral triangle, square, and circle is 396 ft., what is the area of each figure ?
Ex. 689. If the inscribed and circumscribed circles of a triangle are con- centric, the triangle is equilateral.
Ex. 690. If an equilateral triangle is inscribed in a circle, any side will cut oflE one fourth of the diameter from the opposite vertex,
Ex. 691. The square inscribed in a circle is equivalent to one half the square circumscribed about that circle.
Ex. 692. A circle is inscribed in a square whose side is 4 in. How much of the area of the square is without the circle ?
Ex. 693. What is the width of the ring between the circumferences of two concentric circles whose circumferences are 48 ft. and 36 ft. respectively ?
Ex. 694. Of all squares that can be inscribed in a given square, the mini- mum has its vertices at the middle points of the sides.
Ex. 695. Every equiangular polygon circumscribed about a circle is regular.
Ex. 696. In any regular polygon of an even number of sides, the lines joining opposite vertices are diameters of the circumscribing circle.
PLANE GEOMETRY.— BOOK VI.
239
Ex. 697. Given the side of a regular inscribed polygon and the side of a similar circumscribed polygon^ to compute the perimeters of the regular in- scribed and circumscribed polygons of double the number of sides.
Data : AB, the side of a regular inscribed polygon, and C*Z>, the side of a similar circum- scribed polygon, tangent to the arc AB at its middle point E.
Denote the perimeters of these polygons tty P and Q respectively, and the number of sides in each by n ; denote the perimeters of the inscribed and circumscribed polygons which have 2 n sides by S and T respectively.
Required to^compute the value of S and of T.
Solution. Through A and B draw tangents to meet CD in F and Cr respectively ; also draw AE and BE.
Then, § 376, AE and FO are sides of the polygons whose perimeters are iSand r respectively. P
AB =
n
AE
8 T
-^, and FG=^'
2 w 2 n
Draw the radii 00, PO, EG, and BO. Since, § 385, A lies in CO,
by Ex. 221, .-. §292, but
and
But .'. substituting, or whence,
Again, in the
.-. §301,
and
hence,
and substituting
whence, and
FO bisects Z^ OP, or Z COE ; EF:CF=:EO:CO;
P:Q = EO:CO;
P:Q = EF:CF, P+ Q : P = EF + CF : EF = CE :
CE = iCD = ^, andPP=iPO = ^
2n 4n P-\-Q:P = 2Q:T;
Q + P isosceles ^ABE and AEF,
ZABE = ZAEF', A ABE and AEF are similar, AE:AB = EF:AE; AE^ = ABx EF, for AE, AB, and EF their values,
4 n^ n 4 w *
/8'2 = Pxr,
EF.
JS= VPxT,
240
PLANE GEOMETRY. — BOOK VL
Ex. 698. To compute the approximate ratio of a circumference to its diameter.
Solution. If the diameter of a circle is 1, the side of a circumscribed square is 1, and its perimeter is 4 ; the side of an inscribed square is \ \/2, and its perimeter is 2 V2, or 2.82843.
Thus, § = 4, and P = 2 V2 for computing the octagon.
Substituting these values in the formulae, 7" = ^ ^ , S — y/P x T
Q-\-P (Ex. 697), and solving, the results tabulated below are found, showing the
perimeters to five decimal places.
No. OF
Computation of T
Computation of S
8
16
32
64
128
256
512
1024
y^2QxP Q+P
2 X 4 X 2.82843 4 + 2.82843
2 X 3.31371 X 3.06147 3.31371 + 3.06147
2 X 3.18260 X 3.12145
3.18260 + 2 X 3.15172
3.12145 X 3.13655
3.15172 + 3.13655 2 X 3.14412x3.14033
3.14412 + 2 X 3.14222
3.14033 X 3.14128
3.14222 + 2x3.14175
3.14128 X 3.14151
3.14175 + 2 X 3.14163
3.14151 X 3.14157
3.14163 + 3.14157
3.31371 3.18260 3.15172 3.14412 3.14222 3.14175 3.14163 3.14159
|
s=Vp |
xT |
|
V2.82843 X |
3.31371 |
|
V3.06147 X |
3.18260 |
|
V3. 12145 X |
3.15172 |
|
V3. 13655 X |
3.14412 |
|
V3. 14033 X |
3.14222 |
|
V3. 14128 X |
3.14175 |
|
V3. 14151 X |
3.14163 |
V3.14157X 3.14159
3.06147 3.12145 3.13655 3.14033 3.14128 3.14151 3.14157 3.14159
The results of the last two computations show that the circumference of a circle whose diameter is 1 is approximately 3.1416 ; that is, the ratio of the diameter of a circle to its circumference is equal to the ratio of 1 to 3.1416, approximately.
Ex. 699. The sides of an inscribed rectangle are 30cm and 40°™. What is the area of the part of the circle without the rectangle ?
Ex. 700. What is the area of a figure bounded by four semicircumfer- ences described on the sides of a three foot square ?
Ex. 701. A square piece of land and a circular piece of land each con- tain one acre. Which perimeter is the greater, and how much ?
PLANE GEOMETRY. ^BOOK VL 241
Ex. 702. The area of an inscribed equilateral triangle is one half the area of a regular hexagon inscribed in the same circle.
Ex. 703. Of all triangles that have the same vertical angle and whose bases pass through a given point, the minimum is the one whose base is bisected at that point.
Ex. 704. An arc of a circle whose radius is 6 ft. subtends a central angle of 20° ; an equal arc of another circle subtends a central angle of 30°. What is the radius of the second circle ?
Ex. 705. Two tangents make with each other an angle of 60°, and the radius of the circle is 7 in. What are the lengths of the arcs between the points of contact ?
Ex. 706. If the apothem of a regular hexagon is 10™, what is the area of the ring between the circumferences of its inscribed and circumscribed circles ?
Ex. 707. If a circle 18<»» in diameter is divided into three equivalent parts by two concentric circumferences, what are their radii ?
Ex. 708. The square upon the side of a regular inscribed pentagon is equivalent to the sum of the squares upon the radius of the circle and the side of a regular inscribed decagon.
Ex. 709. The radius of a regular inscribed polygon is a mean proportional between its apothem and the radius of the similar circumscribed polygon.
Ex. 710. If the radius of a regular inscribed octagon is r, prove that its
side = r V2 - V2, and its apothem = - V2 + y/2.
2
Ex. 711. If the radius of a regular inscribed decagon is r, prove that its
side = ~ ( VS - 1), and its apothem = ^ VlO + 2 V6. 2 4
Ex. 712. If the radius of a regular inscribed dodecagon is y, prove that its side = r V2-V3, and its apothem = - V2 + V3.
Ex. 713. If the radius of a regular inscribed pentagon is r, prove that its
side = - VlO - 2 V5, and its apothem = \ Ve + 2 V6. 2 4
Ex. 714. The square upon a side of an inscribed equilateral triangle is equivalent to three times the square upon the side of a regular inscribed hexagon.
Ex. 715. The area of an inscribed square is 16"9i». pind the length of a side of a regular inscribed octagon.
Ex. 716. If the radius of a circle is r, prove that a side of a regular circumscribed hexagon is -^ Vs.
Ex. 717. The area of a regular inscribed dodecagon is equal to three •limes the square of the radius. milne's oeom. — 16
242 PLANE GEOMETRY. — BOOK VI.
Ex. 718, Find the side of a regular hexagon circumscribed about a circle whose diameter is 1.
Ex. 719. The apothem of an inscribed regular hexagon is equal to one half the side of the inscribed equilateral triangle.
Ex. 720. The area of a ring bounded by two concentric circumferences is equal to the area of a circle whose diameter is a chord of the outer circumfer- ence and is tangent to the inner circumference.
Ex. 721. If the radius of a circle is r, find the area of a segment whose chord is one side of a regular inscribed hexagon.
Ex. 722. Three equal circles with a radius of 12 ft. are drawn tangent to each other. What is the area between them ?
Ex. 723. The area of an inscribed regular hexagon is equal to three fourths that of a regular hexagon circumscribed about the same circle.
Ex. 724. The altitude of an equilateral triangle is equal to the side of an equilateral triangle inscribed in a circle whose diameter is the base of the first triangle.
Ex. 725. If the radius of a circle is r and the side of a regular inscribed
polygon is a, prove that the side of a similar circumscribed polygon is ,
v4r2— a"^
Ex. 726. If the alternate vertices of a regular hexagon are joined by straight lines, another regular hexagon is formed which is one third as large as the original hexagon.
Ex. 727. The diagonals of a regular pentagon divide each other in extreme and mean ratio.
PROBLEMS OF CONSTRUCTION
Ex. 728. Construct x, ii x = Vab.
Ex. 729. Inscribe a circle in a given sector.
Ex. 730. In a given circle describe three equal circles tangent to each other and to the given circle.
Ex. 731. Divide a circle into two segments such that an angle inscribed in one shall be three times an angle inscribed in the other.
Ex. 732. Construct a circumference equal to the sum of two given circumferences.
Ex. 733. Inscribe a square in a given quadrant.
Ex. 734. Inscribe a square in a given segment of a circle.
Ex. 735. Through a given point draw a line so that it shall divide a. given circumference into two parts having the ratio 3:7.
Ex. 736. Construct a circle equivalent to twice a given circle.
Ex. 737. Construct a circle equivalent to three times a given circle.
SOLID GEOMETRY
BOOK YII
PLANES AND SOLID ANGLES
427. A plane is a surface such that a straight line joining any two of its points lies wholly in the surface. § 14.
A plane is considered to be indefinite in extent, but in a diagram it is usually represented by a quadrilateral segment.
428. The student will be aided in obtaining correct concepts of the truths presented in the geometry of planes by using pieces of cardboard or paper to represent planes, and drawing such lines upon them as are required. Pins may be used to represent the lines which are perpendicular or oblique to the planes.
429. 1. By using cardboard to represent a plane and the point of a pin or pencil to represent a point in space, discover in how many directions the plane may be passed through the point.
2. By using a card as before and the points of a pair of dividers to represent two fixed points in space, discover whether the num- ber of directions that the plane may take is greater or less than when it was passed through one fixed point.
3. Suppose a plane is passed through three fixed points not in the same straight line, how many directions may it take ? How many points, then, determine the position of a plane ?
4. Since two of the points must be in a straight line, what else besides three points determine the position of a plane ?
5. Since a straight line through the other point may intersect the straight line joining the two points, what else will determine the position of a plane ?
243
244
SOLID GEOMETRY. — BOOK VII.
6. Since a straight line may join two of the points and a straight line parallel to that may be drawn through the other point, how else may the position of a plane be determined ?
In what ways, then, may the position of a plane be determined ?
430. A plane is determined by certain points or lines, when it is the only plane which contains those points or lines.
A plane is determined by
1. Three points not in the same straight line.
2. A straight line and a point ivithout that line,
3. Two intersecting straight lines.
4. Two parallel straight lines.
431. The point at which a line meets a plane is called the Foot of the line.
432. A straight line that is perpendicular to every straight line in a plane drawn through its foot is perpendicular to the plane.
In this case the plane is perpendicular to the line.
433. A straight, line that is not perpendicular to every line in a plane drawn through its foot is oblique to the plane.
434. A straight line and a plane which cannot meet, however far they may be produced, are parallel to each other.
435. Two planes which cannot meet, however far they may be produced, are parallel to each other.
436. The locus of the points common to two non-parallel planes is the Intersection of the planes.
437. The foot of the perpendicular, let fall from a point to a plane, is called the Projection of the point on the plane.
438. The locus of the projections on a plane of all points in a line is called the Projection of the line.
The point D is the projection of the point A upon the plane MN, and DEF is the projection of the line ABC on the plane Mir.
SOLID GEOMETRY. — BOOK VIL
245
439. The angle which a straight line makes with a plane is the acute angle between the line and its projection on the plane, and is called the inclination of the line to the plane.
MN is a plane ; AB a straight line meeting MN\ and AD the projection of AB on MN. Then, angle BAD is the angle which AB makes with the plane MN.
440. The distance from a point to a plane is understood to be the perpendicular distance from that point to the plane.
Proposition I
441. Place two planes * so that they intersect. What kind of a line is the line of their intersection ?
Theorem. The intersection of two planes is a straight line.
Data: Any two intersecting planes, as MN and PQ.
To prove the intersection of MN and PQ a straight line.
Proof. Suppose that E and F are any
two of the points in which MN and PQ
intersect. Draw the straight line EF. '*2
Since E and F are points in the plane MN, § 427, the straight line joining them must lie in MN; and since they are also points in PQ, the straight line joining them must lie in PQ.
Hence, EF is common to MN and PQ ;
and since, § 430, only one plane can contain a line and a point with- out that line, no point without EF can be common to MN and PQ; .*. § 436, EF is the intersection of MN and PQ.
But, const., EF is a straight line ;
hence, the intersection of MN and PQ is a straight line. q.e.d.
♦ The student may represent planes and lines as suggested in § 428.
246
SOLID GEOMETRY. — BOOK VII.
Proposition II
442. In a plane draw two intersecting straight lines. If a third straight line is perpendicular to each of these at their point of intersec- tion, what is its direction with reference to the plane?
Theorem. If a straight line is perpervdicular to each of two other straight lines at their point of intersection, it is perpendicular to the plane of the two lines.
Data : Any two straight lines, as SB and CD, intersecting at ^; MN, the plane of these lines ; and HE, a perpendicular to AB and CD at E.
To prove HE perpendicular to MN.
Proof. Through E, in the plane MN^ draw any other straight line, as JK; also draw AC intersecting JK in L.
Produce HE through MN to F, making EF = HE, and draw HAy HL, HC, FA, FL, and FC.
In Aach and ACF, AC is common, § 103, HA = FA, and HC = FC;
.'. § 107, A ACH = A ACF,
and /.HAC=AfaG\
.-. § 100, A ALH = A ALF,
and HL = FL;
.'. §106, LE±HE;
that is, HE±JK.
Consequently, HE is perpendicular to every straight line drawn in MN through E.
Hence, § 432, HE is perpendicular to MN.
Therefore, etc. q.e.d.
443. Cor. A straight line, which is perpendicular to a plane at any point, is perpendicular to every straight line which can be drawn in that plane through that point.
SOLID GEOMETRY. — BOOK VII.
247
Proposition III
444. 1. At any point in a given straight line erect two* perpendicu- lars to the line, and- through them pass a plane. What is the direction of the plane with reference to the given line? Can any perpendiculars be drawn to the given line, at this point, which do not lie in this plane ?
2. Can any other plane be passed through this point perpendicular to the given line V
3. Through a point without a straight line pass as many planes as possible perpendicular to the line. How many such planes can oe passed through the point?
Theorem, Every perpendicular to a straight line at a given point lies in a plane which is perpendicular to the line at that point.
Data: Any straight line, as AB-, and a plane, as MN, perpendicular to AB at E\ also any line, as EF, perpendicular to AB at E.
To prove that EF lies in MN.
Proof. Suppose that the plane of AB and EF intersects MN in the line EH.
Then, § 443, ABA. EH.
Since, § 51, in the plane of AB and EF only one perpendicular can be drawn to AB at E, EF and EH coincide, and EF lies in MK.
Hence, every perpendicular to AB at E lies in the plane MN.
Therefore, etc. q.e.d.
445. Cor. I. At a given point in a straight line one plane per- pendicular to the line can he passed, and only one.
446. Cor. II. Through a given point without a straight line one 2^la7ie perpendicular to the line can be passed, and only one.
Ex. 738. What is the locus of the perpendiculars to a given straight line at a given point in the line ?
248
SOLID GEOMETRY. — BOOK VIL
Proposition IV
447. 1. Erect a perpendicular to a plane ; connect a point in the per- pendicular with points in the plane which are equally distant from the foot of the perpendicular. How do these oblique lines compare in length?
2. Connect the same point in the perpendicular with points in the plane unequally distant from the foot of the perpendicular. How do these oblique lines compare in length ?
3. Represent a perpendicular and several other lines from a point to a plane. Which line is the shortest?
Theorem. If from a point in a perpendicular to a plane oblique lines are drawn to the plane,
1. Those lines which meet the plane at equal distances from the foot of the perpendicular are equal.
2. Of two lines which meet the plane at unequal dis- tances from the foot of the perpendicular, that which meets it at the greater distance is the greater.
Data: A perpendicular to the plane MN, as CD, and the oblique lines CE, CF, and CG, which meet Mlfso that DE=DF, and DG > DE.
|
To prove |
1. |
CE = CF. |
|
2. |
CG > CE. |
|
|
Proof. |
1. |
Data, CD±MN-', |
CD± DE, and CD ± DF, In rt. A EDC and FDC, DE = DF,
CD is common, and, § 52, Z EDC = Z fdG\
.'. § 100, A EDC = A FDCy
and CE = CF.
2. On DG take DH= DE, and draw CH.
Then, CH= CE.
Hence, § 132, CG > CH, or CE.
Why?
Why?
Q.B.D.
SOLID GEOMETBY. — BOOK VIL
•249
448. Cor. I. A perpendicular is the shortest line that can be drawn from a point to a plane.
449. Cor. II. Equal oblique lines from a point in a perpendicii' lar to a plane meet the plane at equal distances from the foot of the perpendicular; and of two unequal oblique lines the greater meets the plane at the greater distance from the. foot of the perpendicular,
450. Cor. III. Tlie locus of a point in space equidistant from all points in the circumference of a circle is a straight line passing through the center and perpendicular to the plane of the circle.
Proposition V
451. Erect a perpendicular to a plane; from the foot of the perpen- dicular draw a straight line at right angles to any other straight line of the plane ; join the point of intersection of these two lines with any point in the perpendicular. What is the direction of this joining line with reference to the line in the plane that does not pass through the foot of the perpendicular ?
Theorem, If from the foot of a perpendicular to a plane a straight line is drawn at right angles to any straight line in the plane, the line drawn from the point of meeting to any point in the perpendicular is perpen- dicular to the line of the plane.
Data: A perpendicular . %tf
to the plane MN, as CD; any straight line in MN, as EF'j DG perpendicular to EF; and CG joining any point in CD with G. / i> / ^
To prove CG perpendicu- lar to EF.
Proof. From C and B draw lines to H and J, two points in EV^ equally distant from Q.
Then, § 103, HD = JD',
.-.§447, HC^JG\
hence, § 106, CGLhJ:,
that is, CG is perpendicular to EF,
Therefore, etc. Q.e.d
M^
250
SOLID GEOMETRY. — BOOK VII.
452. Cor. The locus of a point in space equidistant from the extremities of a straight line is the plane perpendicular to the line at its middle point.
Proposition VI
453. At any two points in a plane erect perpendiculars to the plane. What is the direction of the perpendiculars with reference to each other?
Theorem. Two straight lines perpendicular to the same plane are parallel.
Data : Any two straight lines perpendicular to plane MN, as CD and EF.
To prove CD and EF parallel.
M
Proof. Draw CF and DF, and through^ J^ draw GHl. DF.
Then, §443, EF±GH,
const., DF± GHy
and, § 451, CF±GH;
.'. § 444, EF, DF, and CF lie in the same plane.
Hence, CD and EF lie in the same plane.
But, § 443, CD ± DF, and EF ± DF;
hence, § 71, CD and EF are parallel.
Therefore, etc.
454. Cor. I. If one of two parallel straight lines is perpendicular to a plane, the other is also perpendicular to the plane.
455. Cor. II. Two straight lines that are parallel to a third straight line in another plane are parallel to each other.
Q.E.D.
N
Ex. 739. The length of a perpendicular from a given point to a plane is 5dm, What is the diameter of the circle which is the locus of the foot of an oblique line drawn from the same point to the plane, if the oblique line is 13<»™ long ?
SOLID GEOMETRY. — BOOK VIL 251
Proposition VII
456. 1. At any point in a plane erect as many perpendiculars to the plane as possible. How many can be erected ?
2. Choose a point above or below the plane, and from that point draw as many perpendiculars to the plane as possible. How many such per- pendiculars can be drawn ?
Theorem, From a given point only one perpendicular to a given plane can be drawn.
B\ jC
Data: Any plane, as MN, and any I /
point, as A. I '^
To prove that from A only one per- / ^
pendicular to MN can be drawn.
M
Proof. Case I. When the given point is in the given plane.
Draw AB X MN, and from A draw any other line, as AC.
li AC is perpendicular to MN, § 453, AB and AC are parallel to every line that is perpendicular to MN; but, § 70, this is impossible ;
AC is not perpendicular to MN.
Hence, only one perpendicular to MN can be drawn from A.
Case II. When the given point is without the given plane.
Draw AB _L MN, and from A draw l\
any other line to MN, as AC.
If ^C is perpendicular to MN, § 453, AB and AC are parallel to every line that is perpendicular to MN; but, § 70, this is impossible ; ^^
C
AC is not perpendicular to MN.
Hence, only one perpendicular to MN can be drawn from A. Therefore, etc. q.e.d.
252 SOLID GEOMETRY. — BOOK VII.
Proposition VIII
457. 1. Represent two parallel straight lines, only one of which is in a given plane. What is the direction of the other line with reference to the plane ?
2. Represent a plane and a straight line parallel to it ; pass any plane through the line so that it intersects the given plane. What is the direction of the intersection with reference to the given line ?
3. Represent two intersecting planes and a straight line parallel to their intersection. What is the direction of this line with reference to each of the planes ?
4. Represent any two straight lines in space ; through one of them pass any plane. Can this plane be turned on the line as an axis into a position parallel to the other given line ?
5. Represent any two straight lines in space and a point without them ; represent a line passing through this point and parallel to one of the given lines; represent a second line through the same point and parallel to the other given line ; represent the plane of these lines that intersect at the given point. What is the direction of this plane with reference to each of the given lines ?
Theorem. If a straight line without a plane is parallel to any straight line in the plane, it is parallel to the plane.
Data: Any straight line in
plane MN, as EF, and any / — j -i -jN
straight line without MN and parallel to EF, as CD.
To prove CD parallel to MN.
Proof. Through CD and EF pass the plane ED.
Now, if CD is not parallel to MN, it must meet MN in the inter section of MN and ED, that is, in EF.
But, data, CD cannot meet EF-,
hence, - CD cannot meet MNi
that is, CD is parallel to MN. Q.e.d.
458. Cor. I. If a straight line is parallel to a plane, the intersec- tion of the plane with any plane passed through the line is parallel to the line.
SOLID GEOMETRY. — BOOK VIL 253
459. Cor. II. A straight line parallel to the intersection of two planes is parallel to each of the planes.
460. Cor. III. Through any given straight line a plane may be passed parallel to any other given straight line not intersecting the first; if the lines are not parallel, only one such plane can he passed.
461. Cor. IV. Through a given poiiyt a plane may he passed parallel to any two given straight lines in space; and if the lines are not parallel, only one such plane can he passed.
Proposition IX
462. Represent two planes each perpendicular to the same straight line. What is the direction of the planes with reference to each other ?
Theorem, Two planes perpendicular to the same straight line are parallel.
Data : Any two planes perpendicu- lar to EF, as MN and PQ.
To prove MN and PQ parallel.
7
F
Proof. If MN and PQ are not parallel, they will meet, and thus there will be two planes passing through the same point and per- pendicular to the same line EF.
But, § 446, this is impossible.
Hence, MN and PQ cannot meet;
that is, MN and PQ are parallel.
Therefore, etc. Q-ed-
Ex. 740. If a straight line and a plane are perpendicular to another straight line, they are parallel.
Ex. 741. If a hne is equal to its projection on a plane, it is parallel to the plane.
Ex. 742. If a line makes equal angles with three intersecting lines in the same plane, it is perpendicular to that plane.
Ex. 743. If a plane bisects a straight line at right angles, any point in the plane is equidistant from the extremities of the line.
264 SOLID GEOMETRY.— BOOK VIL
Proposition X
463. 1. Represent two parallel planes each intersected by a third plane. In what direction, with reference to each other, do the lines of intersection extend ?
2. Represent two parallel straight lines included between two parallel planes. How do the lines compare in length ?
Theorem, The intersections of two parallel planes hy a third plane are parallel.
Data: Any two parallel planes, as MN and PQ, intersected by a third m^ plane, as RS, in GH and JK, respec- tively.
To prove GH and JK parallel. p^
Proof. § 435, MN and PQ cannot meet ; .*. GH and JK, which are lines lying in MN and PQ respectively, cannot meet.
But GH and JK lie in the same plane i?S;
hence, GH and JK are parallel.
Therefore, etc. q.e.d.
464. Cor. I. Parallel straight lines included between parallel planes are equal.
465. Cor. II. Two parallel planes are everywhere equally distant.
Ex. 744. Draw a perpendicular to a given plane from any point without the plane.
Ex. 745. Erect a perpendicular to a given plane at a given point in the plane.
Ex. 746. A line parallel to two intersecting planes is parallel to their intersection.
Ex. 747. If two lines are parallel, the intersections of any planes pass- ing through them are parallel.
Ex. 748. If a plane is passed through a diagonal of a parallelogram, the perpendiculars to it from the extremities of the other diagonal are equal.
SOLID GEOMETRY. — BOOK VI I. 255
Proposition XI
466. 1. Represent a straight line perpendicular to one of two parallel planes. What is the direction of the line with reference to the other plane ?
2. Pass as many planes as possible through a point and parallel to a given plane. How many such planes can be passed ?
3. Represent two intersecting straight lines each parallel to a given plane. What is the direction of the plane of these lines with reference to the given plane ?
Theorem, A straight line perpendicular to one of two parallel planes is perpendicular to the other.
Data : Any two parallel planes, as MN and PQ, and any straight line M' perpendicular to MN, as EF.
To prove EF perpendicular to PQ.
7
H
Proof. Through EF pass any two planes, as EH and EK, inter- secting MN in EG and EJ, and PQ in FH and FK, respectively.
Then, § 463, EG II FH, and EJ II FK;
and, § 443, EF±EG smd EJ-,
EF ± FH 3iiid FK. Why?
Hence, § 442, EF is perpendicular to PQ.
» Therefore, etc. q.e.d.
467. Cor. I. Through a given point one plane, and only one, can he passed parallel to a given plane.
468. Cor. II. If two intersecting straight lines are each parallel to a given plane, the plane of those lines is parallel to the given plane.
Proposition XII
469. Draw an angle in one plane, and in another, an angle whose sides are respectively parallel to the sides of the first angle and extending in the same direction. How do the angles compare in size? In what direc- tion do their planes extend with reference to each other?
256
SOLID GEOMETRY. — BOOK VII.
Theorem, If two angles, not in the same plane, have their sides respectively parallel and extending in the same direc- tion, they are equal and their planes are parallel.
Data ; Any two angles, as F and J, in the planes MN and PQ respec- tively, having the sides FE and FG parallel to and extending in the same direction with JH and JK respectively.
To prove Zf=Z.J, and MN II PQ.
Proof. 1. Take FE=:.JH, and FG = JK\ and draw FJ^ EH, GK, EG, and HK.
Then, § 150, fehj and FGKJ are parallelograms ; EH=FJ=z GK, and EH II FJ II GK', EQKH is a parallelogram, and EG = HK. A EFG = A HJKf
Zf = Zj.
FEWpq, and FGWpQ;
MN 11 PQ.
Hence,
and
2. § 457, hence, § 468,
Therefore, etc.
Why? Why? Why?
Q.E.D.
Proposition XIII
470. Represent two straight lines intersected by three parallel planes. If one line is divided into segments which are in the ratio of 2 : 3, what is the ratio of the segments of the other line? If one line is divided into segments which are in any ratio whatever, how does the ratio of the segments of the other line compare with this ratio ?
Theorem, If two straight lines are intersected hy three parallel planes, their corresponding segments are propor- tional.
Data: Any two lines, as AB and CD, intersected by any three paral- lel planes, as MN, PQ, and RS, in the points A, L, B, and C, G, D, respec- tively.
To prove AL:LB=CG: GD.
|
M^ |
||
|
L |
\ \ |
|
|
r/ |
L.\a? / |
|
|
r |
\ \ |
|
|
/ |
'V |
|
|
rI- |
--Dj |
SOLID GEOMETRY. — BOOK VIL 257
Proof. Draw AD intersecting PQ in 0; and draw LO, OG, AC, and JS-D.
Then, §§ 430, 463, LO II bd, and OG II AG; .-. § 289, and hence, AL : LB == CQ : GD,
Therefore^ etc. ' q.e.d.
DIHEDRAL ANGLES
471. The opening between two intersecting planes is called a Dihedral Angle, or simply a Dihedral.
The intersection of the planes is called the edge of the dihedral angle, and the two planes are called its faces.
472. A dihedral angle may be desig- y\ nated by letters at four points, two in /^ \ its edge and one in each face, the two i>^ a \ letters at the edge being written between \ \ \ the other two. / ^— -\-— .^^
When but one dihedral angle is formed / e X -\^
at the same edge it is designated simply ^/ \/
by two letters at this edge.
AB is the edge, and BO and BD are the faces of the dihedral angle AB, or G-AB-D.
473. The angle formed at; any point in the edge of a dihedral angle by two perpendiculars to the edge, one in each face, is called the Plane Angle of the dihedral angle.
EF and GF m BG and BD respectively, both perpendicular to AB at F, form the plane angle EFQ of the dihedral angle G-AB-D.
The plane angle is of the same size at whatever point in the edge it is constructed. (§§ 71, 469).
The size of a dihedral angle does not depend upon the extent of its faces, but upon their difference in direction.
474. Two dihedral angles which can be made to coincide are equal.
milne's geom. — 17 .
268 SOLID GEOMETRY. — BOOK VII.
475. Dihedral angles are adjacent, right, acute, obtuse, comple- mentary, supplementary, or vertical, according as their plane angles conform to the definitions of those terms given in plane geometry.
Two dihedral angles are adjacent, if they have a common edge, and a common face between them ; they are rights if they are formed by two perpendicular intersecting planes ; they are vertical^ if the faces of one are prolongations of the faces of the other.
EXERCISES
476. 1. Eepresent a plane meeting another plane. How does the sum of the two dihedral angles thus formed compare with two right dihedral angles ?
2. Represent two adjacent dihedral angles whose sum is equal to two right dihedral angles. How do their exterior faces lie ?
3. Eepresent two intersecting planes. How do the vertical dihedral angles compare in size ?
4. Eepresent two parallel planes intersected by a third plane. How do the alternate interior dihedral angles compare in size ? How do the corresponding dihedral angles compare ? To how many right dihedral angles is the sum of the two interior dihedral angles on the same side of the intersecting plane equal ?
5. Eepresent two planes intersected by a third plane. In what direction do the two planes extend with reference to each other, if the alternate interior dihedral angles are equal ? If the corresponding dihedral angles are equal? If the sum of the interior dihedral angles on the same side of the intersecting plane is equal to two right dihedral angles ?
6. Eepresent two dihedral angles whose corresponding faces are parallel. How do these dihedrals compare in size, if both corresponding pairs of faces extend in the same direction from their edges ? If both extend in opposite directions ?
Discover whether it is possible for the dihedrals to have their faces parallel and yet not be equal.
7. Eepresent two dihedral angles whose corresponding faces are perpendicular to each other. How do the dihedrals compare in size, if both are acute 9 If both are obtuse ?
Discover whether it is possible for the dihedrals to have their faces perpendicular and yet not be equal.
SOLID GEOMETRY. — BOOK VII.
259
Proposition XIV
477. Represent two dihedral angles whose plane angles are equal. How do the dihedral angles compare in size ?
Theorem, Two dihedral angles are equal, if their plane <ingles are equal.
Data: Any two dihedral angles, as AB and EF, whose plane angles CAB and GEH are equal.
To prove dihedral /. AB = dihedral Z EF.
Proof. Suppose- that dihedral Z AB is applied to dihedral Z EF in such a way that the plane Z CAD coincides with the equal plane Z GEH.
Then, point A coincides with point E,
and, § 430, plane CAD coincides with plane GEH-, .-. § 456, AB, the perpendicular to plane CAD, coincides with EF, the perpendicular to plane GEH.
Since AB coincides with EF, and AC with EG, § 430, plane BC coincides with plane FG.
In like manner it may be proved that
plane BD coincides with plane FH.
Hence, § 474, dihedral Z AB = dihedral Z EF.
Therefore, etc. q.e.d.
Ex. 749. If two planes intersect each other, the vertical dihedral angles are equal.
Ex. 750. If a plane intersects two parallel planes, the alternate interior dihedral angles are equal.
Ex. 751. The line ABC pierces three parallel planes in A^ B, and C, respectively, and the line DEF pierces the same planes in 2), E, and F, respectively. If AB is 6 in., BC 8 in., and DF 12 in., what is the length of DE and of EF?
260
SOLID GEOMETRY, — BOOK VII.
Proposition XV
478. Represent two dihedral angles whose plane angles are in the ratio of 3 to 4, or any other ratio. What is the ratio of the dihedial angles ?
Theorem, Dihedral angles are to each other as their plane angles.
H
Data: Any two dihedral angles, as C-^^-D and G-EF-U^ whose plane angles are CAD and G^fl" respectively.
To prove C-AB-D : G-EF-H=z A CAD : Z GEH.
Proof. Suppose that A CAD and GEH have a common unit of measure, as Z CAJ, which is contained in Z CAD three times and in Z GEH four times.
Then, Z CAD : Z GEH =3:4.
Divide the plane angles CAD and GEH into parts each equal to Z CAJ, and through the several lines of division and the edges AB and EF pass planes..
By § 477, these planes divide C-AB-D into three and G-EF-H into four equal parts ;
C-AB-D : G-EF-H = 3 : 4.
Hence, C-AB-D : G-EF-H = Z CAD : Z GEH.
By the method of limits, exemplified in § 223, the same may be : proved when the dihedral angles are incommensurable.
Therefore, etc. q.e.d.
479. Sch. It is evident that the plane angle of a dihedral may be taken as the measure of the dihedral.
Ex. 752. If the sum of two adjacent dihedral angles is equal to two right dihedral angles, their exterior faces are in the same plane.
SOLID GEOMETRY.^ BOOK VIL 261
Proposition XVI
480. 1. Represent two planes that are perpendicular to each other; in one of them draw a straight line perpendicular to their intersection. What is the direction of the line with reference to the other plane ?
2. Represent two planes perpendicular to each other and a straight line perpendicular to one of them at any ^point of their intersection. How does this line lie with reference to the other plane?
Theorem. If two planes are perpendicular to each other, a straight line drawn in one of thein perpendicular to their intersection is perpendicular to the other.
Data: Any two planes perpen- dicular to each other, as MN and PF, intersecting in EF, and any line in PF perpendicular to EF, as GH.
To prove
GH perpendicular to MN.
Proof. In MN draw HJl. EF.
\L
Then, § 473, the angle GHJ is the plane angle of the right di- hedral angle P-EF-N .'. § 475, Z GHJ is a rt. Z.
Data, Z GHE is a rt. Z ;
hence, GHl. HJ and EF at their point of intersection ; and, § 442, GH is perpendicular to MN.
Therefore, etc. q.e.d.
481. Cor. If tivo planes are perpendicular to each other, a per- peyidicular to one of them at any point of their intersection lies in the other.
Ex. 753. Find the locus of all points equidistant from two parallel planes.
Ex. 754. Parallel lines which pierce the same plane make equal angles with it.
Ex. 755. If two planes intersect each other, the sum of the two adjacent dihedral angles on the same side of either plane is equal to two right dihedral angles.
Ex. 756. If a plane intersects two parallel planes, the interior dihedral angles on the same side of the intersecting plane are supplementary.
262 SOLID GEOMETRY. — BOOK VIL
Proposition XVII
482. 1. Represent a straight line perpendicular to a plane. What is the direction, with reference to this plane, of every plane passing through the perpendicular?
2. Represent a dihedral angle and a plane perpendicular to its edge. What is the direction of this plane with reference to each of the faces of the dihedral ?
Theorem, If a straight line is perpendicular to a plane, every plane passed through the line is perpendicular to that plane.
Data : Any straight liue perpen- dicular to plane MN, as CD, and any plane passing through CD, as EF.
To prove
EF perpendicular to MN.
M'
iN
E
Proof. In MN draw DH A. EG, the intersection of EF and MN.
§443, CD A. EG',
.-. § 473, Z CDH is the plane angle of dihedral Z F-EG-N.
But, § 443, Z CDH is a rt. Z ;
hence, F-EG-N is a right dihedral angle ;
that is, EF is perpendicular to MN.
Therefore, etc. q.e.d.
483. Cor. A plane perpendicular to the edge of a dihedral angle is perpendicular to each of its faces.
Ex. 757. Find the locus of all points in space equidistant from two given points.
Ex, 758. Find the locus of all points at a given distance from a given plane.
Ex. 759. A line and its projection on a plane determine a second plane perpendicular to the fii*st.
Ex. 760. If a line is parallel to one plane and perpendicular to another, the two planes are perpendicular to each other.
Ex. 761. D is any point in the perpendicular AF from A to the side BC of the triangle ABC. If DE is perpendicular to the plane ABC, and GB pas.ainff through E is parallel to BC, then, AE is perpendicular to GH.
SOLID GEOMETRY. — BOOK VII.
263
Proposition XV III
484. 1. Represent two intersecting planes each perpendicular to a third plane. What is the direction of their intersection with reference to the third plane ?
2. What is the direction of the third plane with reference to the inter- section of the other two planes V
3. Represent two planes perpendicular to each other and another plane perpendicular to each of them. What is the direction of the inter- section of any two of these planes with reference to the third plane? What is the direction of each intersection with reference to the other two intersections?
Theorem, If two intersecting planes are each perpendic- ular to a third plane, their intersection is perpendicular to the third plane.
Data : Any two planes, as PQ and RS, intersecting in EF, and perpendicular to a third plane,
as MN.
To prove
EF perpendicular to MN.
Proof. At F, the point common to the three planes, erect a perpendicular to the plane MN.
By § 481, thjs perpendicular lies in both PQ and i?/S, and hence must coincide with their intersection EF.
Consequently, EF is perpendicular to MN.
Therefore, etc. q.e.d.
485. Cor. I. A plane perpendicular to each of two intersecting planes is perpendicular to their intersection.
486. Cor. II. If a plane is perpendicular to two planes which are perpendicular to each other, the intersection of any two of these planes is perpendicular to the third plane, and ea/ih of the three intersections is perpendicular to the other two.
• Ex. 762. If from a point within a dihedral angle perpendiculars are drawn to its faces, the angle contained by these perpendiculars is equal to the plane angle of the adjacent dihedral angle formed by producing one of the planes.
264
SOLID GEOMETRY. — BOOK VII.
Proposition XIX
487. Represent a straight line oblique to a plane. How many pUues can be passed through that line perpendicular to the given plane ?
Theorem. Through any straight line not perpendicular to a plane, one plane perpendicular to that plane can he passed, and only one.
Data: Any plane, as MN, and any straight line not perpendicular to MN, as CD.
To prove that through CD one plane perpendicular to MN can be passed, and only one.
Proof. From any point of CD, as E, draw EF ± MN.
Through CD and EF pass a plane, as GD, intersecting MN.
Then, § 482, GD is perpendicular to MN.
Now, if any other plane perpendicular to MN could be passed through CD, § 484, CD would be perpendicular to MN.
But, data, CD is not perpendicular to MN;
hence, only one plane perpendicular to MN can be passed through CD.
Therefore, etc. q.e.d.
Proposition XX
488. Represent two intersecting planes forming a dihedral angle, and a third plane bisecting this angle ; select any point in the bisecting plane. How do the distances of the point from the faces of the angle compare V
Theorem. Every point in the plane which bisects a dihedral angle is equidistant from the faces of the angle.
Data: Any dihedral angle, as C-AB-D ; the plane bisecting it, as AK', and any point in AK, as H. ^
To prove U equidistant from the faces CB and AD.
SOLID GEOMETRY, — BOOK VII.
265
Proof. Draw HE and HF, the perpendiculars from H to CB and
AD respectively, and through, them pass a plane intersecting the planes CB, AD, and AK in the lines EG, FG, and HG respectively.
Then, § 482, plane EGFHl. CB and AD ; .-. § 485, EGFHl. AB.
Hence, § 443, EG, FG, and HG are perpendicular to AB.
Data, C-AB-K=D-AB-K'j
hence, § 479, Z EGH = Z FGH.
In rt. A EGH and FGH, GH is common, and ZEGH=/.FGH'y
AEGH=AFGH, Why?
and HE=HF',
that is, § 440, H is equidistant from the faces CB and AD.
Therefore, etc. q.e.d.
Proposition XXI
489. Represent a line oblique to a given plane and represent its pro- jection upon that plane. How does the acute angle formed by the line with its projection compare in size with the angle which the line makes with any other line of the plane ?
Theorem, The acute angle formed by a line and its pro- jection upon a plane is the least angle which the line makes with any line of the plane.
Data: Any plane, as MN-, any line, as CD, meeting MN in C; the projection of CD upon MN, as CE; and any other line drawn in MN through C, as CF.
To prove Z DCE less than Z DGF.
Proof. Take CG = GE, and draw DG and DE.
In A CED and CGD, CD is common,
CE = CG, and ED <GD', Why ?
hence, § 130, Z DCE is less than Z DCF.
Therefore, etc q.e.d.
266 SOLID GEOMETRY. — BOOK VII.
Proposition ZXII
490. Represent two straight lines in different planes and a common perpendicular to them. How many such perpendiculars can there be ?
Theorem, Between two straight lines, not in the same plane, one common perpendicular can be drawn, and only one.
c, -^
Data : Any two straight lines |
not in the same plane, as AB !
and CD. /] ' ! V" I^r'T"^
To prove that one common / |
perpendicular can be drawn to / ^__S/k '
AB and CJ5, and only one. / [a_j
Proof. 1. Through any point, as K, in AB draw LK II CB^ and let MN be the plane of LK and AB ; through CD pass a plane, as QD, perpendicular to MN, intersecting MN in GB. and AB in H.
Then, § 457, CD 11 MN and, § 458, GH II CD.
In the plane GD draw ED ± GH.
Then, HdLcd, Why?
and, § 480, HD A. MN;
/. §443, HDA.AB.
2. Now, suppose that any other line, as JK, is perpendicular to AB and CD.
In MN draw KL II CD, and in GD draw JE± QH.
Then, hyp. and § 72, JK±KL; hence, § 442, JK±3IN.
But, §480, JEA.MN.
Hence there are two perpendiculars from J to MN', but, § 456, this is impossible.
Hence, JK is not perpendicular to AB and CD, and HD is the only perpendicular common to those lines.
Therefore, etc. q.e.d.
Ex. 763. The shortest line that can be drawn between two straight lines not in the same plane is their common perpendicular.
SOLID GEOMETRY.^ BOOK VIL
26?
POLYHEDRAL ANGLES
491. The angle formed by three or more planes which meet at but one point is called a Polyhedral Angle, or Polyhedral.
The point in which the planes meet is called the vertex; the intersections of the planes are called the edges; the portions of the planes incliided between the edges are called the /aces; and the angles formed by the edges are called the face angles of the polyhedral angle.
In the polyhedral angle Q-ABCD, Q is the vertex; QAB, QBC, etc., are the faces; QA, QB, etc., are the edges; and angles AQB, BQO^ etc. , are the face angles.
492. The faces of a polyhedral angle are of indefinite extent, but for convenience they are represented as limited by an inter- secting plane called the Base.
ABCD is the base of Q-ABGD.
493. A polyhedral angle whose base is a convex polygon is called a Convex Polyhedral Angle.
494. Polyhedral angles which have their face angles and their dihedral angles equal, each to each, and arranged in the same order, are equal, for they can be made to coincide. Polyhedral angles which have their face angles and their dihedral angles equal, each to each, and arranged in reverse order, are symmetrical, but not generally equal.
The trihedral angles Q-ABG and Q'-A'B'C are symmetrical, when the face angles AQB, BQC, CQA are equal respectively to the face angles A'Q'B', B'Q'G'^ G'Q'A', and the dihedral angles QA, QB, QG are equal respec- tively to the dihedral angles Q'A', Q'B',Q'G>.
Two symmetrical polyhedral angles cannot, generally, be made to coincide.
268 SOLID GEOMETRY. — BOOK VI I.
495. If the edges of one of two polyhedral angles are pro- longations of oho edges of the other through their common ver- tex the angles are called Vertical Polyhedral Angles.
496. A polyhedral angle having three faces is called a trihedral angle; one having four faces is called a tetrahedral angle; etc.
Proposition XXIII
497. Represent two vertical polyhedral angles. How do the face angles of one compare with the face angles of the other? How do the dihedral angles of one compare with the dihedral angles of the other? Are they arranged in the same or in a reverse order in the two polyhe- drals ? What name is given to such polyhedral angles ?
Theorem. Two vertical polyhedral angles are symrnet' rical.
B'
Data: Any two vertical polyhedral angles, as Q-ABGD and Q-A'B'CfD'. To prove Q-ABCD and Q-A^B^CfD' symmetrical.
B
c
Proof. § 59, face Z AQB = face Z A^QB^, and face Z BQC, etc. = face Z b'qc', etc., respectively.
§§ 473, 477, dihedral Zqa = dihedral Z QA', and dihedral Z QB, etc. = dihedral Z QB', etc., respectively.
But the face and dihedral angles of Q-ABCD are arranged in an order which is the reverse of the equal face and dihedral angles of Q-A'B'C'D'.
Hence, § 494, Q-ABCD and Q-A'b'&D' are symmetrical.
Therefore, etc. q.e.d.
Ex. 764. A plane can be passed perpendicular to only one edge and only two faces of a polyhedral angle.
Ex. 765. Every point within a dihedral and equidistant from its faces lies in the plane which bisects that dihedral.
Ex. 766. The sides of an isosceles triangle are equally inclined to any plane through its base.
SOLID GEOMETRY. ^ BOOK VIL 269
Proposition XXIV
498. Represent a trihedral angle. How does the sum of any two of
its face angles compare with the third face angle ?
Theorem, The sum of any two face angles of a tri- hedral angle is greater than the third face angle.
The theorem requires proof only when the q
third angle is greater than each of the others. /K
Data: Any trihedral angle, as Q-ABC, hav- / y\
ing one face angle, as AQC, greater than either / \\\ '
of the other face angles. / \ \ \
To prove i ^^^^^^ \ /\
Zaqb-^Zbqc greater than ZAQa ^b\
Proof. In the face AQC draw QD, making Zaqd = ZAQB; through any point, as n, of QD draw ADO in the plane AQC; take QB = QD, and through line AC and point B pass a plane.
Then, Aaqb = Aaqd, and AB = AD, Why ?
In A ABC, AB-^BC>AD-^ D0\ Why ?
but AB=zAB',
.'. Ax. 5, BC>DC.
In A bqc and DQC, qb = QD,
QC is common, and BC>DC;
Z BQC is greater than Z DQC. Why ?
Const., ZaQB = ZaqD;
.-. Ax. 4, Z AQB + Z BQC is greater than Z AQD 4- Z DQC, or Z AQB + Z BQC is greater than Z AQC.
Therefore, etc. q.e.d.
Proposition XXV
499. Represent any convex polyhedral angle ; around some point in a plane as a common vertex construct in succession angles equal to the face angles of this polyhedral. How do^s their Sum compare with four right angles?
270 SOLID GEOMETRY. — BOOK VIL
Theorem, The sum of the face angles of any convex polyhedral angle is less than four right angles.
Q
Datum : Any convex polyhedral angle, /// 1 \
as Q. , / // 1 \
To prove that the sum of the face angles / 'e\ \
of Q is less than four right angles. /---'/ '1 -- \
\c
Proof. Pass a plane intersecting the edges of Q in A, B, c, etc.
Then, ABODE is a convex polygon.
From 0, any point within ABODE, draw OA, OB, 00, etc.
The number of triangles whose common vertex is Q equals the number whose common vertex is O.
Hence, the sum of the angles of the triangles whose vertex is Q equals the sum of the angles of the triangles whose vertex is O.
But in the trihedral angles whose vertices are A, B, O, etc., § 498, Z QBA + Z QBO is greater than Z ABO, or Z ABO -f Z OBO, and Z QOB + Z QCD is greater than Z BOD, or Z BOO + Z i)(70.
Hence, reasoning in a similar manner regarding the other base angles of the triangles, the sum of the base angles of all the triangles whose vertex is Q is greater than the sum of the base angles of the triangles whose vertex is O.
Therefore, the sum of the face angles at Q is less than the sum of the angles at 0. Why ?
But the sum of the angles at 0 equals four right angles.
Hence, the sum of the face angles of Q is less than four right angles.
Therefore, etc. q.e.d.
Proposition XXVI
500. Represent two trihedral angles having the three face angles of one equal respectively to the three face angles of the other. How do the trihedrals compare? Can there be two trihedrals which fulfill the same conditions and yet not be equal? What name is given to such trihedrals?
SOLID GEOMETRY.— BOOK VTL
271
Theorem. Two trihedral angles are either equal or sym- metrical, if the three face angles of one are equal to th^ three face angles of the other, each to each.
C C
Data : Any two trihedral angles, as Q and Q', having the face angles AQB, BQC, AQC equal to the face angles A'Q'b', b'q'O', A'Q'&y each to each.
To prove Q either equal or symmetrical to Q'.
Proof. On the edges of Q and Q' take the equal distances QAj QB, QC, Q'A', Q'B', Q'C', and draw AB, BC, AC, A'B', B'C', A'C'.
Then, A QAB, QBC, QAC are equal to A Q'a'b', Q'b'c', Q'A'&y each to each. Why ?
Hence, AABO = Aa'b'c', and Z BAC= Z b'a'c'. Why?
On the edge QA take AD and on Q'A' take A'n' = AD. At D and D' construct the plane A HDK and H'd'k' of the dihedrals QA and Q'A' respectively, the sides meeting AB, AC, A'b', and A'c' as at H, K, h' and E^ respectively, inasmuch as A QAB, QAC, etc., are acute, being angles of isosceles A QAB, etc. Draw HE and H'K',
Then, const., AD = A'd',
and Z DAH =ZD'A'n']
rt. AADH= rt. AA'd'h', and AH= A'H* ; similarly, AK= A'k',
and, since Z BAC = Z b'a'& ;
A AHK = A A'H'K', and HK = H'K' ; but DH = D'H", and DK^d'e*',
A HDK = A H'D'K*, and Z HDK = Z H'D'K*. Hence, § 477, dihedral ZQA = dihedral Z Q'A',
Why . Why?
272 SOLID GEOMETRY. — BOOK VIL
In like manner it may be shown that the dihedral angles QB and QC are equal to the dihedral angles Q'^' and Q^& respectively.
Hence, § 494, if the equal angles are arranged in the same order, as in the first two figures, the two trihedral angles are equal; but if they are arranged in the reverse order, as in the first and third figures, the two trihedral angles are symmetrical.
Therefore, etc. q.e.d.
501. Cor. If two trihedral angles have three face angles of the one equal to three face angles of the other, then the dihedral angles of the one are respectively equal to the dihedral angles of the other.
SUPPLEMENTARY EXERCISES
Ex. 767. If a straight line is parallel to a plane, any plane perpendicular to the line is perpendicular to the plane.
Ex. 768. If a straight line intersects two parallel planes it makes equal angles with them.
Ex. 769. If a line is parallel to each of two planes, the intersections which any plane passing through it makes with the planes are parallel.
Ex. 770. The projections of parallel straight lines on any plane are either parallel or coincident.
Ex. 771. Find the locus of points which are equidistant from three given points not in the same straight line.
Ex. 772. From any point within the dihedral angle A-BC-D, EF and EG are drawn perpendicular to the faces AC and BD, respectively, and GH perpendicular to ^O at H. Prove that FH is perpendicular to BG.
Ex. 773. If a plane is passed through the middle point of the common perpendicular to two straight lines in space, and parallel to both lines, it bisects every straight line drawn from any point in one line to any point in ihe other line.
Ex. 774. If the intersections of several planes are parallel, the perpen- diculars drawn to them from any point lie in one plane.
Ex. 775. If two face angles of a trihedral are equal, the dihedral angles opposite them are also equal.
Ex. 776. A trihedral angle, having two of its dihedral angles equal, may be made to coincide with its symmetrical trihedral angle.
Ex. 777. In any trihedral the three planes bisecting the three dihedrals intersect in the same straight line.
Ex. 778. In any trihedral the planes which bisect the three face angles, and are perpendicular to those faces, respectively, intersect in the same straight line.
BOOK VIII
POLYHEDRONS
502. A solid bounded by planes is called a Polyhedron.
The intersections of the planes which bound a polyhedron are called its edges; the intersections of the edges are called its vertices; and the portions of the planes included by its edges are called its faces.
The line joining any two vertices of a polyhedron, not in the same face, is called a diagonal of the polyhedron.
503. A polyhedron having four faces is called a tetrahedron; one having six faces is called a hexahedron; one having eight faces is called an octahedron; one having twelve faces is called a dodecahedron; one having twenty faces is called an icosahedron.
504. If the section made by any plane cutting a polyhedron is a convex polygon, the solid is called a Convex Polyhedron.
Only convex polyhedrons are considered in this work.
PRISMS
505. A polyhedron two of whose faces are equal polygons, which lie in parallel planes and have their homologous sides parallel, and whose other faces are parallelograms, is called a Prism.
The two equal and parallel faces of the prism are called its bases; the other faces are called lat- eral faces; the intersections of the lateral faces are called lateral edges; the sum of the lateral faces is called the lat- eral, or convex surface ; and the sum of the areas of the lateral faces is called the lateral area of the prism.
The lateral edges of a prism are parallel and equal. § 153.
The perpendicular distance between the bases of a prism is its altitude.
milnb's geom. — 18 273
274
SOLID GEOMETRY. — BOOK VIII.
506. A prism is called triangular, quadrangular, hexagonal, etc., according as its bases are triangles, quadrilaterals, hexagons, etc.
507. A prism whose lateral edges are perpendicu- lar to its bases is called a Right Prism.
508. A prism whose lateral edges are not perpendicular to its bases is called an Oblique Prism.
509. A right prism whose bases are regular polygons is called a Regular Prism.
510. A section of a prism made by a plane perpendicular to its lateral edges is called a Right Section.
511. The part of a prism included between one base and a section made by a plane oblique to that base, and cutting all the lateral edges, is called a Truncated Prism.
512. A prism whose bases are parallelograms is called a Parallelepiped.
513. A parallelopiped whose lateral edges are perpendicular to its bases is called a Right Paral- lelopiped.
514. A parallelopiped all six of whose faces are rectangles is called a Rectangular Parallelopiped.
515. A parallelopiped whose six faces are all squares is called a Cube.
516. The quantity of space inclosed by the surfaces which bound a solid is called the Volume of the solid.
A solid is measured by finding how many times it contains some other solid adopted as the unit of measure.
The units of measure for volume are the cubic inch, the cubic foot, the cubic yard, the cubic centimeter, the cubic decimeter, etc.
SOLID GEOMETRY. — BOOK VIII.
275
Suppose that the cube M is the unit of measure and that AB is the rectan- gular parallelopiped to be measured.
Apply an edge qf M to each edge of AB and at the points of division pass - planes respectively perpendicular to those edges. These planes divide AB into cubes, each equal to the unit M.
It is evident that there will be as many layers of these cubes as the edge of M is contained times in the altitude of AB, that each layer will contain as many rows of cubes as the edge ot M is contained times in the width of AB, and that each row will con- tain as many cubes as the edge of M is contained times in the length of ^5; and, therefore, that the product of the numerical measures of the three dimensions of AB is equal to the number of times that M is contained in AB.
In this case the edge of *M is contained 4 times in BE, 3 times in DA, and 5 times in DC; consequently, there are 5 cubes in each cow, 3 rows in each layer, and 4 layers in the parallelopiped ; that ts, M is contained in AB 5 x 3 x 4 = 60 times, or the rectangular parallelopiped contains 60 cubic units.
Therefore, if the edge of 3f is a common unit of measure of the three dimensions of a rectangular parallelopiped, the product of the numerical measures of the three dimensions expresses the num- ber of times that the rectangular parallelopiped contains the cube, and is the numerical measure of the volume of the rectangular parallelopiped.
517. For the sake of brevity, the product of the three dimensions is used instead of the product of the mimerical measures of the three dimensioris.
The product of three lines is, strictly speaking, an absurdity, but since the expression is used to denote the volume of a rectan- gular parallelopiped, it follows that the geometrical concept of the product of three lines is the rectangular parallelopiped whose edges they are.
Thus, DC X DA X DE implies a product, which is a numerical result, but it must be interpreted geometrically to mean the rectan- gular parallelopiped whose edges are DC, DA, and DE.
276 SOLID GEOMETRY. — BOOK VIII.
For similar reasons, the cube of a line must be interpreted geo- metrically to mean the cube constructed upon the line as an edge, and conversely, the cube constructed upon a line may be indicated by the cube of the line.
518. Solids which have the same form are similar; those which have the same volume are equivalent; and those which have the same form and volume are equal.
Proposition I
519. 1. Form * a prism. Since the faces are parallelograms, how does each face coi'npare with a rectangle having the same base and altitude ? Considering a lateral edge as the base of each, how does the sum of the alti- tudes compare with the perimeter of the right section ? Since the lateral edges are equal, how does the lateral surface of a prism compare with the rectangle of its lateral edge and the perimeter of a right section?
2. To what rectangle is the lateral surface of a right prism equivalent?
Theorem, The lateral surface of a prism is equivalent to the rectangle formed hy a lateral edge and the perime- ter of a right section.
Data : Any prism, as AD\ of which AA^ is a lateral edge, and FGHJK any right section. To prove lateral surface of
AD' ^ rect AA' . (FG -i-GH-}- etc.).
Proof. § 505, AB', BCf, cd\ etc., are parallelograms, data, FGHJK is a right section ;
.-. § 443, FG A. AA', GH± BE', HJ A. CC', etc.
Now the lateral surface oi AD' ^ AB' + BC' + etc. ;
* Objective representations of the solids referred to in this and the follow- ing books will aid the student very greatly in acquiring the correct geomet- rical concepts. Solids made from wood or glass may be procured, but it will be far better for the student to form them for himself from some plastic mate- rial, like clay or putty. He can then cut them readily in any desired manner with a thin-bladed knife.
SOLID GEOMETRY.— BOOK VIII. 277
but, § 331, AB' =0= rect. AA' • FO^
BC' ^ rect. BB' - GH, etc.; AB' -\-B(f + etc. =0= rect. AA' - FG-\- rect. BB' • GH + etc., or lat. surf, of AD' ^ rect. aa' • FG + rect. BB' • GH+ etc.
But, § 505, AA' = £^' = cc' = etc.
Hence, lat. surf, of AD' =c= rect. AA' - (FG -{- GH-\- etc.). q.e.d.
520. Cor. The lateral surface of a right prism is equivalent to the rectangle formed by its altitude and the perimeter of its base.
Arithmetical Rules : To be framed by the student. § 339
Proposition II
521. 1. Form a prism; cut it by parallel planes. What figures are the sections made by these planes? How do they compare?
2. How does any section of a prism parallel to the base compare with the base ?
3. How do all right sections of the same prism compare ?
Theorem, The sections of a prism made hy parallel planes are equal polygons.
Data: Any prism, as PR, cut by any par- allel planes, as AD and FJ, making the sec- tions ABODE and FGHJK.
To prove ABODE = FGHJK.
Proof. § 463, AB II FG, BO II GH, etc.;
.-. § 469, Z ABO = Z FGH, Z BOD = Z GHJj etc.
Also, § 151, AB = FG, BO = GH, etc.
Then, ABODE and FGHJK are mutually equiangular and equi- lateral, and one can be applied to the other so that they will ex- actly coincide.
Hence, § 36, ABODE = FGHJK. q.e.d.
522. Cor. I. Any section of a prism parallel to the base is equal to the base.
523. Cor. II. All right sections of the same prism are equals
278
SOLID GEOMETRY. — BOOK VIII.
Proposition III
524. 1. Form two prisms such that three faces including a trihedral angle of one are equal to the corresponding faces of the other and simi- larly placed in each prism. How do the prisms compare ?
2. Form two truncated prisms such that three faces including a tri- hedral angle of one are equal to the corresponding faces of the other and similarly placed in each prism. How do the prisms compare ?
3. Form two right prisms having equal bases and equal altitudes. How do they compare?
Theorem, Two prisms are equal, if three faces includ- ing a trihedral angle of one are equal to three faces in- cluding a trihedral angle of the other, each to each, and these faxes are similarly placed.
Data : Any two prisms, as AJ and A' J', having the faces AG, AD, and AK equal to the faces A'G\ A'n', and A'K', each to each, and similarly placed.
To prove AJ=A'J'. ^*
B C B' C
Proof. Data, the face angles BAE, BAF, and EAF are equal to the face angles B'A'e', B'a'f', and E'a'f', respectively;
.-. § 500, trihedral angle ^-5^J^ = trihedral angle A'-B'e'f'.
Apply prism A'J^ to AJ so that the faces of trihedral Z A' shall coincide with the equal faces of the trihedral Z A.
Then, the points O' and D' fall upon C and B, respectively, and, § 505, C'h' and n'j^ take the direction of CH and DJ, respectively.
Since the points F', G', K* coincide with F, G, K, respectively, § 430, the planes of the upper bases must coincide.
Then, H^ coincides with H, and J' with J.
Hence, the prisms AJ and A^J^ coincide in all their parts ; that is, AJ=A'j'. Q.E.i>.
525. Cor. I. Tico tmncated prisms are equal, if three faces in- cluding a trihedral angle of one are equal to three faces including a trihedral angle of the other, each to each, and these faces are simi- larly placed.
SOLID GEOMETRY. — BOOK VIII. 279
526. Cor. II. Two right prisms are equal, if they have equal bases aiid equal altitudes.
Proposition IV
527. Form any oblique prism ; on a base equal to a right section of the oblique prism form a right prism whose altitude is equal to a lateral edge of the oblique prism. How do these prisms compare in volume ?
Theorem, An oblique prism is equivalent to a right prism which has its base equal to a right section of the oblique prism, and its altitude equal to a lateral edge of the oblique prisma.
Data: Any oblique prism, as AD^ ^ a right sec- tion of it, as FGHJK] and' a lateral edge, as AA'.
To prove AD^ equivalent to a right prism whose base is FGHJK and altitude equal to AA\
Proof. Produce AA^ to F^ making FF^ = AA', and at F' pass a plane perpendicular to FF' cut- ting all the faces of AD' produced and forming the right section f'G'h'j'k' parallel to FGHJK. b c
Then, § 521, section F'G'h'j'k' = section FGHJK, and FJ' is a right prism whose base is FGHJK and altitude equal to AA'.
In the truncated prisms AJ and A'j*, § 505, the bases AD and a'd' are equal.
Const., AA' = FF', and BB' = GG' -,
.-. Ax. 3, AF=A'F', and BG = B'g'.
AB and FG are equal and parallel to A'b' and F'g', respectively; and A FAB, ABG, etc., of the face AG are equal respectively to Af'a'b', A'b'G', etc., of the face A'g'-, ^ Why?
.-. AG and A'g' are mutually equiangular and equilateral, and one can be applied to the other so that they will exactly coincide.
Hence, § 36, AG = A'G'.
In like manner AK may be proved equal to A'K*.
Hence, § 525, prism AJ = prism A'J^.
Adding to each the prism FD', then^ AD'=o=FJ'. 9.E.D.
280
SOLID GEOMETRY. — BOOK VIII.
Proposition V
528. Form any parallelepiped. How do the opposite faces compare ? In what direction do they extend with reference to each other?
Theoretn. The opposite fdces of a parallelopiped are equal and parallel.
Data : Any parallelopiped, as AG, and any opposite faces of AG, 2i^ AF and DG.
To prove AF and DG equal and parallel.
Proof. AB il DC, and BF II CG-, Why ? .-. § 469, Zabf=Zdcg.
Also, AB = DC, and BF=CG', Why ?
AF=zDG, and, § 469, AF II DG.
Why?
Q.E.D.
Proposition VI
629. 1. Form any parallelopiped; pass planes through any three pairs of diagonally opposite edges. What plane figures are formed by these edges and the intersections of these planes with the faces of the parallelopiped? How do the diagonals of these parallelograms corre- spond with the diagonals of the parallelopiped? How do the segments of each diagonal compare in length ?
Theorem. The diagonals of a, parallelopiped bisect each other.
Data : Any parallelopiped, as AG, whose diagonals are AG, BH, CE, and DF.
To prove that AG, BH, CE, and DF bisect each other.
Proof. Through the opposite edges AE and CG pass a plane. § 505, AE and CG are equal and parallel ;
ACGE is a parallelogram ; and, § 154, diagonals AG and CE bisect each other at 0.
In like manner, AG, BH, and AG, DF also bisect each other at 0.
Hence, AG, BH, CE, and DF bisect each other. q.e.d.
530. Cor. The diagonals of a rectangtdar parallelopiped are epial.
SOLID GEOMETRY.— BOOK VIIL
281
Proposition VII
531. 1. Form two rectangular parallelopipeds whose bases are equal and whose altitudes are in the ratio of 2:3, or any other ratio. How does the ratio of their volumes compare with the ratio of their altitudes ?
2. How does the ratio of two rectangular parallelopipeds having two dimensions in common compare with the ratio of their third dimensions?
Theorem. Rectangular parallelopipeds which have equal bases are to each other as their altitudes.
T>A
A " B
Data: Any two rectangular parallelopipeds, as ^ and B, whose bases are equal and whose altitudes are CD and EF respectively.
To prove A:B=CD:BF.
Proof. Suppose the altitudes CD and BF have a common measure which is contained in CD 3 times and in BF 5 times.
Then,
CD:BF=3:5.
I
Divide CD into three and BF into five equal parts by applying this common measure to them, and through the several points of division pass planes perpendicular to these lines.
§ 462, these planes are parallel to each other and to the bases of A and B ;
.-. §§ 523, 526, A is divided by these parallel planes into three, and B into five equal rectangular parallelopipeds ; ^:^ = 3:5.
Hence, A: B= CDiBF.
By the method of limits exemplified in § 327 the same may be proved when the altitudes are incommensurable.
Therefore, etc. q.e.d.
532. Cor. Rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions.
282
SOLID GEOMETRY. — BOOK VIII.
Proposition VIII
533. 1. Form two rectangular parallelepipeds whose altitudes are equal and the areas of whose bases are in the ratio of 2:3, or any other ratio. How does the ratio of their volumes compare with the ratio of their bases ?
2. If two rectangular parallelepipeds have one dimension in common, how does the ratio of their volumes compare with the ratio of the prod- ucts of their other two dimensions?
Theorem, Rectangular parallelopipeds which have equal altitudes are to each other as their bases.
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Data : Any two rectangular parallelepipeds, as A and 5, which have a common altitude, as ^, and the dimensions of whose bases are d, e, and m, n, respectively.
To prove
A:Bz=d X e:m X
Proof. Construct a third rectangular parallelopiped C, having the altitude ^, and the dimensions of its base d and n.
Then, § 532, and hence, § 287,
Therefore, etc.
A'.C=e:ny C:B = d:m; A:B=:d xe:m xn.
Q.B.J).
534. Cor. Rectangular parallelopipeds which have one dimension in common are to each other as the products of their other two dimen- sions.
SOLID GEOMETRY. — BOOK VIII.
283
Proposition IX
535. Form any two rectangular parallelopipeds, and also a third one whose base is equal to the base of the first and whose altitude is equal to that of the second. By comparing each of the first two with the third, discover how the ratio of their volumes compares with the ratio of the products of their three dimensions.
Theorem, Rectangular parallelopipeds are to each other as the products of their three dimensions.
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Data : Any two rectangular parallelopipeds, as A and B, whose dimensions are d, e, /, and I, m, n, respectively.
To prove A:B = dxexf:lxmxn.
Proof. Construct a third rectangular parallelopiped C, having the dimensions d, e, and n.
Then, § 532, A'.C = f:n,
and, § 534, C : B = d x e:l x m-^
hence, § 287, A.B = dxexf'.lxmxn.
Therefore, etc. q.e.d.
Proposition X
536. Form any rectangular parallelopiped, and a cube, whose edge is some linear unit, as a unit of measure. Since the ratio of these solids is equal to the ratio of the products of their three dimensions, find the measure of the volume of the parallelopiped in terras of its three dimen- sions.
Theorem. The volume of a rectangular parallelopiped is equal to the product of its three dimensions.
284
SOLID GEOMETRY.— -BOOK VIII,
Data: Any rectangular parallelo- piped, as A, whose dimensions are d, e, and /.
To prove
volume oi A = dxe xf.
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Proof. Assume that the unit of volume is a cube, if, whose edge is the linear unit.
Then, § 535, ^ :3f = (Z x e x/: 1 X 1 X 1,
or
JJf 1 xl xl "^
But, § 516, the volume of A is measured by the number of times it contains the unit of measure, M\
M
= volume of A,
But
~=dxex/
Hence, volume ot^ = dxex/ Therefore, etc.
Q.E.D.
537. Cor. Tlie volume of a rectangular paraUelopiped is equal to the product of its base by its altitude.
Proposition XI
538. 1. Form any oblique parallelepiped. How does it compare in volume with a rectangular parallelepiped having an equivalent base and the same altitude ?
2. What, then, is the measure of the volume of any parallelepiped in terras of its base and altitude ?
Theorem. Any parallelepiped is equivalent to a rec- tangular parallelepiped having the same altitude and an equivalent base.
SOLID GEOMETRY. — BOOK VIII.
285
Data : Any parallelepiped, as A&, whose base is ABCD. To prove AC^ equivalent to a rectangular parallelopiped having the same altitude and a base equivalent to ABCD.
Proof. On AB produced take EF equal to AB, and through E and F pass planes _L EF, as EH^ and FG'. Produce the faces AC, a'g', AB', and DC' to intersect the planes eh' and FG', forming the right parallelopiped EG'.
Then, § 527, AC' =0= EG'.
On HE produced take MJ equal to HE, and through M and J pass planes ^MJ, as ML' and Jit. Produce the faces EG, E'g', eh', and FG' to intersect the planes ML' and JK', forming* the right parallelopiped JL'.
Then, § 527, EG'=(>JL';
AC'oJL'.
Const., EF = AB, and AF II i)G! ;
.-.§333, EFGHo^ABCD.
Also, const., ifJ" = HE, and ^J II GS';
JELM o ^i^^G^Zf =0= ^5C2). Why ?
Const., plane ^'g'j' II plane AGJ, and hence the three solids have the same altitude.
Const., § 482, faces EH^ and FG' are perpendicular to AGJ-, hence, faces JM^ and ElJ are perpendicular to AGJ.
Also, const., § 482, faces ML' and Jit are perpendicular to AGJ-, the faces of JlJ are rectangles; hence, § 514, JL' is a rectangular para-llelopiped.
But AC' =0= JL', and JKLM =c= ^5CZ>.
286 SOLID GEOMETRY.— BOOK VIII.
Hence, ACf is equivalent to a rectangular parallelopiped having the same altitude and a base equivalent to ABCD.
Therefore, etc. q.e.d.
539. Cor. The volume of any parallelopiped is equal to the product of its base by its altitude.
Proposition XII
540. Form a parallelopiped ; divide it into two triangular prisms by a plane passing through two diagonally opposite edges. How do these prisms compare in volume? What, then, is the volume of a triangular prism in terms of its base and altitude ?
Form any prism ; divide it into triangular prisms by planes through a lateral edgCo What is the volume of each triangular prism? What, then, is the volume of any prism ?
Theorem. The plane passed through two diagonally op- posite edges of a parallelopiped divides it into two equiva- lent triangular prisms.
H — ^
Data: Any parallelopiped, as AG, and ^(^i^ ^ P'/
a plane, as ACGE, passing through two AuzA- ---riiiJr.Ji
diagonally opposite edges, as AE and GG. I j ~^^d/^
To prove prism ABC-F =o prism ADC-H. / dI -L-Jo
A B
Proof. Through the parallelopiped AG pass a plane forming the right section JKLM and intersecting ACGE in JL.
§ 528, AF II DG, and AH II 5G ;
hence, § 463, JK II ML, and JM II KL ;
JKLM is a parallelogram, and JL is its diagonal.
Hence, A JKL = A JML. Why ?
Now, § 527, prism ABC-F is equivalent to a right prism whose base is JKL, and whose altitude is AE;
also prism ADC-H is equivalent to a right prism whose base is JML, and whose altitude is AE.
But, § 526, these two right prisms are equal ; hence, ABC-F =0= ADC-H.
Therefore, etc. q.b.d.
SOLID GEOMETRY. — BOOK VIII. 287
541. Cor. I. A triangular prism is equivalent to one half of a paralldopiped having the same altitude and a base twice as great.
542. Cor. II. The volume of a triangular prism is equal to the product of its base by its altitude. ^^<^^
543. Cor. III. The volume of any prism is equal / //' j to the product of its base by its altitude. j j / I
544. Cor. IV. Prisms are to each other as the 1 1-^- J products of their bases by their altitudes; conse- \f '~~'v quently, prisms which have equivalent bases are to each other as their altitudes; prisms which have equal altitudes are to each other as their bases; and prisms which have equivalent bases and equal altitudes are equivalent.
PYRAMIDS
545. A, polyhedron whose base is a polygon, and whose lateral faces are triangles which have a common vertex, is
called a Pyramid. ^
. The common vertex of the triangles which form the lateral faces of a pyramid is called the vertex / of the pyramid; the lines in which the lateral / face?^ intersect are called the lateral edges; the sum V
of the lateral faces is called the lateral, or convex
surface; and the sum of the areas of the lateral faces is called the lateral area of the pyramid.
The perpendicular distance from the vertex of a pyramid to the plane of its base is its altitude.
546. A pyramid is called triangular, quadrangular, etc., accord- ing as its base is a triangle, quadrilateral, etc. ^
547. A pyramid whose base is a regular polygon, and whose vertex lies in the perpendicular to the base erected at its center, is called a Regular Pyra- mid. I
Since a regular polygon may be inscribed in a ^ ^^
circle, it is evident, from § 450, that the vertex of a regular pyrar mid is equidistant from the vertices of the polygon forming its
288 SOLID GEOMETRY.— BOOK VIII.
base ; and hence the lateral edges of a regular pyramid are equal, and its lateral faces are equal isosceles triangles.
548. The perpendicular distance from the vertex of a regular pyramid to the base of any one of its lateral faces is called the Slant Height of the pyramid.
The slant height is therefore the altitude of the equal isosceles triangles which form the lateral faces of the pyramid.
549. The part of a pyramid included between its base and a plane which cuts all its lateral edges is called a Truncated Pyramid.
550. A truncated pyramid whose bases are par- /T^^f^ \ allel is called a Frustum of a pyramid. / ;. . I A \
The perpendicular distance between the bases of //' I \ a frustum is its altitude. ^^>^^ 1 ^/^
The lateral faces of the frustum of a regular ^^1^
pyramid are equal isosceles trapezoids, and the common altitude of these trapezoids is the slant height of the frustum.
Proposition XIII
551. 1. Form a regular pyramid. Since the lateral faces are triangles, how does each compare with the rectangle having the same base and altitude ? How does the altitude of each compare with the slant height of the pyramid ? How does the sum of the bases of the lateral faces com- pare with the perimeter of the base of the pyramid? How, then, does the lateral surface of a regular pyramid compare with the rectangle of the perimeter of its base and its slant height ?
2. Form the frustum of a regular pyramid. What plane figures are its faces? To what, then, is the surface of each equivalent? How does the sum of the upper bases of the faces compare with the perim- eter of the upper base of the frustum? The sum of the lower bases of the faces with the perimeter of the lower base of the frustum ? How, then, does the lateral surface of the frustum of a regular pyramid com- pare with the rectangle of its slant height and the sum of the perimeters of its bases ?
Theorem, The lateral surface of a regular pyramid is equivalent to one half the rectangle formed hy the perim- eter of its hose and its slant height.
SOLID GEOMETRY. — BOOK VIII. 289
Data: Any regular pyramid, as Q-ABCDE, // '\\
whose slant height is QH. // j \ \
To prove lateral surface of / / 1 [e\ \
Q-ABCDE =G= i rect. (AB + BC + etc.) . QH, ^/' / 7 \ \d
i^ ^
Proof. § 545, lat. surf, of Q-ABCDE ^ A QAB + A QBC -\- etc., and, since, § 548, the altitudes of these triangles = QH,
A QAB ^ i rect. AB • QH, Why ?
A QBC^^ rect. BC • Qfl, etc. ; .-. A QAB + A QBC-\- etc. =0= i rect. AB - QH+^ rect. 5C • Qfl-^- etc., or lat. surf, of Q-ABCDE ^ ^ rect. {AB + BC + etc.) • QH. Therefore, etc. q.e.d.
552. Cor. The lateral surface of a frustum of a regular pyra- mid is equivalent to one half the rectangle formed by its slant height and the sum of the perimeters of its bases.
Arithmetical Rules : To be framed by the student. § 339
Ex. 779. The perimeter of the base of a regular pyramid is 14 ft. and its slant height is 6 ft. What is its lateral area ?
Proposition XIV
553. 1. Form a pyramid ; cut it by a plane parallel to its base. How do the ratios of the segments of the lateral edges compare with each other, and with the ratio of the segments of the altitude?
2. Is the section equal, equivalent, or similar to the base ?
Theorem, If a pyramid is cut hy a plane parallel to its base,
1. The lateral edges and the altitude are divided propor- tionally.
2. The section is a polygon similar to the base,
milnb's geom. — la
290
SOLID GEOMETRY. — BOOK VIIL
Data: Any pyramid, as Q-ABCDE, wtose altitude is QO, and any plane parallel to the base, as MJSf, which cuts the pyramid in the section FGHJK, and the altitude in P.
To prove 1. QF . qa= QG: QB = QP: Q0 = etc. 2. FGHJK and ABCDE similar.
Proof. 1. Through Q pass a plane parallel to ABCDE.
Then, the lateral edges and the altitude are intersected by three parallel planes ; .-. § 470, QF : QA = QG: QB= QP: QO, etc.
2. § 463, FG II AB, GH 11 BC, HJ II CD, etc. ;
. § 469, Z FGH = Z ABC, Z GHJ = Z BCD, etc.
Also, § 306, A QFG, QGH, etc., are similar to A Q^5, QBC, etc., each to each ;
FG:AB=QG: QB, and (?^: BC= QG: QB] hence, FG: AB = GH: BC.
In like manner, GH: BC — HJ: CD, etc.
Hence, FGHJK and ABCDE are mutually equiangular and have their homologous sides proportional ; that is, § 299, FGHJK and ABCDE are similar.
Therefore, etc. q.e.d.
554. Cor. I. Parallel sections of a pyramid are to each other as the squares of their distances from the vertex.
For, § 344, FGHJK : ABCDE = F^ : i? ;
but, § 553, FG : AB = QG: QB = QP : QO;
bence, FGHJK : ABCDE = QF' : off.
SOLID GEOMETRY. — BOOK VIII. 291
655. Cor. II. If two pyramids have equal altitudes, sections 'parallel to their bases and equally distant from their vertices are to each other as the bases.
For, § 554, FGHJK : ABODE = Qp : Q(?, and F'G'H' : A^B^C' = Q^F^ : Q^l
But QP = Q'P', and QO = Q'O' ;
FGHJK : AB ODE = F'G'H' : A'B'&j or FGHJK : F' G'H' = ABODE : A'B'C'. Why ?
556. Cor. III. If two pyramids have equal altitudes and equiva- lent bases, p^.ctions parallel to their bases and equally distant from their vertices are equivalent.
Proposition XV
657. Form two triangular pyramids which have equivalent bases and equal altitudes. How do they compare in volume?
Theorern, Triangular pyramids which have equivalent bases and equal altitudes are equivalent.
Data: Any two triangular pyramids, as Q-ABO and T^DEF, with equivalent bases ABC and DEF, and equal altitudes.
To prove Q-ABO ^ t-def.
Proof. Divide the equal altitudes into equal parts, each n units long, and through the points of division pass planes parallel to ABO and DEF respectively.
By § 556, the corresponding sections of the two pyramids, formed by these planes, are equivalent.
292 SOLID GEOMETRY.— BOOK VIII.
If the pyramids are not equivalent, suppose Q-ABC is the greater. On its base, and on each section as a lower base, construct a prism with lateral edges parallel to ^ Q and altitude equal to n.
B E
On each section of T-def, as an upper base, construct a prism with lateral edges parallel to DT and altitude equal to n.
Then, § 544, each prism in T-DEF is equivalent to the prism next above it in Q-ABC ; consequently, the difference between the two sets of prisms is the lowest prism of the first set.
Now, if ?i is decreased indefinitely, the lowest prism is decreased indefinitely, and the difference between the two sets of prisms may be made less than any assigned volume, however small.
But the sum of all the prisms of the first set is greater than Q-ABC and the sum of all the prisms of the second set is less than T-DEF', therefore, the difference between Q-ABC smd T-DEF is less than the difference between the two sets of prisms, and consequently, less than any assigned volume, however small.
Hence, Q-ABC =o T-DEF.
Therefore, etc. q.e.d.
Proposition XVI
558. 1. Form any triangular prism and divide it into three triangular pyramids. How do the pyramids compare with each other? To what part, then, of the prism is the pyramid, which has the same base and altitude, equivalent ?
2, Form any pyramid ; divide it into triangular pyramids by planes through a lateral edge. What is the volume of each triangular pyramid? What, then, is the vohime of any pyramid?
Theorem.. A triangular pyramid is equivalent to one third of a triangular prism which has tJie same base and altitude.
SOLID GEOMETRY.-^ BOOK VIII.
293
Data: Any triangular pyramid, as Q-ABC, and a triangular prism, as DQE~ABC, which has the same base and altitude.
To prove Q-ABC =0= | D QE-ABC.
Proof. Through QC and QD pass a plane intersecting the parallelogram ACED in CD. Then, D QE-ABC is composed of the three triangular pyramids Q-ABC, Q-ACD, and Q-CDE.
AACD = ACDEy Why?
that is, Q-ACD and Q-CDE have equal bases, and the same altitude ;
.-. § 557, Q-ACD ^ Q-CDE.
Regarding C as the vertex and DQE as the base of Q-CDE, § 505, Q-ABC and C-DQE have equal bases and equal altitudes;
.-. § 557, Q-ABC =0 C-DQE, or Q-CDE',
consequently, Q-ABC =0= Q-ACD =0= Q-CDE.
Hence,
Q-ABC =0= i D QE-ABC.
Q.E.D.
559. Cor. I. The volume of a triangular pyramid is equal to one third the product of its base by its altitude.
560. Cor. II. The volume of any pyramid is equal to one third the product of its base by its altitude.
561. Cor. III. Pyramids are to each other as the products of their bases by their altitudes; con- sequently, pyramids which have equivalent bases
are to each other as their altitudes; pyramids which have equal altitudes are to each other as their bases; and pyramids which have equivalent bases and equal altitudes are equivalent.
294
SOLID GEOMETRY, — BOOK VIIL
Proposition XVII
562. Form the frustum of a triangular pyramid ; divide it into three triangular pyramids one of which shall have for its base the lower base of the frustum, another the upper base, and both the altitude of the frustum for their altitude. It will be shown that the third pyramid is equivalent to a pyramid whose altitude is the altitude of the frustum and whose base is a mean proportional between the bases of the frustum.
To the sum of what triangular pyramids, then, is the frustum of a triangular pyramid equivalent?
Theorem, A frustum of a triangular pyramid is equiv- alent to the sum of three pyramids of the same altitude as the frustum, and whose bases are those of the frustum and a mean proportional between them.
Data : A frustum of any triangular pyramid, as ABC-DEF, whose bases are ABC and DEF. Denote its altitude by H.
To prove ABC-DEF equivalent to the sum of three pyramidc* whose common altitude is H and whose bases are ABC, VEF, ana a mean proportional between them. *
Proof. Through the points A, E, C, and B, E, C pass planes, thus dividing the frustum into the three pyramids E-ABC, C-BEF, and E-ADC.
Then, E-ABC and C-DEF have the bases ABC and DEF, respec- tively, and the common altitude H.
It remains to prove E-ADC equivalent to a pyramid whose alti- tude is H and whose base is a mean proportional between ABC and DEF,
SOLID GEOMETRY. — BOOK VIII. 295
In the face DB, draw EJ II DA, and pass tlie plane EJC-, also draw JD.
§ 457, EJ II plane ACFD ;
hence, iJ and J are equally distant from plane ACFD-,
E-AD C ^ J- AD C. Why ?
Kow, D may be regarded as the vertex, and AJC as the base of J- ADC.
Then, E-ADC is equivalent to D-AJC, whose altitude is H.
Draw JK II ^F.
Then, § 469, Z ^J^ = Z DEF, Z e/^ZT = Z EDF, and, § 151, AJ=DE;
AAJK = Adef, Why?
and AK = DF. Why ?
Now the altitudes of A ^5C and ^JC on AB are equal, and thte altitudes of A ^J'C and AJK on ^c are equal; •. § 336, A ABC : A AJC = AB:AJ=AB: DE,
and AAJC:AAJK=AC:AK=AC:DF,
But, § 553, A ABC and D-E^i^ are similar ;
hence, AB:DE = AC : DF ; Why ?
AABC:AAJC = AAJC: AAJK.
But A AJK = A DEF;
AABG:AAJC = AAJC: ADEF; that is, A ^JC is a mean proportional between A ABC and DEF.
Hence, ABC-DEF is equivalent to the sum of three pyramids, whose common altitude is H and whose bases are ABC, DEF, and a mean proportional between them.
Therefore, etc. q.e.d
563. Cor. I. A frustum of any pyramid /^^v^^^^^-^ is equivalent to the sum of three pyramids of I j ^^\\ the same altitude as the frustum and whose I \ \ bases are those of the frustum and a mean \- {----- \--^\ proportional between them. \j ^^^-^\^y^
564. Cor. II. The volume of a frustum of any pyramid is equal to one third the product of its altitude by the sum of its bases and a mean proportional between them.
296 SOLID GEOMETRY, — BOOK VIIL
Proposition XVIII
565. Form a truncated triangular prism and through one of its upper vertices pass planes dividing it into three triangular pyramids. Since any face of a pyramid may be considered as its base, discover whether each one of these pyramids is equivalent to some one of the three pyra- mids whose common base is the base of the prism and whose vertices are the three vertices of the inclined section.
Theorem, A truncated triangular prism is equivalent to the sum of three pyramids whose comm^on base is the base of the prism, and whose vertices are the three vertices of the inclined section.
Dit
Data : Any truncated triangular prism, as ABC-DEF, and the three pyramids E-ABC, D-ABC, and F-ABC.
To prove ABC-DEF ^ E-ABC -f D-ABC
+ F-ABC. ^^
Proof. Pass planes through E, A, C, and E, D, C, dividing the prism into the pyramids E-ABC, E-ACD, and E-DCF.
D-ABC may be regarded as having ACD for its base and B for its vertex.
Now, § 505, AD II BE II CF', and hence, § 457, BE II plane ACD-,
B-ACD and E-ACD have equal altitudes; hence, § 557, D-ABC =o B-ACD o E-ACD.
F-ABC may be regarded as having ACF for its base and B for its vertex ;
but A ACF <> A DCF, and BE II plane DCF ; Why ?
hence, F-ABC =o= B-A CF <)= E-D CF ; Why ?
E-ABC 4- D-ABC + F-ABC ^ E-ABC + E-ACD + E-DCF.
But ABC-DEF ^ E-ABC + E-ACD + E-DCF ;
hence, ABC-DEF o= E-ABC + D-ABC + F-ABC.
Therefore, etQ. (^.e.v.
SOLID GEOMETRY. — BOOK VIII.
297
566. Cor. I. The volume of a truncated right triangular prism is equal to the product of its base by one third the sum of its lateral edges.
1. What is the direction of the lateral edges AD, BE, CF with reference to the base ABC?
2. How, then, do AD, BE, and CF (compare with the altitudes of the three pyramids whose sum is equivalent to ABC-BEF ?
3. To what is the volume of each of these pyramids equal ?
4. To what, then, is the volume of ABC-DEF equal ?
567. Cor. II. The volume of any truncated triangular prism is equal to the product of its right section by one third the sum of its lateral edges.
1. If GHK is a right section, to what is the volume of GHK-DEF equal ?
2. To what is the volume of GHK- ABC equal ?
3. To what, then, is the volume of ABC-DEF equal?
Ex. 780. What is the lateral area of a right prism whose altitude is 12 in. and the perimeter of whose base is 20 in.?
Ex. 781. Find the ratio of two rectangular parallelopipeds, if their alti- tudes are each 7™ and their bases 3'" by 4"' and 7™ by 9™, respectively.
Ex. 782. Find the ratio of two rectangular parallelopipeds, if their di- mensions are 2dn\ 4dm, 3dm, and 6^™, 7'^", S*!*", respectively.
Ex. 783. What is the volume of a rectangular parallelepiped whose edges are 20.5'", 12.75™, and 8.6™, respectively ?
Ex. 784. The altitude of a regular hexagonal prism is 12 ft., and each side of its base is 10 ft. What is its volume ?
Ex. 785. What is the volume of a pyramid whose altitude is 18<i™ and whose base is a rectangle 10^™ by 6<i™ ?
Ex. 786. What is the volume of a truncated right triangular prism, if each side of its base is 3 ft. and its edges are 3 ft., 4 ft., and 6 ft., respectively ?
Ex. 787. What is the lateral area of the frustum of a square pyramid whose slant height is 13™, each side of the lower base being 3.5™, and of the upper base 2™ ?
298
SOLID GEOMETRY.^BOOK VIIL
Proposition XIX
568. Form two tetrahedrons which have a trihedral angle of one equal to a trihedral angle of the other. Considering homologous faces of these trihedrals as bases of the tetrahedrons, how does the ratio of their volumes compare with the ratio of the products of their bases by their altitudes? (§ 561) How does the ratio of their bases compare with the product of the ratios of the homologous edges which include the basal face angles of the equal trihedrals? (§ 340) How does the ratio of the altitudes of the tetrahedrons compare with the ratio of the third edges of the equal trihedrals? (§ 299) From these equal ratios discover how the ratio of the volumes of the tetrahedrons compares with the ratio of the products of the three edges of the equal trihedral angles.
Theorem, Tetrahedrons which have a trihedral angle of one equal to a trihedral angle of the other are to each other as the products of the edges of the equal trihedral angle^^
Data : Any two tetrahedrons, as Q-ABC and T-DEF, which have the trihedral angles Q and T equal.
To prove Q-ABC : T-DEF = QA X QB X QC : TD X TE X TF.
Proof. Apply T-DEF to Q-ABC SO that the equal trihedral angles T and Q coincide.
Draw CO and FP perpendicular to the plane QAB.
Then, their plane intersects QAB in QPO.
Now, CO and FP are the altitudes of the triangular pyramids G-QAB and F-QDE ; .-. § 561, C-QAB : F-QDE = QAB X CO : QDE X FP. (1)
But, § 340, QAB : QDE = QA X QB : QD X QE.
Substituting in (1),
C-QAB : F-QDE = QA X QB X CO : QD X QE X FP. (2)
SOLID GEOMETRY. — BOOK VIII. 299
Now, rt. A QOC and QPF are similar ; Why ?
.-. § 299, CO : FP = QG : QF.
Substituting in (2),
C-QAB : F-QDE = QA X QB X QC : QD X QE X QF] that is, Q-ABO : T-BFF = QA X QB X QC : TD X TE X TF.
Therefore, etc. q.e.d,
SIMILAR AND REGULAR POLYHEDRONS
569. Polyhedrons which have their corresponding polyhedral angles equal, and have the same number of faces similar each to each, and similarly placed, are called Similar Polyhedrons.
Faces, edges, angles, etc., which are similarly placed in similar polyhedrons are called homologous faces, edges, angles, etc.
570. Since the homologous sides of similar polygons are pro- portional, the homologous edges of similar polyhedrons are proportional.
571. Since similar polygons are proportional to the squares upon any of their homologous lines, the homologous faces of similar polyhedrons are proportional to the squares upon any of their homologous edges.
572. From § 571 it is evident that the entire surfaces of simi- lar polyhedrons are proportional to the squares upon any of their homologous edges.
Proposition XX
573. 1. Form two similar polyhedrons, and if possible divide them into the same number of tetrahedrons, similar each to each.
2. How does the ratio of any two homologous lines in two similar polyhedrons compare with the ratio of any two homologous edges ?
Theorem. Similar polyhedrons may he divided into the same number of tetrahedrons, similar each to each, and similarly placed.
mo
SOLID GEOMETRY. — BOOK VIII.
Data : Any two similar polyhedrons, as AJ and A' J'. To prove that AJ and A'j' may be divided into the same num- ber of tetrahedrons, similar each to each, and similarly placed.
K
Proof. Select any trihedral angle in AJ, as B, and through the extremities of its edges, as A, G, C, pass a plane. Also through the homologous points, A', G', C', pass a plane.
Then, § 310, in the tetrahedrons B-AGC and B'-A'g'C', the faces BACj BAG, BGC are similar to the faces bWc', B'A'G'y b'g'C', each to each.
Hence, AG : A'G' = BG : B'g' = CG : C'G',
and AG:A'G' = AB:A'B' = AO:A'Cf;
AG:A'G'= CGiC'G' =AC:A'C'.
Hence,
face ACG is similar to face A'c'G'
Why?
.'. the homologous faces of these tetrahedrons are similar.
Also, § 500, the homologous trihedral angles of these tetra- hedrons are equal.
Hence, § 569, tetrahedrons B-AGC and b'-A'g'C' are similar.
Now, if these similar tetrahedrons are removed from the simi- lar polyhedrons of which they are a part, the polyhedrons which remain will continue to be 'similar; for the faces and polyhedral angles of the original polyhedrons will be similarly modified.
By continuing to remove similar tetrahedrons from them, the original polyhedrons may be reduced to similar tetrahedrons, and they will then have been divided into the same number of tetra- hedrons, similar each to each, and similarly placed.
Therefore, etc. q.e.d.
674. Cor. Homologous lines in similar polyhedrons are propor- tional to tJieir homologous edges.
SOLID GEOMETRY. — BOOK VJII. 301
Proposition XXI
575. Form two similar tetrahedrons. How do the trihedral angles at the vertices compare ? Then, how does the ratio of the tetrahedrons compare with the product of the ratios of the homologous edges of the corresponding trihedral angles? (§ 568) How do the ratios of these edges compare with each other ? Then, how does the ratio of the tetra- liedrons compare with the ratio of the cubes of any two homologous
Theorem, Similar tetrahedrons are to each other as the tubes of their homologous edges. .
B
Data : Any two similar tetrahedrons, as Q-ABC and T-DEF.
To prove Q-ABC : T-DEF = O? : ^ = etc.
Proof. § 569, trihedral Z Q = trihedral Z T;
/. § 568, Q-ABC : T-DEF = QA X QB X QC : TD X TE X TF,
Q-ABC QAX QB X QC QA QB QC
or = = — X — X — •
T-DEF TD X TE X TF TD TE TF
But QA:TD= QB:TE= QC:TF,OV — = = — -•
TD TE TF
-u- Q-ABC QA ^QA ^ QA QA^ . Hence, = — x — x — - = ^=r. j
T-DEF TD TD TD tD that is, Q-ABC : T-DEF =QA^:T^.
In like manner, the same may be proved for any two homolo- gous edges.
Therefore, etc. Q-e.i>.
302 SOLID GEOMETRY. — BOOK VIIL
576. Cor. Similar polyhedrons are to each other as the cubes of their homologous edges.
1. Into what may two similar polyhedrons be divided ? § 573
2. How do the ratios of these portions compare with the ratios of the cubes of their homologous edges ? § 575
3. How do these ratios compare with the ratios of the cubes of any two homologous edges of the polyhedrons ? § 574
4. How, then, does the ratio of the sums of these portions compare with the ratio of the cubes of any two homologous edges of the polyhedrons ?
577. A polyhedron whose faces are equal regular polygons, and whose polyhedral angles are equal, is called a Regular Polyhedron.
578. 1. What is the least number of faces that a convex poly- hedral angle may have ? How does the sum of the face angles of any convex polyhedral angle compare with 360° ? Since each angle of an equilateral triangle is 60°, may a convex polyhedral angle be formed by combining three equilateral triangles ? Four ? Five ? Six ? Why ? Then, how many regular convex polyhedrons are possible with equilateral triangles for faces ?
2. How many degrees are there in the angle of a square ? May a convex polyhedral angle be formed by combining three squares ? By combining four ? Why ? Then, how many regular convex polyhedrons are possible with squares for faces ?
3. Since each angle of a regular pentagon is 108°, may a convex polyhedral angle be formed by combining three regular penta- gons ? By combining four ? Why ? Then, how many regular convex polyhedrons are possible with regular pentagons 1 3r faces ?
4. Since each angle of a regular hexagon is 120°, may a convex polyhedral angle be formed by combining three regular hexagons ? Why ? By combining three regular heptagons ? Why ? What is the limit of the number of sides of a regular polygon that may be used in forming a convex polyhedral angle, and therefore in forming a regular convex polyhedron ?
What, then, is the greatest number of regular convex poly- hedrons possible ?
SOLID GEOMETRY. — BOOK VIII 303
o79. There are only five regular convex polyhedrons possible, called from the number of their faces the tetrahedron, the hexahe- dron, the octahedron, the dodecahedron, and the icosahedron.
The tetrahedron, octahedron, and icosahedron are bounded by equilateral triangles; the hexahedron by squares; and the dodecahedron by pentagons.
580. The point within a regular polyhedron that is equidistant from all the faces of the polyhedron is called the center of the polyhedron.
The center is also equidistant from the vertices of all the poly- hedral angles of the polyhedron.
Therefore, a sphere may he inscribed in,, and a sphere may he circumscribed about, any regular polyhedron, §§ 640, 641
Proposition XXII
581. Problem, Upon a given edge to construct the regular polyhedrons.
Datum : An edge, as AB. jv
Required to construct the regular polyhe- drons on AB.
Solutions. 1. The regular tetrahedron.
Upon AB construct an equilateral triangle, Ni^-'^
HS ABC. ^
At the center of A ABC erect a perpendicular, and take a point D in this perpendicular such that DA = AB. Draw lines from D to the vertices of A ABC. Then, the polyhedron D-ABC is a regular tetrahedron. q.e.f. Proof. By the student. Suggestion. Refer to §§ 450, 500.
rr
2. The regidar hexahedron. /r y^
Upon AB construct the square ABCD, and e upon its sides construct the squares AF, BG, CH, and BE perpendicular to ABCD.
Then, the polyhedron ^G is a regular hexa- hedron. Q.E.F.
DI
Proof. By the student. Suggestion. Refer to § 600.
304
SOLID GEOMETRY.-- BOOK VUL
3. The regular octahedron.
Upon AB construct the square ABCD, and through its center 0 pass a line perpendicular to its plane.
On this perpendicular take the points E and F such that AE and AF are each equal to AB.
Draw lines from E and F to the vertices of ABCD.
Then, the polyhedron E-ABCD-F is a regular octahedron, q.e.f.
Proof. By the student. Suggestion. Refer to § 460,
4. The regular dodecahedron.
Upon AB construct a regular pentagon ABODE, and to each side of it apply an equal pen- tagon so inclined to the plane of ABODE as to form trihedral angles at A, B, C, D, E.
Then, a convex sur- face FHKMP, composed of six regular pentagons, has been constructed.
Construct a convex surface f'h'k'm'p' equal to FHKMP, and apply one to the other so as to form a single convex surface.
The surface thus formed is that of a regular dodecahedron, q.e.f.
Proof. By the student. Suggestion. Refer to § 600.
5. The regular icosahedron.
Upon AB construct a regular pentagon ABODE; at its center erect a perpendicular ; and take a point Q in this perpendicular such that QA = AB.
Draw lines from Q to the vertices of the pentagon forming a regular pentagonal pyra- mid Q- ABODE.
Complete the pentahedral angles at A, B, 0, D, E by adding to each three equilateral tri- angles, each equal to A QAB,
SOLID GEOMETRY. — BOOK VIII.
Construct a regular pentagonal pyramid Q'-A'b'c'b'e' equal to Q-ABCDE, and join it to the convex surface already formed so as to form a single convex surface.
The surface thus formed is that of a regular icosahedron. q.e.f.
Proof. By the student. Scggestion. Refer to § 460.
582. Sch. The five regular polyhedrons may be made from cardboard in patterns as given below, by cutting half through along the dotted lines, folding, and pasting strips of paper along the edges.
TETRAHEDRON
HEXAHEDRON
OOTABBDBON
DODKOAHEDRON
I006AHBDB0N
583.
FORMULA
Notation
B = base.
5 = ipper base.
p = perimeter of base.
p' = perimeter of upper base.
P" = perimeter of right section.
H = altitude.
milne's geom. — 20
L = slant height.
E = lateral edge.
d]
e = dimensions of a paral
/ lelopiped.
A = lateral area.
r = volume.
306 SOLID GEOMETRY, — BOOK VIIL
Prism.
^ = ^ X P" . . . . § 519
A=HXP (E-iglit Prism) § 520
V = BxH §§542,543
Rectangular Parallelepiped.
V = dxexf §536
r = B XH (True also for any parallelopiped) §§ 537, 539
Pyramid.
A = ^P XL (Regular Pyramid) § 551
V = lBxH §§559,560
Frustum of a Pyramid.
A = ^L(p-{-P') (Regular Pyramid) § 552
v = ^H(B-\-b-\-^Wxl>) §564
SUPPLEMENTAKY EXERCISES
Ex. 788. The edge of a cube is 5^ in. What are its volume and surface ?
Ex. 789. What are the entire area and volume of a right prism 4.5 ft. in altitude, if the bases are equilateral triangles 13 in. on a side ?
Ex. 790. What is the total area of a regular triangular pyramid whose slant height is IS*^™ and each side of whose base is 9^™ ?
Ex. 791. What is the volume of a triangular pyramid whose altitude is 11 ft. and the sides of whose base are 3 ft., 4 ft., and 5 ft.?
Ex. 792. What is the edge of a cube whose volume equals that of a rec- tangular parallelopiped whose edges are 9 in., 12 in., and 16 in.?
Ex. 793. The altitude of a prism is 6^"^ and the area of its base is 2.5'^ ^™. What is the altitude of a prism of the same volume, if the area of its base is
5.75sq dm ?
Ex. 794. The homologous edges of two similar tetrahedrons are as 4 : 6. What is the ratio of their surfaces ? What is the ratio of their volumes ?
Ex. 795. What is the altitude of a pyramid whose volume is 36^" m and the sides of whose triangular base are 6™, 5™, and 4™ ?
Ex. 796. The area of the upper base of the frustum of a pyramid is 48 sq. ft. and that of the lower base is 72 sq. ft. If the altitude of the frus- tum is 60 ft., what is ita volume ?
SOLID GEOMETRY— BOOK Vltl. 307
Ex. 797. What is the altitude of the frastum of a regular hexagonal pyramid, if its volume is IG^um and the sides of its bases are respectively 1.5™ and 2.5'"?
Ex. 798. A pyramid 20 ft. high has 100 sq. ft. in its base ; a section parallel to the base has an area of 55 sq. ft. How far is the section from the base?
Ex. 799. What is the volume of an oblique truncated triangular prism whose edges are 5"\ 7'", and 9'", and the area of whose right section is IQ^™ ?
Ex. 800. What is the edge of a cube whose entire area is 1*^™ ?
Ex. 801. The base of a pyramid contains 121 sq. ft. ; a section parallel to the base and 3 ft. from the vertex contains 49 sq. ft. What is the altitude of the pyramid ?
Ex. 802. What is the lateral area of a regular hexagonal pyramid whose base is inscribed in a circle whose diameter is 15 ft., the altitude of the pyra- mid being 8 ft. ? What is the volume of the pyramid ?
Ex. 803. Any lateral edge of a right prism is equal to the altitude.
Ex. 804. The square of a diagonal of a rectangular parallelepiped is equal to the sum of the squares of its three edges.
Ex. 805. If the edges of a tetrahedron are all equal, the sum of the angles at any corner is equal to two right angles.
Ex. 806. The section of a triangular pyramid made by a plane parallel to two opposite edges is a parallelogram.
Ex. 807. The lateral faces of right prisms are rectangles.
Ex. 808. The section of a prism made by a plane parallel to a lateral edge is a parallelogram.
Ex. 809. Thife diagonal of a cube is equal to the product of its edge and V8.
Ex. 810. The volume of a regular prism is equal to the product of its lateral area and one half the apothem of the base.
Ex. 811. Any straight line passing through the center* of a parallelo- piped and terminated by two faces is bisected at the center.
Ex. 812. If any two non-parallel diagonal planes of a prism are perpen- dicular to the base, the prism is a right prism.
Ex. 813. The base of a pyramid is 16 sq. ft. and its altitude is 7 ft. What is the area of a section parallel to the base, if it is 2 ft. 6 in. from the apex ?
Ex. 814. The edges of a rectangular parallelepiped are 3 in., 4 in., and 6 in. What is the area of its diagonal planes and the length of its diagonal line?
* The center of a parallelepiped is the intersection of its diagonals.
308 SOLID GEOMETRY— BOOK VIIL
Ex. 815. A portion of a railway embankment is 18 ft. by 380 ft. at the top, and 40 ft. by 380 ft. at the bottom. If its height is 12 ft., how many cubic yards, or loads, of earth does it contain ?
Ex. 816. If the four diagonals of a four-sided prism pass through a com- mon point, the prism is a parallelopiped.
Ex. 817. If a pyramid is cut by a plane parallel to its base, the pyramid . cut off is similar to the given pyramid.
Ex. 818. The lateral area of a right prism is less than the lateral area of any oblique prism having the same base and altitude.
Ex. 819. If a section of a pyramid made by a plane parallel to the base bisects the altitude, the area of the section is one fourth the area of the base, and the pyramid cut off is one eighth of the original pyramid.
Ex. 820. The volume of a right triangular prism is equal to one half the product of any lateral face by its distance from the opposite edge.
Ex. 821. If the diagonals of three unequal faces of a rectangular paral- lelopiped are given, compute the edges.
Ex. 822. What is the lateral area of a regular pyramid whose slant height is 10 ft., the base being a pentagon inscribed in a circle whose radius is 6 ft. ? What is the volume ?
Ex. 823. The volume of a rectangular parallelopiped is 336c" '", its total area is 320sqm^ and its altitude is 4™. What are the dimensions of its base ?
Ex. 824. A pyramid weighs 30^8, and its altitude is 12^"'. A plane parallel to the base cuts off a frustum which weighs 15^^. What is the alti- tude of the frustum ?
Ex. 825. Each side of the base of a regular triangular pyramid is 3 in., and its altitude is 8 in. What are its lateral edge and lateral area ?
Ex. 826. The volume of a regular tetrahedron is equal to the product of the cube of its edge and ^^ \/2.
Ex. 827. The volume of a regular octahedron is equal to the product of the cube of its edge and \ V2.
Ex. 828. Any plane passing through the center of a parallelopiped divides it into two equal solids.
Ex. 829. The lateral area of a regular pyramid is greater than its base.
Ex. 830. The lateral edge of the frustum of a regular triangular pyramid is 4^ ft., a side of one base is 5 ft., and of the other 4 ft. What is the volume ?
Ex. 831. The sum of the perpendiculars from any point within a regular tetrahedron to each of its four faces is equal to its altitude.
Ex. 832. In a regular tetrahedron an altitude is equal t;Q tiUfee times the perpendicular from its foot on a face.
BOOK IX
CYLINDERS AND CONES
584. A surface, generated by a moving straight line which always remains parallel to its original posi- tion and continually touches a given curved line, is called a Cylindrical Surface.
The moving straight line is called the gen- eratrix, and the given curved line is called the directrix.
The generatrix in any position is called an element of the surface.
585. A solid bounded by a cylindrical surface and two parallel planes which cut all its elements is called a Cylinder.
The plane surfaces are called the bases and the cylindrical sur- face is called the lateral surface of the cylinder.
All elements of a cylinder are equal. § 464.
The perpendicular distance between its bases is the altitude of the cylinder.
586. A section of a cylinder made by a plane perpendicular to its elements is called a Right Section.
587. A cylinder whose elements are perpendicular to its base is called a Right Cylinder.
588. A cylinder whose elements are not perpendicular to its base is called an Oblique Cylinder.
589. A cylinder whose bases are circles is a Circular Cylinder. The straight line joining the centers of the bases of a circular
cylinder is called the axis of the cylinder.
590. A right circular cylinder is called a Cylinder of Revolution, because it may be generated by the revolu- tion of a rectangle about one of its sides.
Cylinders of revolution generated by similar rectan- gles revolving about homologous sides are similar.
300
310 SOLID GEOMETRY. — BOOK IX.
591. A plane which contains an element of a cylinder and does not cut the surface is a Tangent Plane to the cylinder. The element is called the element of contact.
692. Any straight line that lies in a tangent plane and cuts the element of contact is a Tangent Line to the cylinder.
593. When the bases of a prism are inscribed in the bases of a cylinder and its lateral edges are elements of the cylinder, the prism is said to be inscribed in the cylinder.
594. When the bases of a prism are circumscribed about the bases of a cylinder and its lateral edges are parallel to the ele- ments of the cylinder, the prism is said to be drcumscnhed about the cylinder.
Proposition I
595. 1. Form a cylinder and cut it by any plane through an element of its surface (§ 519 n.). What plane figure is the section made by the cutting plane?
2. If the cylinder is a right cylinder, what plane figure does such a plane make?
Theorem, Any section of a cylinder made hy a plane passing through an element is a parallelogram.
Data : Any section of the cylinder EF, as ABCD, made by a plane passing through AB, an element of the surface.
To prove ABCD a parallelogram.
Proof. The plane passing through the ele- ment AB cuts the circumference of the base in a second point, as D. Draw BO II AB. ^{
Then, § 63, DC is in the plane BAD ; ^
and, § 584, DC is an element of the cylinder.
Hence, DC, being common to the plane and the lateral surface of the cylinder, is their intersection.
Also, § 463, AD II BC;
hence, § 140, ABCD is a parallelogram.
Therefore, etc. q.e.d
SOLID GEOMETRY. — BOOK IX.
311
596. Cor. Any section of a right cylinder made by a plane pass- ing through an element is a rectangle.
Proposition II
597. 1. Form a cylinder. How do its bases compare?
2. Cut the cylinder by parallel planes. which cut all its elements. How do the sections thus made compare with each other ?
3. How does a section made by a plane parallel to the base compare with the base?
Theorem, The hases of a cylinder are equal.
Data : Any cylinder, as MG, whose bases are HG and MN.
To prove HG = MN.
Proof. Take any three points in the perim- eter of the upper base, as D, E, F, and from them draw the elements of the surface DA, EB, FC, respectively.
Draw AB, BC, AC, BE, EF, and DF.
§§ 585, 584, AB, BE, and CF are equal and parallel ;
.-. § 150, AE, AF, and BF are parallelograms; and AB = BE, AC = DF, BC=EF'j
hence, A ABC = A BEF.
Apply the upper base to the lower base so that BE shall fall upon AB.
Then, F will fall upon C.
But F is any point in the perimeter of the upper base, there- fore, every point in the perimeter of the upper base will fall upon the perimeter of the lower base.
Hence, § 36, HG = MN.
Why? Why?
Therefore, etc.
Q.E.D.
598. Cor. I. The sections of a cylinder made by parallel planes
cutting all its elements are equal.
599. Cor. II. Hie axis of a circular cylinder passes through the centers of all the sections parallel to the bases.
812 SOLID GEOMETRY. — BOOK IX,
Proposition III
600. To what is the lateral surface of any prism equivalent? (§ 519) If the number of its lateral faces is indefinitely increased, what solid does the prism approach as its limit ? How, then, does the lateral sur- face of any cylinder compare with the rectangle formed by an element and the perimeter of a right section ?
Theorem, The lateral swrfax^e of a cylinder is equivo/- lent to the rectangle formed by an element of the surface and the perimeter of a right section.
Data: Any cylinder, as FK\ any right section of it, as ABODE ; and any element of its surface, as FG.
Denote the lateral surface of FK by S, and the perimeter of its right section by P.
To prove S ^ rect. FG • P.
Proof. Inscribe in the cylinder a prism ; denote its lateral surface by s' and the perimeter of its right section by P'.
Then, § 593, each lateral edge is an element of the cylinder, and, § 585, the elements are all equal ; .-. § 519, S' 0= rect. FG • P'.
Now, if the number of lateral faces of the inscribed prism is indefinitely increased,
§ 393, P' approaches P as its limit ;
S' approaches S as its limit.
But, however great the number of faces, S' =0= rect. FG . P'.
Hence, § 326, S =0= rect. FG • P.
Therefore, etc. q.e.d.
601. Cor. The lateral surface of a cylinder of revolution is equivalent to the rectangle formed by its altitude and the circum- ference of its base.
Arithmetical Rules : To be framed by the student. § 339
SOLID GEOMETRY. — BOOK IX.
313
6U2. formulae : Let A denote the lateral area, T the total area H the altitude, and R the radius of the base of a cylinder f revolution.
Then, § 395, ^ = 2 Tri? x fl",
and, § 398, 7= 2 Tri? x H-\-'2 ttR' = 2 iri?(fl--f- R),
Proposition IV
603. Compute the areas of any two similar cylinders of revolution, as those whose altitudes are 4" and 2" and whose radii are 2" and 1", respectively. How does the ratio of their lateral areas, or of their total areas, compnre with the ratio of the squares of their altitudes, or with the ratio ol the squares of their radii ?
Theoretn, The lateral areas, or the total areas, of simU lar eylinders of revolution are to each other as the square& of their altitudes, or as the squares of their radii.
Data: Any two similar cylinders of revolution, whose altitudes are H and S^, and radii R and i?', respectively.
Denote their lateral areas by A and A\ and their total areas by T and T\ respec- tively.
To prove 1. A\A^ = H'^ : B.^ = i?' : i?'^. 2. r;r' = ^2.^/2^^.^f2
Proof. 1. Since the generating rectangles are similar, R'~ ~
H
§§ 299, 279, .-. § 602,
H' R' + B ^x^
7»
H'
or
J___ 2 7rRH
A'~2'7rR'H' R' IT IT A:A' = H^:H'^=R':R".
9 8^09 r^ 2'kR{R^-H) ^R[ R + H\^ie __H\
or T:T' = H^:H"'=R^:R"'.
Therefore, etc. Q.e.b
604. Cor. The lateral areas, or the total areas, of similar cylin- ders of revolution are to each other as the squares of any of their like dimensions.
314 SOLID GEOMETRY. — BOOK IX.
Proposition V
605. To what is the volume of any prism equal? (§ 543) If tha number of its lateral faces is indefinitely increased, what solid does the prism approach as its limit ? To what, then, is the volume of anj cylinder equal ?
Theorem. The volume of any cylinder is equal to the product of its base by its altitude.
Data : Any cylinder, as A, whose base is B and altitude H.
Denote its volume by F.
To prove V= B x H.
Proof. Inscribe in the cylinder a prism, and denote its volume by F' and its base bys'.
Then, the altitude of the prism is H,
and, § 543, F' = 5' X H.
Now, if the number of lateral faces of the inscribed prism is indefinitely increased,
§ 393, 5' approaches B as its limit ;
F' approaches F as its limit. But, however great the number of faces,
F' = 5' X H. Hence, § 222, V=BxH.
Therefore, etc. q.e.d.
606. Formula : Let B denote the radius of the base of a cylinder of revolution.
Then, §398, B = 7rlf',
V=TTiexH.
Proposition VI
607. Compute the volumes of any two similar cylinders of revolution, as those whose altitudes are 4" and 2" and whose radii are 2" and 1" respectively. How does the ratio of their volumes compare with the ratio of the cubes of their altitudes, or with the ratio of the cubes of their radii?
SOLID GEOMETRY. — BOOK IX.
315 .
Theorem, The volumes of similar cylinders of revolution are to each other as the cubes of their altitudes, or as the cubes of their radii.
Data : Any two similar cylinders of rev- olution, whose altitudes are H and H', and radii R and i2' respectively.
Denote their volumes by V and r' re- spectively.
To prove V-.V^ = H^\ H^ = :^: r'^
Proof. Since the generating rectangles are similar,
H
299, . § 606,
|
R H R' H*' |
|
|
V |
ttR^H R^ H -kR^'H' R" ^ H' |
|
V: V' —H^i H'^ -. |
ir'3 R"
or V: V = H'' : H"" = I^ : R^.
Therefore, etc. q.e.d.
608. Cor. The volumes of similar cylinders of revolution are to each other as the cubes of any of their like dimensions.
CONES
A surface, generated by a moving straight line which passes through a fixed point and continually touches a given curved line, is called a Conical Surface.
The moving straight line is called the generatrix, the fixed point the vertex, and the given curved line the directrix.
The generatrix in any position is called an element of the surface.
If the generatrix extends on both sides of the vertex, the whole surface consists of two portions which are called the lower and upper nappes respectively.
Q-ABCD and Q-A'B'C'D' are the lower and upper nappes respectively of a conical surface of which A A' is the generatrix, Q the vertex, ABGD the directrix, and AA' , BB'^ CC, DD', etc., are elements.
. 316 SOLID GEOMETRY. — BOOK IX.
610. A solid bounded by a conical surface and a plane which cuts all its elements is called a Cone.
The plane surface is called the base and the conical surface is called the lateral surface of the cone.
The perpendicular distance from its vertex to the plane of its base is the altitude of the cone.
611. A cone whose base is a circle is called a Circular Cone. The straight line joining the vertex and the center of the base
of a cone is called the axis of the cone.
612. A cone whose axis is perpendicular to its base is called a Right Cone.
613. A cone whose axis is not perpendicular to its base is called an Oblique Cone.
614. A right circular cone is called a Cone of Revolution, be- cause it may be generated by the revolution of a right triangle about one of its perpendicular sides.
All the elements of a cone of revolution are equal, and any one of them is called the slant height of the cone.
Cones of revolution, which are generated by similar right triangles revolving about homologous K~"l}~^ J perpendicular sides, are similar.
616. A plane which contains an element of a cone and does not cut the surface is a Tangent Plane to the cone.
The element is called the element of contact.
616. Any straight line that lies in a tangent plane and cuts the element of contact is a Tangent Line to the cone.
617. The portion of a cone included between its base and a section parallel to the base and cutting all the elements is called a Frustum of a cone.
The base of the cone is called the lower base of the frustum, and the parallel section the upper base.
The perpendicular distance between its bases is the altitude of the frustum.
The portion of an element of a cone of revolution included between the parallel bases of a frustum is the slant height of the frustum.
SOLID GEOMETRY,— BOOK IX.
317
618. When the base of a pyramid is inscribed in the base of a cone and its lateral edges are elements of the cone, the pyramid is said to be inscribed in the cone.
619. When the base of a pyramid is circumscribed about the base of the cone and its vertex coincides with the vertex of the cone, the pyramid is said to be circumscribed about the cone.
Proposition VII
620. Form a cone and cut it by any plane through its vertex, plane figure is the section made by the cutting plane ?
What
Theorem, Any section of a cone made hy a plane pass- ing through its vertex is a triangle.
Data : Any cone, as Q-ABD, and any section of it, as CQD, made by a plane passing through the vertex Q.
To prove CQD a triangle.
Proof. Draw the straight lines QC and QD.
Then, § 609, QC and QD are elements of the cone, and, § 427, since QC and QD each have two points in common with the plane CQD, they lie in that plane ;
.*. QC and QD are the intersections of the cutting plane and the lateral surface.
Also, § 441, CD is a straight line.
Hence, § 85, CQD is a triangle.
Therefore, etc. q.e.d.
Ex. 833. What is the lateral area of a cylinder of revolution whose alti- tude is 18 ft. and the diameter of whose base is 6 ft. ?
Ex. 834. What is the volume of a cylinder of revolution whose altitude is 7 ft. and the circumference of whose base is 5 ft. ?
Ex. 835. How many cubic feet are there in a cylindrical log 14 ft. long and 2.5 ft. in diameter ?
Ex. 836. The altitude of a cylinder of revolution is 16^"" and the diameter of its base is 11dm. What is its total area ? What is its volume ?
318
SOLID GEOMETRY.— BOOK IX,
Proposition VIII
621. !• Form a circular cone and cut it by any plane parallel to its base. What plane figure is the section made by the cutting plane?
2. In what points will the axis of the cone pierce all the sections that are parallel to the base ?
Theorem. Any section of a circular cone made hy a plane parallel to the base is a circle.
Data : Any circular cone, as Q-ABC and any section of it, as DEF, made by a plane parallel to the base.
To prove DEF a circle.
Proof. Draw QO, the axis of the cone piercing BEF in P.
Through QO and any elements, QA, QB, etc., pass planes cutting the base in the radii OA, OB, etc., a\ and the parallel section in the straight lines P2>, PE, etc.
Data, planes DEF and ABC are parallel ;
.-. § 463, PB II OA, and PE II OB.
Consequently, A QPD and QOA are mutually equiangular and therefore similar ; likewise A QPE and QOB are similar.
Hence, § 299, PB : OA = QP : QO,
and PE:OB= QP: QO;
PB : OA = PE : OB.
But OA = OB', Why?
.-. § 272, PB = PE',
that is, all the straight lines drawn from the point P to the perim- eter of the section BEF are equal.
Hence, § 173, BEF is a circle.
Therefore, etc. q.e.d.
622. Cor. The axis of a circular cone passes through the centers of all the sections that are parallel to the base.
Ex. 837. The total area of a cylinder of revolution is 659"<i' altitude is IS**'". What is the diameter of its base ?
and its
SOLID GEOMETRY. — BOOK IX.
319
Proposition IX
623. To what is the lateral surface of any regular pyramid equivalent ? If the number of its lateral faces is indefinitely increased, what solid does the pyramid approach as its limit ? How, then, does the lateral sur- face of a cone of revolution compare with the rectangle formed by the circumference of its base and its slant height?
Theoretn, The lateral surface of a cone of revolution is equivalent to one half the rectangle formed by the circum- ference of its base and its slant height.
Data: Any cone of revolution, as Q-ABFB, whose slant height is L, and the circumference of whose base is C.
Denote its lateral surface by S.
To prove S^^ rect. C - L.
Proof. Inscribe in the cone a regular pyramid of any number of faces and denote its lateral sur- face by S', its slant height by L', and the perimeter of its base by P.
Then, § 618, each lateral edge of the pyramid is an element of the cone ; and, § 551, s' o= i rect. P • L'.
Now, if the number of lateral faces of the inscribed pyramid is indefinitely increased,
§ 392, p approaches C as its limit ;
L' approaches L as its limit,
and s' approaches s as its limit.
But, however great the number of faces, S' ^ 1 rect. P ' L'.
Hence, § 326, S^^ rect. C • L.
Therefore, etc. Q.E.D.
Arithmetical Rule : To be framed by the student.
624. Formulae : Let R denote the radius of the base of a cone of revolution, A its lateral area, and T its total area.
Then, § 395, A = ^(2 ttR x L) = ttRL,
and, § 398, T= ttRL + irPi^ = ttR^L + R),
320 SOLID GEOMETRY. — BOOK IX,
Proposition X
625. ^ Compute the areas of any two similar cones of revolution, as those whose altitudes are 8" and 4", slant heights 10" and 5", and the radii of whose bases are 6" and 3", respectively. How does the ratio of their lateral areas, or of their total areas, compare with the ratio of the squares of their altitudes, or with the ratio of the squares of the radii of their bases ?
Theorem, The lateral areas, or the total areas, of similar cones of revolution are to ea/ih other as the squares of their altitudes, or as the squares of the radii of their bases.
Data : Any two similar cones of revo- a
lution, whose altitudes are H and ^, /!\
slant heights L and i', and the radii of / i \ /t\
whose bases are R and R\ respectively. / P \ / i '\
Denote their lateral areas by A and /''p t~~~""A /"'^~~\ A\ and their total areas by T and r', re- (;-— -^ 1 ^.-.j^^ spectively.
To prove 1. A-.A'^h'^: H'^ = i?- : R'^
2. T:Tf = H^:H'^ = R'':R".
Proof. 1. Since the generating triangles are similar,
§§299,279, E.=.L = ^=±±1.;
' ' H' L' R' L'-\-R''
a aoA -^ ttRL R L R^ H
or A:A' = H^:H'^=I^:R'K
2 Sfi24. T- 7rR(L-\-R) ^R ( L -{- R\_Iil' _^ ^ ' r' TrR\L' + R') R'\L'-^R'J R" H'^
or T:T' = H^:H'^ = R^: R'^.
Therefore, etc. q.e.d.
626. Cor. TTie lateral areas, or the total areas, of similar cones of revolution are to each other as the squares of their like dimensions.
Ex. 838. What is the lateral area of a cone of revolution whose slant height is 13 ft. and the diameter of whose base is 6 ft. ?
Ex. 839 What is the ratio of the lateral surface of a right circular cylinder to that of a right circular cone having the same base and altitude, if the altitude is 5 times the radius of the base t
SOLID GEOMETRY.— BOOK IX. 321
Proposition XI
627. 1. To what is the lateral surface of a frustum of any regular pyramid equivalent? If the number of its lateral faces is indefinitely increased, what solid does the frustum of the pyramid approach as its limit? How, then, does the lateral surface of the frustum of any cone of revolution compare with the rectangle formed by its slant height and the sum of the circumferences of its bases ?
2. How does the circumference of a section equidistant from the bases compare with half the sum of the circumferences of the bases ?
Theorem, The lateral surface of a frustujn of a cone of revolution is equivalent to one half the rectangle formed by its slant height and the sum of the circumferences of its bases.
Data: Any frustum of a cone of revolution, as A, whose slant height is L, and the circumferences of whose lower and upper bases are C and C', respec- tively.
Denote the lateral surface of the frustum hj S.
To prove S^^ rect. L - (c -\- C').
Proof. Inscribe in the frustum of tlie cone the frustum of a regular pyramid, and denote its lateral surface by s\ its slant iieight by L\ and the perimeters of its lower and upper bases by P and P', respectively.
Then, § 552, S' ^ i- rect. L' - {P ^ P').
Now, if the number of lateral faces of the inscribed frustum is indefinitely increased, § 392, P and P' approach C and C', respectively, as their limits;
L' approaches i as its limit, and S' approaches S as its limit.
But, however great the number of faces, S' o= i rect. z' . (P + P').
Hence, § 326, S^^ rect. i • (C + C').
Therefore, etc. q.e.d.
628. Cor. The lateral surface of a frustum of a cone of revolu- tion is equivalent to the rectangle formed by its slant height and the circumference of a section equidistant from its bases.
Arithmetical Rules : To be formed by the student. milne's geom. ^21
822 SOLID GEOMETRY, — BOOK IX.
Proposition XII
629. To what is the volume of any pyramid equal? If the number of its lateral faces is indefinitely increased, what solid does the pyra- mid approach as its limit? To what, then, is the volume of any cone equal?
Theorem, The volume of any cone is equal to one third the product of its base hy its altitude.
Data: Any cone, as Ay whose base is B and altitude H.
I>enote its volume by V.
To prove v=i:lbxh.
Proof. Inscribe in the cone a pyramid, and denote its volume by V' and its base by B\
Then, the altitude of the pyramid is H,
and, § 560, r = ^B'xH.
Now, if the number of lateral faces of the inscribed pyramid is indefinitely increased,
§ 393, jB' approaches B as its limit ;
F' approaches V as its limit.
But, however great the number of faces,
Hence, § 222, V=\BxH.
Therefore, etc. q.e.d.
630. Formula : Let R denote the radius of a cone of revolution Then, §398, B = tt^',
V=^TrI^ XB.
Ex. 840. The slant height of a right circular cone is 21 ft. and its altitude is 16 ft. What is its total area ?
Ex. 841. The slant height of a right circular cone is 6™ and the radius of its base is 6"™. What is its lateral area ? What is its volume ?
SOLID GEOMETRY. — BOOK IX. 323
Proposition XIII
631. Compute the volumes of any two similar cones of revolution, as those whose altitudes are 8" and 4", and the radii of whose bases a^e 6" and 3" respectively. How does the ratio of their volumes compare with the ratio of the cubes of their altitudes, or with the ratio of the cubes of the radii of their bases ?
Theorem, The volumes of similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases.
Data : Any two similar cones of revo- A
lution, whose altitudes are H and H', / \\
and the radii of whose bases are R and / j \ A
R\ respectively. / ' \ / lA
Denote their volumes by V and F*, re- . /-''^ i ""A 1''e'\~'\ spectiv^y. V" J \^I_-^
To prove V: r' = H^: h'^ = I^i B'K
Proof. Since the generating triangles are similar,
^299, | = |;
or V'. r = H^:H'^ = Ii^:Ii'\
Therefore, etc. q.e.d.
632. Cor. The volumes of similar cones of revolution are to each other as the cubes of any of their like dimensions.
Ex. 842. What is the volume of a cone whose altitude is 13 ft. and the
circumference of whose base is 9 ft. ?
Ex. 843. What is the total area of the frustum of a cone of revolution whose slant height is n^"^ and the radii of whose bases are 5^™ and 3^^™ ?
Ex. 844. How far from the base must a cone of revolution whose altitude is 15 in. be cut by a plane parallel to the base so that the volume of the frus- tum shall be one half that of the entire cone ?
Ex. 845. At what distances from the vertex must a cone of revolution be cut by planes parallel to the base to divide it into three equivalent solids ?
Ex. 846. A plane parallel to the base of a cone of revolution cuts the altitude at a point | of the distance from the vertex. What is the ratio of the volume of the cone cut off to that of the original cone ?
324 SOLID GEOMETRY. — BOOK IX.
Proposition XIV
633. To what is the volume of a frustum of any pyramid equal? If the number of its faces is indefinitely increased, what solid does the frustum of the pyramid approach as its limit? To what, then, is the volume of the frustum of any cone equal ?
Theorem. The volume of a frustum of any cone is equal to one third the product of its altitude hy the sum of its bases and a mean proportional between them.
Data : Any frustum of any cone, as A, whose altitude is H, and whose lower and upper bases are\B and b respectively.
Denote the volume of the frustum by V.
To prove V=^\H(b + b +V-B x h).
•
Proof. Inscribe in the frustum of the cone the frustum. of a pyramid, and denote its volume by F' and its lower and upper bases by B' and &', respectively.
Then, the altitude of the frustum of the pyramid is Hy and, § 564, r = \ h(b' + b' + ^W^Cb^).
Now, if the number of lateral faces of the inscribed frustum is indefinitely increased,
§ 393, B^ and 6' approach B and h respectively as their limits ; F' approaches V as its limit.
But, however great the number of faces,
r = iH(B' + &' + V^' X 6').
Hence, § 222, F = ^ h(b + & + V^ X b).
Therefore, etc. _ q.e.d.
634 Formula : Let R and R' denote the radii of the baees of a frustum of a cone of revolution.
Then, § 398, B^ttI^, b = irR^\ and -VBxb = vRR' ;
Ex. 847. What is the volume of the frustum of a cone whose altitude is 21 ft. and the circumferences of whose bases are 17 ft. and 13 ft. respectively ?
SOLID GEOMETRY. — BOOK IX.
325
|
635. FORMULA |
||
|
Notation |
||
|
B = base. |
R |
= radius of base. |
|
h z=z upper base. |
R' |
= radius of upper base. |
|
C = circumference of base. |
H |
= altitude. |
|
C' = circumference of upper |
L |
= slant height. |
|
base. |
A |
= lateral area. |
|
C" = circumference of mid- |
T |
= total area. |
|
section. |
V |
= volume. |
Cylinder of Revolution. A=C X H. A = 2 ttRH.
T = 2 ttR (H -^ R).
V— B X H. (True also for any cylinder.) V = ttR^H. %
Cone of Revolution.
A = ^C X L.
A = ttRL.
T=zttR{L-\- R).
V=\B X H. (True also for any cone.)'
F = i TT^H.
Frustum of a Cone of Revolution.
^ = |Z((7+C'). A = LX C".
V=^H(B + b H-V5 X b).
r=l7rH(R'+R'^ + RR').
(True also for any frustum.)
SUPPLEMENTARY EXERCISES
Ex. 848. What is the lateral area, total area, and volume of a cylinder of revolution whose diameter is 8 in. and altitude 12 in. ?
Ex. 849. What is the lateral area, total area, and volume of a cone of revolution whose base is 10^™ in diameter and whose altitude is 12^™ ? • Ex. 850. What is the lateral area, total area, and volume of a frustum of a cone of ravolution, the radii of whose bases are 6 in. and 4 in., respec- tively, and whose altitude is 9 in. ?
326 SOLID GEOMETRY.^BOOK IX.
Ex. 851. On a cylindrical surface only one straight line can be drawn through a given point.
Ex. 852. The intersection of, two planes tangent to a cylinder is parallel to an element.
Ex. 833. How many square yards of canvas are required for a conical tent 18 ft. high and 10 ft. in diameter?
Ex. 854. How many cubic feet are there in a piece of round timber 40 ft. long, whose ends are respectively 3 ft. and 1 ft. in diameter ?
Ex. 855. A cylindrical vessel is 40<im long and 20^™ in diameter. What is the weight, in grams, of the water that it will hold ?
Ex. 856. What is the weight, in grams, of a piece of lead, if when put Under water in a cylindrical tank 24<^i" in diameter it causes the level of the water to rise S^m, the specific gravity of lead being 11.4 ?
Ex. 857. A cylindrical cistern is 6.4™ deep and 5.2™ in diameter. How long will it take to fill it, if 2^1 flow into it per minute ?
Ex. 858. A plane through a tangent to the base of a circular cylinder and the element drawn to the point of contact, is tangent to the cylinder,
Ex. 859. A plane through a tangent to the base of a circular cone and the element drawn to the point of contact, is tangent to the cone.
Ex. 860. The volumes of two similar cylinders of revolution are as 27 : 64. If the diameter of the first is 3 ft., what is the diameter of the second ?
Ex. 861. A cylindrical vessel holds 1728 grams of water. What are the dimensions of the vessel, if the diameter is one third of the altitude ?
Ex. 862. Show that any lateral face of a pyramid circumscribed about a circular cone is tangent to the cone.
Ex. 863. What is the height of a cylinder 4.8<^™ in diameter, if it is equivalent to a cone of revolution 6.6<i™ in diameter and 6.4^™ high ?
Ex. 864. A cylindrical vessel is 12*=™ in diameter and 20"™ high. How many grams of mercury would it hold, the specific gravity of mercury being 13.6 ?
Ex. 865. The volumes of two similar cones of revolution are to each other as 512 : 729. What is the ratio of their lateral areas ?
Ex. 866. How many centigrams of alcohol will a cylindrical bottle hold, if the bottle is 8<=™ in diameter and 24^™ high, the specific gravity of alcohol being .79 ?
Ex. 867. The specific gravity of marble is 2.8. What is the weight, in kilograms, of a conical piece of marble, if the radius of its base is 20^™ and its height 50c™ ?
Ex. 868. The slant height of a cone of revolution is 3™. How far from the vertex must the elements be cut by a plane parallel to the base in order that the lateral surface may be divided into two equivalent parts ?
Ex. 869. If the altitude of a cylinder of revolution is equal to the diameter of its base, the volume is equal to the product of its total area by one third of its radius.
BOOK X
SPHERES
^36. A solid bcOiiided by a surface, every point of which is equally distant from a point within, is called a Sphere.''^
The point within is called the center.
A sphere may be generated by the revolu- tion of a semicircle about its diameter as an axis.
637. A straight line drawn from the center to any point of the surface of a sphere is called a radius.
A straight line which passes through the center of a sphere, and whose extremities are in the surface, is called a diameter.
638. A line or plane which has one, and only one, point in common with the surface of a sphere is tangent to the sphere.
The sphere is then said to be tangent to the line or plane.
639. Two spheres whose surfaces have one, and only one, point in common are tangent to each other.
640. When all the faces of a polyhedron are tangent to a sphere, the sphere is said to be inscribed in the polyhedron.
641. When all the vertices of a polyhedron lie in the surface of a sphere, the sphere is said to be circumscribed about the poly- hedron.
* In teaching Spherical Geometry, the class-room should be furnished with a spherical blackboard, on which the student should draw the diagrams required. It is also advised that each student be provided with some sort of a blackened or slated sphere for use in the preparation of lessons in cases where figures are to be drawn on its surface. A hemispherical cup to fit the sphere will enable him to draw great circles on the sphere.
327
328 SOLID GEOMETRY.— BOOK X.
642. A-x. 18. All radii of the same sphere, or of equal spheres, are equal.
19. All diameters of the same sphere, or of equal spheres, are
20. Two spheres are equal, if their radii or diameters are equal.
Proposition I
643. 1. Form a sphere and cut it by any plane (§ 519 n.). What plane figure is the section thus formed?
2. If a line joins the center of the sphere with the center of a circle of the sphere, what is its direction with reference to the plane of the circle ?
3. Cut a sphere by planes which are equally distant from its center. How do the sections thus formed compare ?
4. If the cutting planes are unequally distant from the center, which circle is the larger ?
Theorem, Any section of a sphere -made hy a plane is a circle.
Data : A sphere ; its center 0 ; and any section, as ABD. To prove ABD a circle.
Proof. Draw OC perpendicular to the plane ABD ; draw the radii OA and OD to any two points in J;he perimeter of the sec- tion ; and draw CA and CD.
Since 0 is a point in the perpendicular 0(7,
and, Ax. 18, OA = OD,
§ 449, CA = CD.
But A and D are any two points in the perimeter of section ABD.
Hence, § 173, ABD is a circle whose center is C. q.e.d.
SOLID GEOMETRY. — BOOK X. 329
644. Cor. I. 77ie line Joining the center of a sphere to the center of a circle of the sphere is perpendicular to the plane of the circle.
645. Cor. II. Circles of a sphere made by planes equally distant from the center are equal.
646. Cor. III. Of two circles of a sphere made by planes un- equally distant from the center, the nearer is the larger.
647. A section of a sphere made by a plane which passes through the center is called a Great Circle of the sphere.
648. A section of a sphere made by a plane which does not pass through the center is called a Small Circle of the sphere.
649. The diameter, which is perpendicular to the plane of a circle of a sphere, is called the Axis of the circle.
650. The ends of the axis of a circle of a sphere are called the Poles of the circle.
651. 1. Form a sphere and cut it by any plane, thus forming a circle of the sphere. Through what point of the circle does its axis pass ?
2. Form a sphere and cut it by two parallel planes, thus form ing two parallel circles. How are their axes situated with refer- ence to each other? How, then, are the poles of one of these circles situated with reference to the poles of the other ?
3. By passing planes form any two great circles of the same, or of equal spheres. How do they compare with each other ?
4. Form a sphere and divide it into two parts by a great circle. How do these parts compare with each other ?
5. By passing planes form any two great circles of a sphere. Since their intersection passes through the center and is a diame- ter of each circle, how do two great circles divide each other ?
6. Form two great circles of a sphere by passing two planes through it perpendicular to each other. Where do these circles pass with reference to each other's poles? If two great circles pass through each other's poles, what is the direction of their planes with reference to each other ?
7. Form a sphere and pass a plane through its center and any two points on its surface. What kind of a circle is the section thus formed ? What kind of an arc, then, may be drawn through any two points on the surface of a sphere ?
330 SOLID GEOMETRY. — BOOK X.
8. Form a sphere and pass a plane through any three points on its surface. What plane figure is the section thus formed? How many planes may be passed through the three points? How many circles, then, may be drawn through any three points on the surface of a sphere ?
652. Tlie axis of a circle passes through the center of that circle.
653. Parallel circles have the same axis and the sayne poles.
654. Cfreat circles of the same sphere, or of equal spheres, are equal.
655. Any great circle of a sphere bisects the sphere.
656. Ttvo great circles of the same sphere bisect each other.
657. Two great circles whose planes are perpendicular pass through each other^s poles; and conversely.
658. Through tico given points on the surface of a sphere an arc of a great circle may be draivn.
659. Through three given points on the surface of a sphere one circle may be drawn, and only one.
Proposition II
660. Form a sphere and select any two points on its surface ; through these points and the center of the sphere pass a plane. What kind of a circle is this section? Then, what kind of an arc joins the given points?
Join them by any other line on the surface. Which line represents the shortest distance between the given points ?
Theorem, The shortest distance on the surface of a sphere hetween any two -points on that surface is the arc, not greater than a semicircujnference, of the great circle which joins them. , ^-^-^ — ^
Data : Any two points on the surface of a sphere, as A and B^ joined by the arc of a great circle, as AB, not greater than a semi-
SOLID GEOMETRY.— BOOK X. 331
circumference ; also any other line on the surface joining A and B, as AECB.
To prove AB less than AECB.
Proof. Take any point in AECB, as D, and pass arcs of great circles through A and D, and B and D. Draw OA, OB, and OD from 0, the center of the sphere.
■ Then, Aaob, ADD, and BOD are the face A of the trihedral Z whose vertex is at 0 ;
.-. § 498, Z AOD -f Z jBOJD is greater than Z ^05.
But, § 224, arcs AD, BD, and AB are the measures of A AOD, BOD, and ^05 respectively ;
arc AD + arc BD > arc ^5.
In like manner, joining any point in AED with A and D, and any point in DCB with Z> and B by arcs of great circles, the sum of these arcs will be greater than arc AD -\- arc BD, and therefore greater than arc AB.
If this process is indefinitely repeated, the points common to AECB and the path from ^ to ^ on the great circle arcs will approach as near each other as we please, and the sum of these arcs will continually increase and approach AECB as a limit.
But the sum of the great circle arcs is always greater than AB.
Therefore, AB is less than AECB. q.e.d.
661. By the distance between two points on the surface of a sphere is meant the shortest distance; that is, the arc of a great circle joining them.
66^. The distance from the nearer pole of a circle to any point in its circumference is called the Polar Distance of the circle.
Proposition III
663. 1. Form a sphere and cut it by ^any plane; pass planes through the axis of the circle thus formed and any points in its circumference. What kind of arcs, then, connect the pole of the circle and the points of its circumference? How do these arcs compare? Then, how do the distances from the pole of a circle of a sphere to all the points in its circumference compare?
382
SOLID GEOMETRY. — BOOK X.
2. If the circle is a great circle, what part of a circumference is its polar distance ?
3. By passing planes form two equal circles of the same or of equal spheres. How do their polar distances compare ?
4. Select a point on the surface of a sphere which is at a quadrant's distance from each of two other points. Where is this point situated with reference to a pole of a great circle that passes through the other two points ?
Theorem, All points in the circumference of a circle of a sphere are equally distant from a pole of the circle.
■p'
Data : Any circle of a sphere, as ABC, and its poles, P and P'.
To prove all points in the circumference of ABC equally distant" from P and also from P\
Proof. Draw great circle arcs from P to any points in the circumference of ABC, as A, B, and C.
§§ 649, 652, PP' A. ABC at its center ;
.*. § 450, chords PA, PB, and PC are equal ;
hence, § 196, arcs PA, PB, and PC are equal.
In like manner, arcs P^A, P'B, and P'C may be proved equal.
But A, B, and C are any points in the circumference of ABC.
Hence, § 661, all points in the circumference of ABC are equally distant from P and als^ from P'.
Therefore, etc. q.e.d.
664. Cor. I. TJie polar distance of a great circle is a quadrant*
665. Cor. II. The polar distances of equal circles on the same^ or on equal spheres, are equal.
* In Spherical Geometry the term quadrant generally means the quadrant of a great circle.
SOLID GEOMETRY. — BOOK X. 333
666. Cor. III. A point, which is at the distance of a quadrant from each of iwo other points on the surface of a sphere^ is a pole of the great circle passing through those points.
667. Sch. I. By using the facts demonstrated in § 663 and in § 664 we may draw the circumferences of
small and great circles on the surface of a .-Jl
material sphere. , IV'^
To draw the circumference of a circle, take a cord equal to its polar distance, and, placing one end of it at the pole, cause a pencil held at the other to trace the cir- cumference, as in the fignire.
To describe the circumference of a great circle, a quadrant must be used as the arc.
668. Sch. II. By means of § QQQ we are enabled to pass the circumference of a great circle through any
two points, as A and B, on the surface of a material sphere in the following manner :
From each of the given points, as poles, and with a quadrant arc, draw arcs to inter- \^ sect, as at 0. The circumference described \ from this intersection with a quadrant arc will be the circumference required.
Proposition IV
669. 1. If a plane is perpendicular to a radius of a sphere at its ex- tremity, how many points do the sphere and the plane have in common ? What name is given to such a plane ?
2. How many points do a straight line and a sphere have in common, if the line is perpendicular to a radius of the sphere at its extremity? What name is given to such a line?
3. What is the direction of every plane or line, that is tangent to a sphere, with reference to the radius drawn to the point of contact?
4. If a straight line is tangent to any circle of a sphere, how does it lie with reference to a plane tangent to the sphere at the point of contact?
5. If a plane is tangent to a sphere, what is the relation to the sphere of any line drawn in that plane and through the point of contact?
6. If two straight lines are tangent to a sphere at the same point, what is the relation of the plane of those lin^s tQ the sphere ?
334
SOLID GEOMETRY. — BOOK X.
Theorem, A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.
'ml
J
Data : Any sphere, and a plane, as MN, perpendicular to a radius, as OP, at its extremity P.
To prove MN tangent to the sphere.
Proof. Take any point except P in MN, as A, and draw OA.
Then, §448, OP<OA',
point A is without the sphere. But A is any point in MN except P ;
every point in MN except P is without the sphere. Hence, § 638, MN is tangent to the sphere at P. q.e.d.
670. Cor. I. Any straight line perpendicular to a radius of a sphere at its extremity is tangent to the sphere.
671. Cor. II. A.iy plane or line tangent to a sphere is perpen- dicular to the radius drawn to the point of contact.
672. Cor. III. A straight line tangent to any circle of a sphere lies in the plane tangent to the sphere at the point of contact.
673. Cor. IV. Any straight line drawn in a tangent plane and through the point of contact is tangent to the sphere at that point.
674. Cor. V. Any two straight lines tangent to a sphere at the same point determine the tangent plane at that point.
Proposition V
675. Select any four points not in the same plane; form a tetrahe- dron of which these points are the vertices ; and at the centers of the circles circumscribed about any two of its faces erect perpendiculars. How do the distances from any point in either perpendicular to the vertices of the face to which it is perpendicular compare ? If these per-
SOLID GEOMETRY. — BOOK X. 335
pendiculars intersect, how, then, do the distances from their intersection to the four given points compare ? Is there any other point that is equi- distant from the four given points ? What surface, then, may be passed through these four points? How many such surfaces may be passed through them?
Theorem, Through any four points not in the same plane one spherical surface may be passed, and only one.
Data : Any four points not in the same plane, as A, B, c, D.
To prove that one spherical surface may be passed through A, B, C, B, and only one.
Proof. Suppose H and G to be the centers of circles circumscribed about the triangles BOB and ACB, respectively.
Draw HK± plane BOB, and GE ± plane ACB.
§ 450, every point in HK is equidistant from points B, C, B, and every point in GE is equidistant from points A, C, B.
Erom H and G draw lines to L, the middle point of CB.
Then, § 106, HL ± CB, and GL±CB',
.*. § 444, the plane through HL and GL is perpendicular to CB, and, § 483, this plane is perpendicular to planes BCB and ACB.
Const., GE J. Tpl^ne ACB Sit G',
.-. § 481, ' GE lies in the plane HLG.
In like manner it may be shown that HK lies in plane HLG.
Hence, tlie perpendiculars HK and GE lie in the same plane, and, being perpendicular to planes which are not parallel, they must intersect at some point, as at O.
Since O is an the perpendiculars HK and GE, it is equidistant from B, C, B, and from A, C, B.
Hence, O is equidistant from A, B, C, B, and the surface of the sphere, whose center is O and radius OB, will pass through the points A, B, C, B.
Now, § 450, the center of any sphere whose surface passes through the four points A, B, C, B must be in the perpendiculars HK and GE.
Hence, 0, the intersection of HK and GE, must be the center of the 07ily sphere whose surface can pass through Aj B, G, B.
Therefore, etc. q.e.d.
336 SOLID GEOMETRY. — BOOK X.
676. Cor. I. A sphere may he circumscribed about any tetra- hedron.
677. Cor. II. The four perperidiculars to the faces of a tetrahe- dron through their centers meet at the same point.
Proposition VI
678. Form any tetrahedron and pass planes bisecting any three of its dihedral angles which have one face in common. How do the distances from the point of intersection of these bisecting planes to the faces of the tetrahedron compare ? What figure, then, may be inscribed in any tetra- hedron ?
Theorem, A sphere may be inscribed in any tetrahedron. Ad aP
Datum : Any tetrahedron, as D-ABC.
To prove that a sphere may be inscribed in D-ABC.
Proof. Bisect any three of the dihedral angles which have one face common, as AB, BC, AC, by the planes GAB, OBC, OAC, respec- tively.
By § 488, every point in the plane OAB is equidistant from the faces ABC and ABD.
Also every point in plane OBC is equidistant from the faces ABC and BCD; and every point in the plane OAC is equidistant from the faces ABC and ACD.
Therefore, point 0, the intersection of these three planes, is equidistant from the four faces of the tetrahedron.
SOLID GEOMETRY. — BOOK X, 337
Hence, § 638, a sphere with 0 as a center and with a radius equal to the distance from 0 to any face will be tangent to each face, and, § 640, it will be inscribed in the tetrahedron.
Therefore, etc. q.e.d.
679. Cor. The six planes which bisect the six dihedral angles of a tetrahedron intersect in the same point. .
Proposition VII
680. Problem, To find the radius of a material spliere.
p
p
/
Datum: Any sphere, as APP'.
Required to find the radius of APP^.
Solution. From any point P as a pole describe any circumfer- ence on the surface. § 667. From any three points in this cir- cumference, as A, B, C, measure the chord distances AB, BC, AC. (Use compasses with curved branches.)
Construct the Aa'b'c' having its sides equal respectively to AB, BC, AC, and circumscribe about it a circle.
With the radius OB' as a side, and PB' equal to the chord of the arc joining P and B, as hypotenuse, construct the rt. Apb'O.
Draw a line from B' perpendicular to B'P, and produce it to meet PO produced in P'.
Then, PP' thus determined is equal to the diameter of the sphere, and its half, PG, is the required radius. q.e.f.
Proof. By the student.
Ex. 870. Equal straight lines whose extremities are in the surface of a sphere are equally distant from the center of the sphere.
Ex. 871. The six planes which bisect at right angles the six edges of a tetrahedron all intersect at the same point. milne's geom. — 22
338 SOLID GEOMETRY. — BOOK X.
SPHERICAL ANGLES AND POLYGONS
681. The angle betiveen two intersecting curves is the angle con- tained by the two tangents to the curves at their intersection.
682. The angle between two intersecting arcs of great ciicles is called a Spherical Angle.
683. A portion of the surface of a sphere bounded by three or more arcs of great circles is called a Spherical Polygon.
The bounding arcs are the sides of the poly- gon; the angles which they form are the angles of the polygon ; and the points of inter- section are the vertices of the polygon.
An arc of a great circle joining any two non-adjacent vertices of a spherical polygon is a diagonal.
684. The planes of the sides of a spherical polygon form a polyhedral angle whose vertex is the center of the sphere, and whose face angles are measured by the sides of the polygon.
O- ABODE is a polyhedral angle whose vertex O is the center of the sphere and whose face angles JE'OD, DOC, etc., are measured by the sides ED, DC, etc., of the spherical polygon ABODE.
685. A spherical polygon whose corresponding polyhedral angle is convex is called a Convex Spherical Polygon.
Spherical polygons will be regarded as convex unless otherwise specified.
686. A spherical polygon of three sides is called a Spherical Triangle.
A spherical triangle is right, oblique, equilateral, isosceles, etc., under the same conditions that plane triangles are right, oblique, etc.
687. The sides of a spherical polygon, being arcs, are usually measured in degrees, minutes, and seconds.
688. Any two points on the surface of a sphere may be joined by two arcs of a great circle, one of which will usually be greater and the other usually less than a semicircumference.
Unless otherwise stated the less arc is always meant.
SOLID GEOMETRY. — BOOK X 839
Proposition VIII
689. 1. Form a sphere and construct on it a spherical angle (§ 667) ; pass planes through the sides of the angle and the center of the sphere ; on the part thus cut out draw a great circle arc with the vertex of the angle as a pole, and draw the radii to the extremities of this arc. How does the angle between the radii compare with the angle contained by the tangents to the sides of the spherical angle at their intersection? What arc measures the angle between the radii? Then, what arc measures the given spherical angle?
2. How does a spherical angle compare with the dihedral angle formed by the planes of its sides?
Theorem, A spherical angle is measured hy the are of a great circle described from its vertex as a pole and irv- eluded hy its sides, produced if necessary.
Data: Any spherical angle, as APB, and the arc of a great circle, as AB, de- scribed from the vertex P, and included between the sides AP and BP.
To prove Z APB measured by arc AB.
Proof. Draw PT and PT' tangent to AP and J5P, respectively at P J also draw the radii OA, OB, and the diameter PPK
Const., § 205, PT±PP' in plane PAP*,
and since, data, PA is a quadrant,
O^ ± PP' in plane P^IP*; Why ?
.. § 71, PTWOA, '
and similarly, PT' II OB ;
.-.§469, Ztpt' = Zaob.
But, § 224, Z^O^'is measured by arc AB; *
Z TPT' is measured by arc AB; that is, § 681, Z APB is measured by arc AB.
Therefore, etc. ^ QJLD,
840 SOLID GEOMETRY, — BOOK X.
690. Cor. I. A spherical angle has the same measure as the di- hedral angle formed by the planes of its sides.
691. Cor. II. If two sides of a spherical triangle are quadrants, the third side measures the angle opposite.
692. Cor. III. If each side of a spherical triangle is a quadrant, each angle is a right angle.
693. Cor. IV. If two arcs of great circles cut each other, their vertical angles are equal.
694. Cor. V. The angles of a spherical polygon are equal to the dihedral angles between the planes of the sides of the polygon.
695. Sch. Since, §§ 684, 694, the sides and angles of a spheri- cal polygon have respectively the same measures as the face and dihedral angles of the corresponding polyhedral angle, we may, from any property of polyhedral angles, infer an analogous property of spherical polygons ; and conversely.
Proposition IX
696. 1. Forin a sphere and draw on it a spherical triangle; pass planes through the sides of the triangle and the center of the sphere, and thus cut out the corresponding trihedral angle. How does the sum of any two face angles of the trihedral compare with the third face angle? How, then, does the sum of any two sides of the spherical tri- angle compare with the third side ?
2. How does any side compare with the difference of the other two sides ?
Theoretn. The sum of any two sides of a spherical tri- angle is greater than the third side.
Data: Any spherical triangle, as ABC, on the sphere whose center is 0. ^
SOLID GEOMETRY. — BOOK X. 341
To prove the sum of any two sides, as AC -\- BCy greater than the third side AB.
Proof. In the corresponding trihedral angle 0-ABC, § 498, Z AOC + Z BOG is greater than Z AOB ;
.-. § 695, AO -{- BC is greater than AB.
Therefore, etc. - q.e.d.
697. Cor. I. A7iy side of a spherical triangle is greater than the difference of the other tivo sides.
Proposition X
698. Form a sphere and draw on it a spherical polygon ; pass planes through the sides of the polygon and the center of the sphere, and thus cut out the corresponding polyhedral angle. How does the sum of the face angles of the polyhedral compare with four right angles? How, then, does the sum of the sides of the spherical polygon compare with the circumference of a great circle ?
Theorem, The sutu of the sides of a spherical polygon is less than the circumference of a great circle.
Data: Any spherical polygon, as ABCD, on the sphere whose center is 0.
To prove AB + BC -{■ CD -{- DA < the circumference of a great circle.
Proof. In the corresponding polyhedral angle 0-ABCD, § 499, Z AOB + Z BOC + Z COD + Z DOA < 4 rt. ^ ;
•. § 695, AB + BC+CD-\-DA< the circum. of a great circle. .
Therefore, etc. q.e.d.
342 SOLID GEOMETRY. — BOOK X.
699. If from the vertices of a spherical triangle as poles arcs of great circles are described, these arcs form by their intersections a second triangle which is called the Polar Triangle of the first.
If ^ , ^, and C are the poles of the great circle arcs B'C, A'C, and^'i?', respectively, then, ^'5' C is the polar triangle of ABC.
If from A, B, and C as poles entire great circles instead of arcs are described, these circles will divide the surface of the sphere iuto eight spherical triangles.
Of these eight triangles, that one is the polar of ABC whose vertex A' corresponding to A lies on the same side of BC as the vertex A ; and in the same way the other corresponding vertices may be determined.
Proposition XI
700. On the surface of a sphere draw a spherical triangle ; draw also its polar triangle. Test the first triangle to see if it is the polar triangle of the second.
Theorem, If one of two spherical triangles is the polar triangle of the other, then the other is the polar triangle of that one.
Data : Any spherical triangle, as ABC, and •
its polar triangle, a'b'C'.
To prove ABC the polar triangle of A'b'C'.
Proof. § 699, A is the pole of arc B'c'-,
. . § 664, 5' is at a quadrant's distance from A.
Also, C is the pole of arc A'b'-,
B' is at a quadrant's distance from C.
Hence, § 666, B' is the pole of arc AC.
In like manner it may be shown that,
a' is the pole of arc BC, and C' the pole of arc AB.
Hence, § 699, ABC is the polar triangle of A^B'c*.
Therefore, etc. q.e.d.
SOLID GEOMETRY. — BOOK X. 343
Proposition XII
701. On the surface of a sphere, draw a spherical triangle and its polar triangle ; select an angle of one of these and extend its sides, if necessary, to meet the opposite side of the other triangle. What part of a circumference is the distance from each point of meeting to the farthest extremity of that opposite side ? Then, how^does the arc intercepted on that side by the sides of the given angle compare with two quadrants less the whole side, or with 180° less the whole side? How, then, does the measure of the given angle compare with the supplement of the opposite side of the polar triangle ?
Theorem. In two polar triangles any angle of the one is measured by the supplement of the opposite side of the other.
Data : Any two polar triangles, as ABC and a'b'c', and any angle of either triangle, as A.
To prove Z A measured by 180° — B'c'.
Proof. Produce the arcs AB and AC until they meet B'c' at the points G and H, respec- tively.
Since, data, A is the pole of arc GH,
§ 689, Z ^ is measured by GH.
Since, data, B' and C' are the poles of arcs AH and AG, respec- tively,
arcs b'h and C'G are quadrants.
Jlence, B'H+ c'G = ?i semicircumference ; that is, B'c' -^ GH=a, semicircumference = 180° ;
GH=1S0° -B'C'. But Z ^ is measured by GH.
Hence, Z ^ is measured by 180° — B'C'.
Therefore, etc. q.e.d.
702. Two polar triangles are also called Supplemental Triangles. For, if we denote the number of angle degrees in each angle by
the letter at the vertex, and the number of arc degrees in each opposite side by the corresponding small letter, we have th(' following relations :
,Za = 180° -a', Zb = 180° -b', Zc = 180° - c', 'z^' = 180°-a, Z^' = 180°-6, Z C' = 180° - c
344 SOLID GEOMETRY. — BOOK X,
Proposition XIII
703. On the surface of a sphere draw a spherical triangle ; draw also its polar triangle. What is the measure of each angle of the given tri- angle ? What, then, is the sum of the measures of its three angles ? Since the sum of the arcs which form the sides of the polar triangle is greater than an arc of 0° and (§ 698) less than an arc of 360°, what is the greatest number and also the least number of degrees that there can be in the sum of the angles of the given triangle or any spherical triangle? Express your conclusions in terms of right angles.
Theorem, The sum of the angles of a spherical triangle is greater than two and less than sijc right angles.
Data: Any spherical triangle, as ABC, j^
whose angles are A, B, and C.
To prove 1. Z ^ + Z if + Z C> 2 rt. zi.
2. Z ^ + Z 5 + Z C < 6 rt. Z.
Proof. 1. Construct A'B'&y the polar trian- gle oiABC, and denote the number of degrees in B'C', A'c', and A'B'hy a', h\ and c', respectively.
Then, § 701, Z J = 180°-a', Z 5=180°-5', and Z (7=180' .-. Ax. 2, Z^ + Z^ + Zc=540°-(a'4-&' + c').
But, § 698, 5'C' + ^'C' + A^B^ < the circum. of a great circle ; that is, a/ + 6' + c'<360°;
Z^-fZ^ + Zo 180°, or 2 rt. A.
2. a'-f-6'-hc'>0°;
hence, Z^ + Z^ + Z(7< 540°, or 6 rt. A
Therefore, etc. q.e.Dc
704. Cor. A spherical triangle may have tivo, or even three, right angles; or it may have two, or even three, obtuse angles.
Ex. 872. The sides of a spherical triangle are 65°, 86°, and 98°. What are the angles of its polar triangle ?
Ex. 873. The angles of a spherical triangle are 53°, 77°, and 92°. What are the sides of its polar triangle ?
Ex. 874. The angles of a spherical triangle are 65°, 80°, and 110°.- What are the sides of its polar triangle ?
kO
SOLID GEOMETRY. — BOOK X.
345
705. A spherical triangle having two right angles is said to be hirectangidar ; and one having three right angles is said to be trirectangular.
706. The excess of the sum of the angles of a spherical tri- angle over two right angles is called tKe Spherical Excess of the •triangle.
707. The excess of the sum of the angles of a spherical polygon over two right angles, taken as many times as the polygon has sides less two, is called the Spherical Excess of the polygon.
If a polygon has n sides, its spherical excess is equal to the sum of the spherical excesses of the n — 2 spherical triangles into which the polygon may be divided by diagonals from any vertex.
708. Spherical triangles in which the sides and angles of the one are equal respectively to the sides and angles of the other, but arranged in the reverse order, are called Symmetrical Spherical Triangles.
Two spherical triangles are symmetrical, when the vertices of one are at the ends of the diameters from the vertices of the other.
Triangles ABC and A'B'C are symmetrical spherical triangles.
Symmetrical spherical triangles are mutually equilateral and equiangular, and the equal sides are oppo- site the equal angles, yet they cannot generally be made to coincide.
To make the symmetrical triangles ABC and A'B'C coincide, any arc BC must be made to coincide with its equal B'C. This can be done in only two ways — with B either on B' or on C. When superposed with B on C, unless the triangles are isosceles, angles B and C are unequal and the triangles will not coincide ; with B on B', A and A' fall on opposite sides of B'C and the triangles will not coincide.
Symmetrical spherical triangles which are isosceles can b^ made to coincide.
346 SOLID GEOMETRY. — BOOK X.
Proposition XIV
709. On the surface of a sphere draw two symmetrical triangles. How do they compare in area? Are the triangles equal or equivalent?
Theorem. Two symmetrical spherical triangles are equivalent.
Data: Any two symmetrical spherical tri- angles, as ABC and A'b'c\
To prove A ABC ^ A a'b'C'.
Proof. Case I. When they are isosceles.
If isosceles, they may be made to coincide ; 2iYesi ABC = Siresi a' b'C'.
Case II. When the triangles are not isosceles.
Suppose P and P' to be the poles of the small circles passing lihrough the points A^ B, C, aud A', B', C', respectively.
Data, arcs AB, AC, 5C= arcs a'b', a'c', b'c', respectively; •. § 196, chords of arcs AB, AC, BC = chords of arcs A'B', A'c', B'C', ■respectively ; hence, § 107, the plane triangles formed by these chords are equal ; .-. § 208, the small circles through A, B, C, and A', b', C' are equal.
Draw the great circle arcs PA, PB, PC, p'a', P'b', p'C'.
Then, § 665, these arcs are equal.
Now, §§ 695, 501, the angles of APAB are equal to the angles of A P'a'b', respectively, and the equal parts of the triangles are in reverse order ;
.-. §§ 708, 686, A PAB and P'a'b' are symmetrical and isosceles, ind. Case I, area PAB = area p'a'b'.
In like manner, area PBC= area P'B'c', and area PAC= area P'A'c' ; area PAB + PBC + PAC = area p'a'b' + p'b'C' + P'A'c', IT area ABC = area A' B'c' ; that is, A ABC <^ A a'b'c'.
If the pole P should be without A ABC, then P' would be with- out A a'b'c', and each triangle would be equivalent to the sum of two isosceles triangles diminished by the third ; consequently, the result would be the same as before.
Therefore^ etc. 9.E.D.
SOLID GEOMETRY. — BOOK X, 347
Proposition XV
710. 1. On the same sphere, or on equal spheres, draw two spherical triangles having two sides and the included angle of one equal to the corresponding parts of the other, and arranged in the same order. Can the triangles be made to coincide? Then, how do they compare?
2. Draw two spherical triangles as befoi'e, but. with the given equal parts arranged in the reverse order ; draw another triangle symmetrical to one of these. How does it compare with the other? Then, are the given triangles equal or equivalent?
Theorem. Two triangles on the same, or on equal spheres, having two sides and the included angle of one equal to two sides and the included angle of the other, each to each, are either equal or equivalent.
*Data: Two spherical triangles, as ABC and DBF, in which AB = DE, AC — DF, and angle A = angle D.
Case I. When the given equal parts of the two triangles are arranged in the same order.
To prove A ABC = A DEF.
Proof. The A ABC can be applied to the A DFF, as in the cor- responding case of plane triangles, and they will coincide.
Hence, § 36, A ABC = A DFF.
Case II. When the given equal parts of the two triangles are arranged in reverse order, as in triangles ABC and A'b'c' in which AB = A'b', AC = A'C', and ZA=zZa'.
To prove A ABC o A a'b'c'.
Proof. Suppose the A DFF to be symmetrical with respect to the A A'b'c'.
Then, § 708, the sides and angles of A DFF are equal respec- tively to those of A A'B'C' ;
.-. in the A ABC and DFF, Z A = Z D, AB = DF, and AC = DF, and the equal parts are arranged in the same order ; .-. Case I, A ABC = A DFF.
But, § 709, A A'B'C' =0= A DFF.
Hence, A ABC =7- A a'b'c'. q.e.d.
348 SOLID GEOMETRY. — BOOK X.
Proposition XVI
711. 1. On the same sphere, or on equal spheres, draw two spherical triangles having a side and two adjacent angles of one equal to the cor- responding parts of the other, and arranged in the same order. Can these triangles be applied to each other so that they will coincide? Then, how do they compare ?
2. Draw two spherical triangles as before, but with the given equal parts arranged in the reverse order ; draw another triangle symmetrical to one of these. How does it compare with the other? Then, are the given triangles equal or equivalent ?
Theorem, Two triangles on the same sphere, or on equal spheres, having a side and two adjacent angles of one equal to a side and two adjacent angles of the other, each to each, are either equal or equivalent.
Proof. One of the triangles may be applied to the other, or to its symmetrical triangle, as in the corresponding case of plane triangles.
Therefore, etc. q.e.d.
Proposition XVII
712. On the same sphere, or on equal spheres, draw two mutually equilateral spherical triangles. How do the angles of one compare with the angles of the other? If the equal parts are arranged in the same order in each, how do the triangles compare ? If the equal parts are in reverse order, are the triangles equal or equivalent? >
Theorem, Two mutually equilateral triangles on the same sphere, or on equal spheres, are mutually equiangu- lar, and are either equal or equivalent.
Proof. By §§ 695, 501, the triangles are mutually equiangular; they are equal or symmetrical. Why?
If they are symmetrical, then, § 709, they are equivalent.
Hence, they are either equal or equivalent.
Therefore, etc. ' Q.e.d.
SOLID GEOMETRY. — BOOK X. ' 349
Proposition XVIII
713. 1. On the surface of a sphere, draw an isosceles spherical trian- gle ; draw the arc of a great circle from its vertex to the middle of the opposite side. In the two triangles thus formed, how do the sides of one compare with the sides of the other? Then, how do the angles of one compare with the angles of the other ? In the original triangle, how do the angles opposite the equal sides compare with each other ?
2. How does the great circle arc from the vertex to the middle of the base of an isosceles spherical triangle divide the vertical angle? What is its direction with reference to the base ? Into what kind of triangles does it divide the given triangle ?
Theorem, In an isosceles spherical triangle, the angles opposite the equal sides are equal.
Data: An isosceles spherical triangle, as ABC, in which AB = AC.
To prove /.B = /.c.
Proof. Draw the arc of a great circle, as AB, from the vertex A, bisecting the side BC.
Then, in A ABB and ACB,
AD is common, AB = AC,BB = BC; Why?
that is, the triangles are mutually equilateral ; .-. § 712, A ABB and ACB are mutually equiangular. Hence, Zb = ZC.
Therefore, etc. q.e.d.
714. Cor. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle of the base bisects the vertical, angle, is perpendicular to the base, and divides the triangle into two symmetrical triangles.
Ex. 875. ■ If the sides of a spherical triangle are 50°, 75°, and 110°. what are the angles of its polar triangle ?
Ex. 876. If the sides of a spherical triangle are 54°, 89°, and 103°, what is the spherical excess of its polar triangle ?
350 SOLID GEOMETRY. — BOOK X
Proposition XIX
715. On the surface of a spliere, draw two mutually equiangular trian- gles ; draw also their polars. How do the sides of their polars compare, each to each ? Then, how do the angles of the polars compare, each to each? How, then, do the sides of the given triangles compare, each to each ? If the equal parts in the given triangles are arranged in the same order in each, how do the triangles compare? If the equal parts are arranged in reverse order, are the triangles equal or equivalent ?
Theorem, Two mutually equiangular triangles on the same sphere, or on equal spheres, are mutually equilateral, and are either 'equal or equivalent.
Data : Two spherical triangles, as A and B, that are mutually- equiangular.
To prove A A and B mutually equilateral, and either equal or equivalent.
Proof. Suppose A ^' to be the polar of A A, and A 5' the polar ^iAB.
Data, A A and B are mutually equiangular ;
• •'. § 701, their polar A, A' and B\ are mutually equilateral; hence, § 712, A A' and B^ are mutually equiangular ; .-. § 701, A A and B are mutually equilateral.
Hence, § 712, A A and B are either equal or equivalent.
Therefore, etc. , q.e.d.
716. Cor. I. If two angles of a spherical triangle are equal, the sides opposite these angles are equal, and the triangle is isosceles.
717. Cor. II. If three planes are passed through /^VrPX the center of a sphere, each perpendicular to the other /:-"'h'i~~ "A two, they divide the surface of the sphere into eight \~tJ^ — / equal tri rectangular triangles. § 695 x^ilX^
SOLID GEOMETRY, — BOOK X. 351
Proposition XX
718. 1. On the surface of a sphere draw a spherical triangle, two of whose angles are unequal. How do the sides opposite these angles com- pare? Which one is the greater?
2. Draw a spherical triangle, two of whose sides are unequal. How do the angles opposite these sides compare? ^Which one is the greater?
Theorem, If two angles of a spherical triangle are un- equal, the sides opposite are unequal, and the greater side is opposite the greater angle; conversely, if two sides are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side.
Data: A spherical triangle, as ABC, in which the angle ACB is greater than the angle ABC.
To prove AB > AC.
Proof. Draw CD, the arc of a great circle, making Z BCD — /.h-
Then, § 716, DB = CD.
Now, § 696, AD^CD> AC,
AD-\-DB> AC,
or ■ AB>AC.
Conversely: Data: A spherical triangle, as ABC, in which the side AB is greater than the side AC.
To prove Z.ACB greater than Z B,
Proof. If /.acb = Zb,
then, § 716, AB = AC,
which is contrary to data.
If Z. ACB is less than Z B,
then, Z £ is greater than Z ACB,
and AC>AB,
which is also contrary to data.
Therefore, both hypotheses, namely, that ^ACB = Ab and that /.ACB is less than /.B, are untenable.
Consequently, Z ACB is greater than Z B.
Therefore, etc. q.b.j>.
352 SOLID GEOMETRY. — BOOK X,
SPHERICAL MEASUREMENTS
719. The portion of the surface of a sphere included between two parallel planes is called a Zone.
The perpendicular distance between the planes is the altitude of the zone, and the circumferences of the sections made by the planes are called the bases of the zone.
If one of the parallel planes is tangent to the sphere, the zone is called a zone of one base.
ABCD is a zone of the sphere.
720. The portion of the surface of a sphere bounded by two semicircumferences of great circles is called a Lune.
The angle between the semicircumferences jj^
which form its boundaries, is called the angle of the lune.
ABCD is a lune of which BAD is the angle.
721. Lunes on the same sphere, or on equal spheres, having equal angles may be made to coincide, and are equal.
722. A convenient unit of measure for the surfaces of spherical figures is the spherical degree^ which is equal to -^^ of the surface of a hemisphere.
Like the unit of arcs, it is not a unit of fixed magnitude, but depends upon the size of the sphere upon which the figure is drawn.
It may be conceived of as a birectangular spherical triangle whose third angle is an angle of one degree.
The distinction between the three different uses of the term degree should be kept clearly in mind; an angular degree is a difference of direction between two lines, and it is the 360th part of the total angular magnitude about a point in a plane (§ 35) ; an arc degree is a line, which is the 360th part of the circumfer- ence of a circle (§ 224) ; a spherical degree is a surface, which is the 360 th part of the surface of a hemisphere, or the 720th part of the surface of a sphere.
SOLID GEOMETRY. — BOOK X. 353
Proposition XXI
723. Represent an axis and a li iie oblique to it, but not meeting it ; draw lines from the extremities and middle point of this line perpendicular to the axis ; from the nearer extremity draw a line parallel to the axis ; also a line perpendicular to the given line at its middle point and terminating in the axis. If the given line revolves about the axis, what kind of a surface will it generate ? To what is this surface equivalent ? (§ 628) By means of the proportion of lines from similar right tri- angles, express the surface in terms of the projection of the given line on the axis arid the circumference of a circle whose radius is the perpen- dicular from the middle point of the given line. Would this result hold true, if the line should meet the axis or be parallel to it ?
Theorem, The surface generated hy a straight line re- volving about an axis in its plane is equivalent to the rectangle formed hy the projection of the line on the axis and the circumference whose radius is a perpendicular erected at the middle point of the line and terminated hy the axis.
Data: Any line, as AB, revolving about an u
axis, as MN\ its projection upon MN, as C7); a __
and EO perpendicular to AB at its middle point /l
and terminating in the axis. "7^^ —
To prove surface AB ^ rect. CD • 2 ttEO. Z__.i__^
Proof. Draw EF ± MN and AK II MN. ^ ^
If AB neither meets nor is parallel to 'MN it generates the lateral surface of a frustum of a cone of revolution whose slant height is AB and axis CZ)j .-. § 628, surface AB ^ rect. AB • 2 ttEF.
§ 307, A ABK and EOF are similar,
and AB: AK=EO: EF;
rect. AB' EF^ rect. AK - EO.
But, § 151, AK= CD;
hence, rect. AB • EF o= rect. CD • EO,
and rect. AB • 2 irEF ^ rect. CD • 2 ttEO ;
that is, surface AB =o rect. CD • 2 -n-EO.
If AB meets axis MN, ot is parallel to it, a conical or a cylindri cal surface is generated, and the truth of the theorem follows.
Therefore, etc. q.e.d.
milne's gbom. — 23
C
N
354 SOLID GEOMETRY. — BOOK X,
Proposition XXII
724. Draw a semicircumference and inscribe in it a regular semi- polygon. How does the sum of the projections of the sides of the poly- gon on the diameter of the semicircle compare with the diameter ? How do the perpendiculars to the sides of the polygon at their middle points compare in length, if they terminate in the diameter ? If the figure is revolved about the diameter as an axis, to what is the surface generated by the perimeter of the semipolygon equivalent ? How does the perim- eter of the semipolygon at its limit compare with the semicircumference, if the number of its sides is indefinitely increased? What is the limit of the perpendicular to the middle point of a side of the semipolygon? How, then, does the surface of a sphere compare with the rectangle formed by its diameter and the circumference of a great circle ?
Theoi'eiti, The surface of a sphere is equivalent to the rectangle formed hy its diameter and the circunfiference of ^ great circle.
Data: A sphere, whose center is 0, generated by
the revolution of the semicircle ABCB about the -^
diameter AD. /
Denote the surface of the sphere by s, and its 1
radius by i?. ^
To prove S =0= rect. AD - 2 irE.
Proof. Inscribe in the semicircle half of a regular polygon of An even number of sides, as ABCD, and let S' denote the surface generated by its sides.
Draw BE and CF ± AD, and the perpendiculars from 0 to the chordsf AB, BC, and CD.
§§ 202, 200, these perpendiculars are equal, and bisect the chords.
Then, § 723, surface AB =0 rect. AE - 2 ttOH,
surface BC ^ rect. EF • 2 irOH,
and surface CD ^ rect. FD • 2.TT0H.
But the sum of the projections AE, EF, and FD equals the diam- eter AD ;
S' <> rect. ^z> . 2 ttO^.
Now, if the number of sides of the inscribed semipolygon is indefinitely increased,
D
SOLID GEOMETRY. — BOOK X. 355
§ 392, the semiperimeter will approach the semicircumference as its limit ;
OH will approach R as its limit,
and s' will approach S as its limit.
But, however great the number of sides of the semipolygon,
S' =0 rect. AD'2 7rbH.
Hence, § 326, s ^ rect. AD - 2 irB.
Therefore, etc. q.e.d.
725. Cor. I. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle.
726. Cor. II. § 725, area = AB x 27rR = 2 R x 2 7rR = 4:7rR^',
that is, the area of the surface of a sphere is equal to the area of four great circles.
727. Cor. III. Let R and i?' denote the radii, D and D' the diameters, and A and A' the areas of the surfaces of two spheres.
Then, § 726, ^ = 4 ttR'', and ^' = 4 irR'^ •
A: A' = 4.7rR:' lA-rrR" = R' : R" = B'-. D" ; '
that is, the areas of the surfaces of two spheres are to each other as the squares of their radii, or as the squares of their diameters.
728. Cor. IV. Area of a zone, as BC =:EF x2tvR\
that is, the area of a zone is equal to the product of its altitude by the circumference of a great circle.
729. Cor. V. Zones on the same sphere, or on equal spheres^ aru to each other as their altitudes.
730. Cor. VI. § 728, area of a zone of one base, as
AB = AE X 2 ttR = ttAE X AB. Draw BB. Then, § 313, AE x AB = Iff ;
area of zone AB = irAB ; that is, the area of a zone of one base is equal to the area of a circle whose radius is the chord of the generating arc.
356 ^ SOLID GEOMETRY. — BOOK X,
Proposition XXIII
731. Divide the surface of a sphere into hemispheres by a great circle ; on one of the hemispheres form two opposite triangles by draw- ing two great circle arcs to intersect; complete the circumferences of which these arcs are parts. By comparing one of these opposite triangles with a triangle on the other hemisphere that completes a lune of which the other of the given triangles is a part, discover how the sum of the given triangles compares with a lune whose angie is the angle between the given arcs.
Theorein. If two arcs of great, circles intersect on the surface of a hemisphere, the sum of the two opposite tri- angles thus formed is equivalent to a lune whose angle is the angle between the given arcs.
Data: Opposite A, as AEB and DUCy formed by two great circle arcs, as AED and BEC, on the hemisphere E-ABDC.
To prove A AEB -\- A DEC ^ lune AEBF.
Proof. Produce arcs AED and BEC around the sphere intersecting as at F.
§ QbQj arc BE = arc AE (each being the supplement of arc AE), arc CE = arc BE (each being the supplement of arc BE), and, § 693, Z DEC = Z AEB = Z AEB ;
.-.§710, ADEC =o=AAFB.
Adding A AEB to each side of this expression of equivalence,
A AEB + A DEC ^ A AEB -f- A AFB'. Hence, A AEB + A DEC =o lune AEBF. q.e.d.
Proposition XXIV
732. On the surface of a sphere draw a lune whose angle is to four right angles as 3 : 12 ; from the vertex of its angle as a pole describe the circumference of a great circle. ^Vhat is the ratio of the arc included between the sides of the lune to the whole circumference? Divide the circumference into 12 equal parts and through the points of division and the poles pass great circle arcs. Into how many equal lunes do these arcs divide the surface of the sphere ? The given lune ? How, then, does the ratio of the given lune to the surface of the sphere compare with the ratio of the angle of the lune to four right angles ?
SOLID GEOMETRY. — BOOK X, 357
Theorem s A lune is to the surface of a sphere as the angle of the lune is to four right angles.
Data: A lime, as ACFD, whose angle is OAl), on the sphere whose center is O. ^
Denote the lune by L and the surface /^''^^^^X
of the sphere by S.^ /I
To prove L: S = Z GAD : 4 rt. A. J'""""" J
Proof. With ^ as a pole, describe the \^ I
circumference of a great circle BCEH. \ I
Then, § 689, arc CD measures Z CAD, and ^---_j^5^
circumference BCEH measures 4 rt. A. ^
Suppose arc CD and BCEH have a common unit of measure, as CJ, contained in CD m times and in BCEH n times.
Then, CD : BCEH = m : n,
or Z CAD : 4 rt. A = m:n.
Beginning at C, divide BCEH into parts, each equal to the unit of measure CJ, and through the points of division and the poles, A and F, of this circumference pass great circles.
By §§ 689, 721, these circles divide the whole surface of the sphere into n equal lunes of which the given lune contains m.
Then, L-. S = m'.n.
Hence, L: S = Z CAD : 4 rt. A.
By the method of limits exemplified in § 223, the same may be proved, when arc CD and BCEH are incommensurable.
Therefore, etc. q.e.d.
733. Cor. I. Let A denote the degrees in the angle of a lune. Then, L : S = A : 360°.
Since, § 722, S contains 720 spherical degrees,- i : 720 = ^ : 360 ; whence, L = 2 A-,
that is, the numerical measure of a lune expressed in spherical degrees is twice the numerical measure of its angle expressed in angular degrees.
734. Cor. II. Lunes on the same sphere, or on equal spheres, are to each other as their angles.
358 SOLID GEOMETRY.^ BOOK X.
Proposition XXV
735. Oil a sphere draw a spherical triangle and complete the great circles whose arcs are its sides. How many triangles having a common vertex with the given triangle occupy the sm-face of a hemisphere? Since the given triangle plus any one of the others is equivalent to a lune whose angle is equal to one of the angles of the given triangle, or to twice as many spherical degrees as that angle contains angular degrees, how does three times the given triangle plus the other three compare with twice the number of spherical degrees that there are angular de- grees in the angles of the given triangle? How many spherical degrees are there in the four triangles occupying the surface of the hemisphere? Then, discover how the number of spherical degrees in the given triangle compares with the sum of the angular degrees in its angles less 180°, that is, w4th its spherical excess.
Theorem, A spherical triangle is equivalent to as many spherical degrees as there are angular degrees in its spJieri- eal excess.
Data : A spherical triangle, as ABC, whose spherical excess is E degrees.
To prove AABCoE spherical degrees.
Proof. Complete the great circles whose arcs are sides of A ABC.
These circles divide the surface of the sphere into eight spherical triangles, any four of which having a common vertex, as A, form the surface of a hemisphere, whose measure is 360 spherical degrees.
§ 731, A ABC -\- Aab'c' ^2i lune whose angle equals angle A ; ,-. § 733, A ABC + A AB'C' <i= 2 A spherical degrees. (1)
In like manner, A ABC + A AB'C o= 2 5 spherical degrees, (2) and A ABC + A ABC' ^2c spherical degrees. (3)
Adding (1), (2), and (3), 3 A ABC -f A AB'C' 4- A AB'C -f- A ABC' =o 2 (^ -f J5 + C7) sph. deg.
But A ^J5C+A^5'C''4-A^fi'a4-A^£C'= 360 spherical degrees; .-. 2 A^5C7-|-360 spherical degrees =c= 2(^14- 5 4-C) spherical degrees ; hence, A ABC o (A + B -\- C — 180) spherical degrees ;
that is, § 706, Aabc^e spherical degrees. q.e.p.
SOLID GEOMETRY.^ BOOK X. 359
Proposition XXVI
736. Draw any spherical polygon- and divide it into spherical tri Angles by diagonals from any vertex. To how many spherical degrees is each triangle equivalent ? How does the number of spherical degrees in the sum of the triangles compare with the number of angular de- grees in the sum of the spherical excesses* of the triangles? To how many spherical degrees, then, is any spherical polygon equivalent?
Theorem^ Any spherical -polygon is equivalent to as many spherical degrees as there are angular degrees in its spherical excess.
Data: Any spherical polygon, as ABCDF, / whose spherical excess is E degrees. ^^
To prove ABCDF =c= E spherical degrees.
Proof. Divide the polygon into spherical triangles by diagonals /rom any vertex, as A.
By § 735,- each triangle is equivalent to as many spherical de- grees as there are angular degrees in its spherical excess.
Hence, the polygon is equivalent to as many spherical degrees as there are angular degrees in the sum of the spherical excesses of the triangles ; that is, § 707, in the spherical excess of the polygon.
Hence, J^CDi^^-E spherical degrees.
Therefore, etc. q.e.d.
•
737. Cor. The area of any spherical polygon is to the area of (he surface of the sphere as the number which expresses its spherical excess is to 720.
Ex. 877. The angle of a lune is 40°. What part of the surface of the sphere is the lune ?
Ex. 878. What is the area of a spherical triangle whose angles are 86°, 120°, and 110°, if the radius of the sphere is lO^m ?
Ex. 879. The area of the surface of a sphere is 160 sq. in. ; the angles of a spherical triangle on this sphere are 93°, 117°, and 132°. What is the area of the triangle ?
Ex. 880. Two spherical triangles on the same sphere, or on equal spheres, are equivalent, if the perimeters of their polar triangles are ec^ual.
360
SOLID GEOMETRY. — BOOK X.
738. A solid bounded by a spherical polygon and the planes of its sides is called a Spherical Pyramid.
The center of the sphere is the vertex of the ^ ^"^.-4 pyramid, and the spherical polygon is its base.
0- ABODE is a spherical pyramid whose vertex is 0 and base ABODE.
739. The portion of a sphere contained between two parallel planes is called a Spherical Segment.
The sections made by the parallel planes are the bases of the spherical segment ; the perpendicular distance between its bases is the altitude of the segment.
If one of the parallel planes is tangent to the sphere, the seg- ment is called a segment of one base.
740. The portion of a sphere bounded by a lune and the planes of its sides is called a Spherical Wedge, or Ungula.
741. The portion of a sphere generated by the revolution of a circular sector about a diameter^ of the circle is called a Spherical Sector.
The zone generated by the arc of the circu- lar sector is the base of the spherical sector.
a
742. MNEFAB is a semicircle, AD and BC are lines from the semicircumference perpendicular to the diameter MN, and OE and OF are radii. Then, if the semicircle is revolved about MN as an axis,'it generates a sphere.
The arc AB generates a zone whose altitude is DC, and whose bases are the circumferences generated by the points A and B.
The figure ABCD generates a spherical segment whose altitude is DC and whose bases are the circles generated by AD and BC.
The arc BM generates a zone of one base, and the figure BCM a spherical segment of one base.
The circular sector OEF generates a spherical sector whose bound- ing surfaces are its base, the zone generated by the arc EFj and the conical surfaces generated by tha radii OE and OF.
SOLID GEOMETRY. — BOOK X,
361
Proposition XXVII
743. Represent a polyhedron circumscribed about a sphere. If pyra- mids are formed having the faces of the polyhedron as bases and the center of the sphere as a common vertex, how will the altitudes of these pyramids compare with each other and with the radius of the sphere ? What is the volume of each pyramid? What, then, is the volume of the sum of the pyramids? If the number of faces of the polyhedron is indefinitely increased, how will its volume compare with the volume of the sphere ? To what, then, is the volume of a sphere equal ?
Theorem, The volume of a sphere is equal to the prod- uct of its surface by one third of its radius.
D
^^^m.
/
I
H B
Data : A sphere whose center is 0, surface S, and radius E.
Denote its volume by F.
To prove V=S x\R.
Proof. Circumscribe about the sphere any polyhedron, as D-ABCj and denote its surface by s' and its volume by v'.
Form pyramids, as 0-ABC,,etG., having the faces of the poly- hedron as bases and the center of the sphere as a common vertex.
Then these pyramids will have a common altitude equal to R, and, § 560, the volume of each pyramid = its base x I- R.
V'
S' x^R.
Now, if the number of pyramids is indefinitely increased by passing planes tangent to the sphere at the points where the edges of the pyramids cut the surface of the sphere,
-s' approaches S as its limit} .-. v' approaches V as its limit.
362 SOLID GEOMETRY.-^ BOOK X.
But, however great the number of pyramids,
r' = s' x^R. Hence, § 222, V = s x^B. q.e.d.
744. Formulae : F=sx^i? = | ttR^ = | TrD\
745. Cor. I. Let R, R' denote the radii, n, d' the diameters, and V, r' the volumes, respectively, of two spheres.
Then, § 744, F = | ttR^ and f' = | ttR'^ ;
F: V' = ^ttR^ : ^ttR'^ = R^ : R'^ = D^ : D'^;
that is, the volumes of tivo spheres are to each other as the cubes of their radii, or as the cubes of their diameters.
746. Cor. II. The volume of a spherical pyramid is equal to the product of its base by one third of the radius of the sphere.
747. Cor. III. The volume of a spherical sector is equal to the product of the zone which forms its base by one third of the radius of the sphere.
748. Formula: Let R denote the radius of a sphere, G the circumference of a great circle, F the volume of a spherical sector, and H and z the altitude and area, respectively, of the corre- sponding zone.
Then, since G = 2irR, and z = 2 ttRH, f = | ttR^H,
Proposition XXVIII
749. Draw a semicircle ; to the extremities of any arc of the semicir- cumference draw radii, and from their extremities draw lines perpendicular to the diameter. If this figure is revolved about the diameter as an axis, what kind of a solid is generated by the part included between the per- pendiculars and the given arc ? Between the radii and the given arc ? Between each radius and the perpendicular from its extremity? Find an expression for the volume of the spherical segment in terms of the spherical sector and the two cones.
Theorem, The volume of a spherical segment is equal to one half the product of its altitude hy the sum of its hoses, plus the volume of a sphere of which that altitude is a diameter.
SOLID GEOMETRY. — BOOK X,
363
Data: A spherical segment, as that generated by the revolution of ABCD about MN as an axis.
Denote the volume of the segment by F; its altitude CD, by H\ the radii of its bases AD and BC, by r and r', respectively ; and the radius of the semicircle by R.
To prove F = | H(Tr7^ + Trr'^) + i- 7rH\
Proof. Draw the radii OA and OB.
The volume of the segment generated by ABCD equals the volume of the spherical sector generated by OAB plus the volume of the cone generated by OBC minus the volume of the cone generated by OAD-,
.' §§748,629, V=^TrR''H^\irr^''OC-\7r7^0D.
But
H= OC — OD,
DC , and R^ — 7^= OD
Then, F= -l-7r)2 R\OC - 0D)-\-{R'^ - OC^)OG-(R^ - OD^)OD\ = ^Tr\2 I^(OC - 0D)+ RXOC - OD)-(OC^ -OD^)l = I 7rH\3 R:'—(0a^ + 0(7 X OD -f 0D^)\.
But, § 348, (OC ~ ODf= 0& - 2 OC y.0D^0D^=H^\
0C^+ OC X OD -\- OD^ = f{0(f -\- OD^)--^
= 3i2«-f(r^^-r«)-^.
Hence,
Therefore, etc.
Q.E.D,
750. Cor. Let the segment be a segment of one base, as that generated by MBAD.
Then, the radius / = 0,
and V=\Trr^H+^TrH^]
that is, the volume of a spherical segment of one base is equal to one half the volume of the cylinder having the same base and the same altitude plus the volume of a sphere of which that altitude is the diameter.
364
SOLID GEOMETRY. — BOOK X,
751.
B = base.
D = diameter.
R = radius.
r = radius of lower base.
r' = radius of upper base:
Sphere.
FORMULA
Notation
A=4.7rR' .
r=sxiR
Zone.
A = 2 ttRH
H= altitude.
S = surface.
A — area (or area of surface).
V = volume.
Spherical Pyramid.
V=B x^R
Spherical Sector.
V=B x^R
V=^7r^H .....
Spherical Segment.
§726 §743 §744 §744
§728
§746
§747 §748
§749
SUPPLEMENTARY EXERCISES
Ex. 881. What is the volume of a sphere whose radius is 9 in. ?
Ex. 882. The circumference of a great circle of a sphere is 36 ft. What is the area of the surface of the sphere ?
Ex. 883. The diameter of a sphere is 13^™. How many cubic decimeters does it contain ?
Ex. 884. The volume of a sphere is ISTO^" ™. What is its radius ?
Ex. 885. The area of the surface of a sphere is 69 sq. ft. What is its diameter ?
SOLID GEOMETRY. — BOOK X. 366
Ex. 886. The edge of a cube is 16cm. what is the volume of the circum- scribed sphere ?
Ex. 887. If the sides of a spherical triangle are 75°, 93°, and 110°, what is the spherical excess of its polar triangle ?
Ex. 888. The angles of a spherical triangle are 98°, 110°, and 160°. What is the area of a symmetrical triangle on the same sphere, the radius being 12'»?
Ex. 889. Find the volume of a triangular spherical pyramid, the angles of the base being 58°, 116°, and 145°, and the diameter of the sphere being 20 in.
Ex. 890. What is the area of a lune whose angle is 60° on the surface of a sphere whose radius is 6 in. ?
Ex. 691. What is the area of a zone whose altitude is 3^™ on the surface of a sphere whose radius is %^^ ?
Ex. 892. What is the volume of a spherical sector whose altitude is 3.5", if the radius of the sphere is 11™ ?
Ex. 893. What is the volume of a spherical wedge whose angle is 72°, the volume of the sphere being 1728 cu. in. ?
Ex. 894. What is the area of a zone of one base, if the chord of its generating arc is 13*^*^^ ?
Ex. 895. The area of a zone of a sphere 20^™ in diameter is 1508<i*i™ What is the altitude of the zone ?
Ex. 896. The angles of the base of a triangular spherical pyramid are 90°, 121°, and 135°. What is the volume of the pyramid, the volume of the sphere being 194 cu. in. ?
Ex. 897. Spherical polygons are to each other as their spherical excesses.
Ex. 898. The base of a spherical pyramid is a trirectangular triangle. What part of the sphere is the pyramid ?
Ex. 899. The surface of a sphere is equivalent to the lateral surface of the circumscribing cylinder.
Ex. 900. What is the spherical excess of a triangle whose area is 261 .8 sq. in. , if the radius of the sphere is 10 in. ?
Ex. 901. A lune and a trirectangular spherical triangle on the same sphere are to each other as the angle of the lune is to an angle of 45°.
Ex. 902. Trirectangular triangles on equal spheres are equal.
Ex. 903. The diameters of two spheres are 12 in. and 14 in. respectively. What is the ratio of their surfaces ? What is the ratio of their volumes ?
Ex. 904. The areas of the surfaces of two spheres are as 144 to* 24. What is the ratio of their diameters ? What is the ratio of their volumes ?
Ex. 905. The diameters of the sim and earth are in the ratio of 109 : 1. What is the ratio of their volumes ?
366 SOLID GEOMETRY. — BOOK X.
Ex. 906. How many quarts of water will a hemispherical kettle hold, if its inside diameter is 12 in. ?
Ex. 907. If lines are drawn from any point in the surface of a sphere to the ends of a diameter, they are perpendicular to each other.
Ex. 908. What is the circumference of a small circle of a sphere whose diameter is 9^™, if the circle is at a distance of S^m from the center ? •
Ex. 909. The 'dihedral angles of a spherical pyramid are 40°, 80°, and 120°, and its edge is 9 ft. What is the volume of the pyramid ?
Ex. 910. Tlie dihedrals of a trihedral angle whose vertex is at the centei of a sphere are 75°, 90°, and 130°. What is the volume of the part of the sphere included by the faces of the trihedral, the radius of the sphere being
Ex. 911. What is the radius of a sphere which is equivalent to the sum of two spheres whose radii are respectively 4 in. and 7 in. '?
Ex. 912. How many cubic decimeters does a segment of a single base contain, if it is cut from a sphere 12<i«i in diameter, the altitude of the seg- ment being ^^^ ?
Ex. 913. In a sphere whose diameter is 20 ft. , what is the volume of a segment, the bases of which are on the same side of the center, one 3 ft. and the other 5 ft. from it ?
Ex. 914. Find the area of the surface of a sphere inscribed in a cube whose surface is 726 sq. ft.
Ex. 915. A trirectangular triangle and a lune on the same sphere are in the ratio of 14 : 9. What is the angle of the Imie, and what part of the sur- face of the sphere is the lun€ ?
Ex. 916. Find the area of a spherical quadrilateral whose angles are 120°, 130°, 140°, and 150°, the volume of the sphere being 1000 cu. ft.
Ex. 917. The base of .a cone of revolution is the gi'eat circle of a sphere, and its altitude is the radius of the sphere. What is the ratio of the surface of the sphere to the lateral surface of the cone ?
Ex. 918. The base of a cone is equal to a great circle of a sphere, and its altitude is equal to the diameter of the sphere. What is -the ratio of their volumes ?
Ex. 919. Find the altitude of a zone whose area is equal ti that of a great circle of the sphere, the radius of the sphere being 8^"'.
Ex. 920. How many spherical bullets^ in. in diameter can be molded from a spherical piece of lead \ ft. in diameter?
Ex. 921. A cannon ball put into a cylindrical tub 2 ft. in diameter causes the water in the tub to rise 2 in. What is the diameter of the cannon ball ?
Ex. 922. A sectiion parallel to the base of a hemisphere bisects its altitude. What is the ratio of the volumes of the spherical segments thus formed ?
Ex. 923. The volume of a sphere is to that of the circumscribed cube as ir is to 6.
SOLID GEOMETRY. — BOOK X. 367
Ex. 924. The volume of a sphere is to that of the inscribed cube as ir
is to — .
V3
Ex. 925. The surface of a sphere is to the total surface of the circum- scrihing cylinder as 2 is to 3.
Ex. 926. The volume of a sphere is to the volume of a circumscribing cylinder as 2 is to 3.
Ex. 927. A sphere is cut by five parallel planes at the distance of 2*'", 3dm^ 4dm, and 5<im from each other, respectively. What are the relative areas of the zones included between the planes ?
Ex. 928. The sides opposite the equal angles of a birectangular spherical triangle are quadrants.
Ex. 929. The slant height of a cone of revolution is equal to the diameter of its base. What is the ratio of its total area to that of the inscribed sphere ?
Ex. 930. The smallest circle whose plane passes through a given point within a sphere is that one whose plane is perpendicular to the radius through the given point.
Ex. 931. The intersection of the surfaces of two spheres is the circum- ference of a circle whose plane is at right angles to the line joining the centers of the spheres, and whose center is on that line.
Ex. 932. \Yhat is the area of the circle of intersection of two spheres whose radii are respectively 5^"! and 8<i™, if their centers are 10^™ apart ?
Ex. 933. What is the weight of an iron ball, the area of whose surface is 28q m^ the specific gravity of iron being 7.5 ?
Ex. 934. If the exterior diameter of a spherical shell is 12 in., what should be the thickness of its wall in order that it may contain 696. 9 cu. in. ?
Ex. 935. What is the weight of a hollow iron shell whose wall is 2 in. thick, if it will hold 31| pounds of water, the specific gravity of iron being 7.5 ?
Ex. 936. An equilateral triangle revolves about its altitude. Compare the volumes of the solids generated by the triangle, the inscribed circle, and the circumscribed circle.
Ex. 937. From a sphere whose surface is 69 sq. ft. a segment of one base is cut, which has an altitude of 3 ft. What is the convex surface of the segment ?
Ex. 938. What is the radius of a sphere inscribed in a regular tetrahedron whose entire area is 16 sq. ft. ?
Ex. 939. What is the area of the surface of a sphere inscribed in a regu- lar tetrahedron whose edge is 6 in. ?
Ex. 940. How much of the surface of the earth could a man see, if he were at the distance of a diameter above it ?
Ex. 941. • How far from the surface of the earth must a man be in order that he may see one fifth of it ?
368 SOLID GEOMETRY. — BOOK X.
Ex. 942. All arcs of great circles drawn through the pole of a given great circle are perpendicular to its circumference.
Ex. 943. The sum of the squares of three chords perpendicular to each other at any point in the surface of a sphere is equal to the square of the diameter.
Ex. 944. If a zone of one base is a mean proportional between the remaining surface of the sphere and its total surface, how far is the base of the zone from the center of the sphere ?
Ex. 945. If any number of lines in space meet in a point, the feet of the perpendiculars drawn to these lines from another point lie in the surface of a sphere.
PROBLEMS OF CONSTRUCTION
Ex. 946. Bisect an arc of a great circle.
Ex. 947. Bisect a spherical angle.
Ex. 948. At a given point in a given arc of a great circle construct a spherical angle equal to a given spherical angle.
Ex. 949. Construct a spherical triangle, the poles of the respective sides being given.
Ex. 950. Construct a spherical triangle, having given two sides and the included angle.
Ex. 951. Construct a spherical triangle, having given a side and two adja- cent angles.
Ex. 952.* Construct a spherical triangle, having given the three sides.
Ex. 953. Construct a spherical triangle, having given the three angles.
Ex. 954. Draw an arc of a great circle perpendicular to a given spherical arc from a point without.
Ex. 955. Draw an arc of a great circle perpendicular to a given spherical arc at a point in it.
Ex. 956. Pass a plane tangent to a sphere at a given point on the surface of the sphere.
Ex. 957. Pass a plane tangent to a sphere through a given straight line without the sphere.
Ex. 958. Cut a given sphere by a plane passing through a given straight line so that the section shall have a given radius.
Ex. 959. Through a given point on a sphere draw a great circle tangent to a given small circle.
Ex. 960. Through a given point on a sphere draw a great circle tangent to two equal small circles whose planes are parallel.
Ex. 961. Describe a circle to pass through three'given points on the sur- face of a sphere.
Ex. 962. Circumscribe a circle about a given spherical triangle.
GEOME TR Y. — RE VIE W. 369
EXERCISES FOR REVIEW
Ex. 1. The perpendicular erected at the middle point of one side of a triangle meets the longer of the other two sides.
Ex. 2. Of the bisectors of two unequal angles of a triangle, produced to the point of intersection, the bisector of the smaller angle is the longer.
Ex. 3. The straight lines which join the 'middle points of the opposite sides of any quadrilateral bisect each other.
Ex. 4. If a line is drawn from the middle point of one base of a trapezoid to pass through the intersection of the diagonals, it will bisect the other base.
Ex. 5. If the opposite sides of a pentagon are produced to intersect, the um of the angles at the vertices of the triangles thus formed is equal to two right angles.
Ex. 6. The sum of the four lines drawn from the vertices of any quad- rilateral to any point except tlie intersection of the diagonals is greater than the sum of the diagonals.
Ex. 7. If the internal bisector of one base angle of a triangle and the external bisector of the other base angle are produced until they meet, the angle included between them is equal to half the vertical angle of the tri- angle.
Ex. 8. The angle contained by the bisectors of two exterior angles of any triangle is equal to half the sum of the two adjacent interior angles.
Ex. 9. If each of two angles of a quadrilateral is a right angle, the bisec- tors of the other angles are either perpendicular or parallel to each other.
Ex. 10. If the side CB of the triangle ABC is greater than the side GA, and CA is produced to D and CB to E^ making AD and BE equal, AE is greater than DB.
Ex. 11. In the triangle ABC a straight line AD is drawn perpendicular to" the straight line BD which bisects angle B. Prove that a straight line through D parallel to BC bisects AC.
Ex. 12. If one side of a triangle is greater than the other, any line from the vertex of the included angle to the base is less than the longer side.
Ex. 13. Lines drawn from one vertex of a parallelogram to the middle points of the opposite sides trisect a diagonal.
Ex. 14. No two straight lines drawn from two vertices of a triangle and terminated by the opposite sides can bisect each other.
Ex. 15. The base of a triangle whose sides are unequal is divided into two parts by a straight line bisecting the vertical angle. Prove that the greater part is adjacent to the greater side.
Ex. 16. If two exterior angles of a triangle are bisected, and from the point of intersection of the bisectors a straight line is drawn to the vertex of the third angle, this line bisects that angle. milne's geom. — 24
370 GEOME TRY. — RE VIE W.
Ex. 17. ABC is a triangle ; D is the middle point of 50, and E of AD ; BE produced meets AC in F. Prove that ^40 is trisected in F.
Ex. 18. In the triangle ABC the sides AB^ BC, and CA are trisected at the consecutive points D and E, F and G^, and H and /i respectively. Prove that the lines EF^ GH, and KD, when produced, form a triangle equal to ABC.
Ex. 19. If one of the equal sides of an isosceles triangle is produced below the base to a oertain length, if an equal length is cut off above the base from the other equal side, and if the two points are joined by a straight line, this line is bisected by the base.
Ex. 20. ABC is a triangle, and BE and CF are drawn perpendicular to AG, a line through ^ ; 2> is the middle point of BC. Show that FD equals ED.
Ex. 21. The angle contained by the bisectors of the base angles of any triangle is equal to the vertical angle of the triangle plus half the sum of the base angles.
Ex. 22. The bisectors of two angles of an equilateral triangle intersect, and from their point of intersection lines are drawn parallel to any two sides. Prove that these lines trisect the third side.
Ex. 23. The opposite sides of a regular hexagon are parallel.
Ex. 24. If in a quadrilateral the diagonals are equal and two sides are parallel, the other sides are equal.
Ex. 25. If from any point in the base of an isosceles triangle perpendicu- lars are drawn to the equal sides, their sum is equal to the perpendicular drawn from either extremity of the base to the opposite side.
Ex. 26. The sum of the perpendiculars from any point within an equi- lateral triangle to its sides is equal to the altitude.
Ex. 27. If from the vertex of any triangle two lines are drawn, one of which bisects the angle at the vertex and the other is perpendicular to the base, the angle between these lines is half the difference of the angles at the base of the triangle.
Ex, 28. In any triangle, the sides of the vertical angle being unequal, the median drawn from the vertical angle lies between the bisector of that angle and the longer side.
Ex. 29. In any triangle, the sides of the vertical angle being unequal, the bisector of that angle lies between the median and the perpendicular drawn from the vertex to the base.
Ex. 30. Lines are drawn through the extremities of the base of an isosceles triangle, making angles with it, on the side opposite the vertex, each equal to one third of a base angle of the triangle, and meeting the sides produced. Prove that the three triangles thus formed are isosceles.
GEOMETRY. — REVIEW. 371
Ex. 31. If two circumferences are tangent internally and the radius of the larger is the diameter of the smaller, any chord of tiie larger drawn from the point of contact is bisected by the circumference of the smaller.
Ex. 32. If perpendiculars are drawn to any chord at its extremities and produced to intersect a diameter of the circle, the points of intersection are equally distant from the center.
Ex. 33. If perpendiculars are drawn from the ends of a diameter of a circle upon any secant, their feet are equally distant from the points in which the secant intersects the circumference.
Ex. 34. Given an arc of a circumference, the chord subtended by it, and the tangent at one extremity. Prove that the perpendiculars dropped from the middle point of the arc upon the tangent and chord, respectively, are equal.
Ex. 35. The bisectors of the angles contained by the opposite sides (pro- duced) of an inscribed quadrilateral intersect at right angles.
Ex. 36. If two opposite sides of an inscribed quadrilateral are equal, the other two sides are parallel.
Ex. 37. In a given square, inscribe an equilateral triangle having its vertex in the middle of a side of the square.
Ex. 38. Find, in a side of a triangle, a point from which straight lines, drawn parallel to the other sides of the triangle and terminated by them, are equal.
Ex. 39. Construct a triangle,' having given the base, one of the angles at the base, and the sum of the other two sides.
Ex. 40. Construct a triangle, having given the base, one of the angles at the base, and the difference of the other two sides.
Ex. 41. Construct a triangle, having given the perpendicular from the vertex to the base, and the difference between each side and the adjacent segment of the base.
Ex. 42. If two circles are each tangent to two parallel lines and a trans- versal crossing them, the line of centers is equal to the segment of- the transversal intercepted between the parallels.
Ex. 43. If through the point of contact of two circles which are tangent to each other externally any straight line is drawn terminated by the circum- ferences, the tangents at its extremities are parallel to each other.
Ex. 44. If two circles are tangent to each other externally and parallel diameters are drawn, the straight line joining the opposite extremities of these diameters will pass through the point of contact.
Ex. 45. Construct the three escribed * circles of a given triangle.
* A circle tangent to one side of a triangle and to the other two sides produced is called an escribed circle.
372 GEOMETRH. ^REVIEW,
. Ex. 46. Construct an isosceles right triangle, having given the sum of the hypotenuse and one side.
Ex. 47. Construct a right triangle, having given the hypotenuse and the sum of the sides.
Ex. 48. Construct a right triangle, having given the hypotenuse and the difference of the sides.
Ex. 49. A and B are two fixed points on the circumference of a circle and CD is any diameter. What is the locus of the intersection of CA and
Ex. 50. Construct a triangle, having given a median and the two angles into which the angle is divided by that median.
Ex. 51. Construct a triangle, having given the base, the difference between the sides, and the difference between the angles at the base.
Ex. 52. Construct an isosceles triangle, having given the perimeter and altitude.
Ex. 53. The circles described on two sides of a triangle as diameters intersect on the third side, or the third side produced.
Ex. 54. ABC is a triangle having AG equal to BC', D is any point in AB. Prove that the circles circumscribed about triangles ADC and DBC are equal.
Ex. 55. Construct a triangle, having given two sides and the median to the third side.
Ex. 56. Construct a triangle, having given its perimeter, and having its angles equal to the angles of a given triangle.
Ex. 57. Construct a triangle, having given one side and the medians to the other sides.
Ex. 58. Construct a circle of given radius to touch a given circle and a given straight line. How many such circles may there be ?
Ex. 59. Construct a circle of given radius which shall be tangent to two given circles. How many solutions may there be ?
Ex. 60. If an equilateral triangle is inscribed in a circle and from any point in the circumference lines are drawn to the vertices, the longest of these lines is equal to tb^. sum of the other two.
Ex. 61. If two circles intersect each other, two parallel lines passing through the points of intersection and terminated by the exterior arcs are equal.
Ex. 62. An isosceles triangle has its vertical angle equal to an exterior angle of an equilateral triangle. Prove that the radius of the circumscribed circle is equal to one of the equal sides of the isosceles triangle.
Ex. 63. If a chord of a circle is extended by a length equal to tlie radius, and from the extremity a secant is drawn through the center of the circle, the length of the greater included arc is three times the length 9l \\x^ less.
GEOME TR Y. — RE VIE W. 373
Ex. 64. Construct three circles having equal diameters and being tangent to each other.
Ex. 65. Construct two circles of given radii to touch each other and a given straight line on the same side of it,
Ex. 66. Construct a triangle, having given the base, the vertical angle, and the point at which the base is cut by the bisector of the vertical angle.
Ex. 67. Construct a circle to touch a given circle and also to touch a given straight line at a given point.
Ex. 68. Construct a circle to touch a given straight line, and to touch a given circle at a given point.
Ex. 69. Construct a circle to touch a given circle, have its center in a given line, and pass through a given point in that line.
Ex. 70. If a :h = c :d, prove that
a'^ -\- ab -{- b'^ : a^ - ab + b^ = c^ + cd -{■ d^ : c^ - cd + d^.
Ex. 71. Given three lines a, b, and c. Construct as = —•
4 r "'
Ex. 72. Construct x, having given - = -.
Ex. 73. The diagonals of a trapezoid divide each other into segments which are proportional.
Ex. 74. If one side of a right triangle is double the other, the perpen- dicular from the vertex upon the hypotenuse divides the hypotenuse into parts which are in the ratio of 1 to 4.
Ex. 75. ABCD is an inscribed quadrilateral. The sides AB and DC are produced to meet at E. Prove triangles ACE and BDE similar.
Ex. 76. If AB is a chord of a circle, and CD is any chord drawn from the middle point C of the arc AB cutting the chord AB at E^ prove that the chord ^C is a mean proportional between CE and CD.
Ex. 77. If perpendiculars are drawn from two vertices of a triangle to the opposite sides, the triangle cut off by the line jbining the feet of the per- pendiculars is similar to the original triangle.
Ex. 78. AB is the hypotenuse of the right triangle ABC. If perpen- diculars are drawn to AB at A and B, meeting BC produced at E, and AC produced at Z>, the triangles ACE and BCD are similar.
Ex. 79. If two circles intersect in the points A and J5, and a secant through B cuts the circumferences in C and D respectively, the straight lines AC and AD are in the same ratio as the diameters of the circles.
Ex. 80. Inscribe a square in a given right isosceles triangle.
Ex. 81. From the obtuse angle of a triangle draw a line to the base, which shall be a mean proportional between the external segments into wnich It divides the base.
374 , GEOMETRY.— REVIEW,
Ex. 82. Through a given point draw a straight line, so that the parts of it intercepted between that point and perpendiculars drawn to it from two other given points may have a given ratio.
Ex. 83. Show that the diagonals of a trapezoid, one of whose bases is double the other, cut each other at a point of trisection.
Ex. 84. A tangent to a circle at the point A intersects two parallel tan- gents whose points of^ contact are D and E, in B and C respectively. BE and CD intersect at F. Prove that the line AF is parallel to the tangents BD and CE.
Ex. 85. The angle C of the triangle ABC is bisected by CD, which cuts the base AB at Z) ; 0 is the middle point of AB. Prove that OD has the same ratio to OA that the difference of the sides has to their sum.
Ex. 86. A and B are two points on the circumference of a circle of which 0 is the center ; tangents at A and B meet at E ; and from A the line AD is drawn perpendicular to OB. Prove BE : B0 = BD : AD.
Ex. 87. AB is a diameter of a circle, CD is a chord at right angles to it, and E is any point in CD ; AE and BE are drawn, and produced to cut the circumference in F and G respectively. Prove that CFDG has any two of its adjacent sides in the same ratio as the remaining two.
Ex. 88. Two circles whose centers are O and P intersect in A, and the tangent to each at A meets the circumference of the other in C and B respec- tively. Prove that AB:AC = OA: PA.
Ex. 89. 0 is the center of the circle inscribed in the triangle ABC ; AG meets BC in D. Trove AO : DO = AB -\- AC : BC.
Ex. 90. ABC is an isosceles triangle; the perpendicular to AC bX C meets the base AB, or the base produced, at ^ ; Z> is the middle point of AE. Prove that ^C is a mean proportional between AB and AD.
Ex. 91. AB and CD are two parallel straight lines ; E is the middle point of CD ; AC and BE meet in F, and AE and BD meet in G. Prove that FG is parallel to AB.
Ex. 92. If two circles are tangent to each other, either internally or externally, any two straight lines drawn through the point of contact will be cut proportionally by the circumferences.
Ex. 93. From one of the points of intersection of two intersecting circles a diameter of each circle is drawn. Prove (1) that the line joining the extremities of these diameters passes through the other point of intersection ; and (2) that this line is parallel to the line of centers of the circles.
Ex. 94. If in a right triangle a perpendicular is drawn from the vertex of the right angle to the hypotenuse, and circles are inscribed in the triangles thus formed, the diameters are proportional to the sides of the given right angle.
Ex. 95. The distance from the center of a circle to a chord 8<*™ long is 4dm, What is the distance from the center to a chord 6<^™ long ?
GEOMETRY. — REVIEW. 875
Ex. 96. If a chord 18^"^ long is bisected by another chord 22^°^ long, what are the segments of the latter?
Ex. 97. If two intersecting chords divide the circumference of a circle into parts whose lengths taken in order are as 1, 1, 2, and 5, what angles do the chords make with each other ?
Ex. 98. The square on the hypotenuse of a right isosceles triangle is equivalent to four times the triangle.
Ex. 99. If the sides of a triangular field are respectively llHm^ gnm^ and 8^1" long, how many hektares are there in the area of the field ?
Ex. 100. The sides of a triangle are respectively 39, 42, and 45 inches in length. What is the radius of the inscribed circle ?
Ex. 101. The sides of a triangle are respectively 5 ft., 5 ft., and 6 ft. What is the diameter of the circumscribed circle ?
Ex. 102. A triangular field has its sides respectively 16 rd., 24 rd., and 36 rd. long. What is the length of a line from the middle of the longest side to the opposite comer ? What is the area of the field ?
Ex. 103. If a chord IQcm long is 5"n distant from the center of a circle, what is the radius of the circle, and the distance from the end of the chord to the end of the radius that is perpendicular to the chord ?
Ex. 104. How many square meters are there in the area of the quadri- lateral ABCD, if AB = 6™, BG = 11"", CD = 4% AD = 13™, and the diago- nal ^(7=15°^?
Ex. 105. If two equivalent triangles have a comm'on base, and lie on opposite sides of it, the base, or the base produced, will bisect the line join- ing their vertices.
Ex. 106. ABC is a given triangle. Construct an equivalent triangle, having its vertex at a given point in BC, and its base in the same straight Hue as AB.
Ex. 107. Through the vertex A of the parallelogram ABCD draw a line meeting the side CB produced in F^ and the side CD produced in E. Prove that the rectangle of the produced parts of the sides is equivalent to the rectangle of the sides.
Ex. 108. ABC is a right triangle having its right angle at B. At A and C perpendiculars to ^C are erected to meet CB and AB produced in E and F respectively, and EF is drawn. Prove that the triangles BEF and ABC are equivalent.
Ex. 109. The square upon the altitude of an equilateral triangle is equivalent to three times the square upon half of one of the sides of the triangle.
Ex. 110. If from a point D in the base AB of the triangle ABC straight lines 'are drawn parallel to the sides AC and BC respectively, so as to meet BC m F and ^O in ^, triangle EFG is a mean proportional between tri- angles ADE and DBF.
876 GEOMETRY.— REVIEW.
Ex. 111. Two sides Of a triangle are 70™ and 65"^, and the difference of the segments of the other side made by a perpendicular from the opposite vertex is 9*". What is the length of the other side ?
Ex. 112. The sum of two sides of a triangle is 128 ft. , and a perpendicular from the vertex opposite the other side divides that side into segments of 60 ft. and 28 ft. What are the sides of the triangle ?
Ex. 113. Two sides of a triangle are in proportion to each other as 6 is to 5, and the adjacent segments of the other side made by a perpendicular from the opposite vertex are 36 ft. and 14 ft. respectively. What are the sides ?
Ex. 114. The difference of the two sides of an oblique triangle, obtuse- angled at the base, is 9"^, and the segments of the base produced made by a perpendicular from the opposite vertex are 30™ and 9™. What are the sides ?
Ex. 115. A flag pole 140 ft. long, standing on an eminence 30 ft. high, broke so that the top struck the level ground at a distance from the base of the pole equal to the length of the part standing. What was the length of the part broken off ?
Ex. 116. If from the extremities of the base of any triangle lines are drawn bisecting the other two sides, these lines intersect within the triangle and form another triangle on the same base equivalent to one third of the original triangle.
Ex. 117. Upon the sides of a right triangle, as homologous sides, three similar polygons of any number of sides are constructed. Prove that the polygon upon the hypotenuse is equivalent to the sum of the polygons upon the other two sides.
Ex. 118. ABCD is a rectangle, and BD is its diagonal ; a circle whose center is 0 is inscribed in the triangle DEC ; EO and FO are drawn perpen- dicular to AB and AB respectively. Then, the rectangle AFOE is equiva- lent to one half the rectangle ABCD.
Ex. 119. If squares are described upon the three sides of a right triangle, and the extremities of the adjacent sides of any two squares are joined, the triangle so formed is equivalent to the given triangle.
Ex. 120. Inscribe a circle in a given rhombus.
Ex. 121. A segment whose arc is 60° is cut off from a circle whose radius is 15 ft. What is the area of the segment ?
Ex. 122. If the bisectors of all the angles of a polygon meet in a point, a circle may be inscribed in the polygon.
Ex. 123. If the area of a certain circle is 154«<J™, how many degrees are there in an angle at the center, if it is subtended by an arc of the circumfer- ence 6. 5"" long ?
Ex. 124. Prove that an equiangular polygon inscribed in a circle is regular, if the number of its sides is odd.
GEOMETRY. — REVIEW, 3T7
Ex. 125. Two parallel chords* in a circle are the sides of regular in- scribed polygons, one a hexagon and the other a dodecagon. If the radius of the circle is 11 in., how far apart are the chords ?
Ex. 126. What is the length of the side of a square equivalent to a circle in which a chord of 30<i™ has an arc whose height is S^^™ ?
Ex. 127. If a 4-inch pipe will fill a cistern in 2 hr. 30 min., how long will it take a 2-inch pipe to fill it ?
Ex. 128. If an equilateral triangle and a regular decagon each has a perimeter of Q^^ what is the difference in area between them ?
Ex. 129. If an equilateral triangle is inscribed in a circle, the line joining the middle points of the arcs cut off by two of its sides will be trisected by those sides.
Ex. 130. If the area between three equal circles, each tangent to the other two, is 40sq ™, what are the radii of the circles ?
Ex. 131. Construct a circle equal to three fourths of a given circle.
Ex. 132. If a circle is circumscribed about a right triangle, and on each of its sides as a diameter a semicircle is described exterior to the triangle, the sum of the areas of the crescents thus formed is equal to the area of the triangle.
Ex. 133. The area of an inscribed regular octagon is equal to that of a rectangle whose sides are respectively equal to the sides of the inscribed and circumscribed squares.
Ex. 134. If the radius of a circle is r, prove that the area of a regular inscribed octagon is 2 r^\/2.
Ex. 135. The area of a circle is a mean proportional between the areas of any two similar polygons, one of which is circumscribed about the circle and the other is isoperimetric with the circle. (^Oalileo^s TJieorem.)
Ex. 136. Prove that the sum of the perpendiculars drawn to the sides of a regular polygon from any point within is equal to the apothem of the polygon multiplied by the number of its sides.
Ex. 137. If upon the sides of a regular hexagon squares are constructed outwardly, the exterior vertices of these squares are the vertices of a regular dodecagon.
Ex. 138. A horse is tethered to a hook on the inner side of a fence which bounds a circular grass plot. His tether is so long that he can just reach the center of the plot. The area of so much of the plot as he can graze over is 3/(4 IT — 3\/3) sq. rd. ; find the length of the tether and the circumference of the plot. {Harvard.)
Ex. 139. If equal straight lines are drawn from a given point to a given plane they make equal angles with the plane.
Ex. 140. Two planes which are each perpendicular to a third plane are parallel, if their intersections with the third plane are parallel.
378 ' GEOMETRY. — REVIEW.
Ex. 141. The line joining the extremities of two equal lines which are perpendicular to a plane, on the same side of it, is parallel to the plane.
Ex. 142. Through a given line in a given plane pass a plane to make a given angle with the given plane.
Ex. 143. Through a given line parallel to a given plane pass a plane to make a given angle with the given plane.
Ex. 144. Through the edge of a given dihedral angle pass a plane to bisect that angle.
Ex. 145. Find the locus of the points in space which are equidistant from two parallel lines.
Ex. 146. If a straight line is perpendicular to a plane, its projection on any other plane is perpendicular to the intersection of the two planes.
Ex. 147. If two planes are perpendicular, a straight line drawn from any point of one plane perpendicular to the other will lie in the first plane.
Ex. 148. The projection of a straight line upon a plane is a straight line.
Ex. 149. Find the locus of the points in space which are equidistant from three given planes.
Ex. 150. Find the locus of the points in space equidistant from two in- tersecting straight lines.
Ex. 151. Two trihedrals are equal or symmetrical, if two face angles, and the dihedral between their faces, in one are equal, each to each, to the corresponding parts in the other.
Ex. 152. Find the locus of the points in space which are equidistant from three given straight lines in the same plane.
Ex. 153. Find the locus of the points in a given plane which are equi- distant from two given points without the plane.
Ex. 154. The angles AOB and AOG in different planes are equal. Prove that the plane bisecting the dihedral angle between their planes is perpen- dicular to the plane BOG.
Ex. 155. The planes through any two pairs of lines that pass through a point intersect in a line which passes through the same point.
Ex. 156. In a given plane find a point equidistant from three given points without the plane.
Ex. 157. Through a given point in 'space, draw a straight line which shall cut two given straight lines not in the same plane.
Ex. 158. Find the locus of the points in space which are equidistant from two given planes and also equidistant from two given points.
Ex. 159. Two planes are perpendicular respectively to two non-parallel lines which are not in the same plane. Prove that their intersection is perpendicular to any plane that is parallel to both lines.
GEOMETRY. — REVIEW. 379
Ex. 160. From the vertex of a trihedral angle a line is drawn within the angle. Prove that the sum of the angles which this line makes with the edges is less than the sum, but greater than half the sum, of the face angles.
Ex. 161. Two trihedrals are equal or symmetrical, if two dihedrals and the included face angle of one are equal, each to each, to the corresponding parts of the other.
Ex. 162. A plane parallel to two sides of a- quadrilateral in space (that is, a quadrilateral whose sides do not all lie in the same plane) divides the other two sides proportionally.
Ex. 163. In any trihedral, the three planes, passed through the edges perpendicular to the opposite faces respectively, intersect in the same straight line.
Ex, 164. In any trihedral, the three planes, passed through the edges and the bisectors of the opposite face angles respectively, intersect in the same straight line.
Ex. 165. What is the edge of a cubical vessel that holds one half ton of water ?
Ex. 166. Represent the base edge of a regular four-sided pyramid by e, its altitude by h, and its total surface by T. Compute the base edge in terms of h and T.
Ex. 167. What is the difference in volume between the frustum of a pyramid and a prism, each 12'i'" high, if the bases of the frustum are squares whose sides are lO'i'" and S^™ respectively, and the base of the prism is a eection of the frustum parallel to its bases and midway between them ?
Ex. 168. What is the volume of a regular tetrahedron whose edge is
10dm?
Ex. 169. The altitude of a regular hexagonal pyramid is 13 in., and its slant height is 16 in. What is its lateral edge ?
Ex. 170. The lateral faces of a regular quadrangular pyramid are equi- lateral triangles, and its altitude is 9™. What is the area of the base ?
Ex. 171. The altitude of a frustum of a regular quadrangular pyramid is 10<=m, and the sides of its bases are respectively IG"^"* and 6'='". What is the lateral area of the frustum ?
Ex. 172. If the altitude of a pyramid is h, at what distance from the vertex will it be cut by a plane parallel to the base and dividing the pyra- mid into two parts which are in the ratio of 3 : 4 ?
Ex. 173. At what distances from the vertex will a lateral edge of a pyra- mid be cut by two planes parallel to the base, if they divide the pyramid into three equivalent parts, the length of the edge being m ?
Ex. 174. If the base edge of a regular square pyramid is m, and its total surface is 7', what is its volume ?
Ex. 175. The perimeter of the base of a regular quadrangular pyramid is p, and the area of a section through two diagonally opposite edges is A. What is the lateral area of the pyramid ?
380 GEOME TRY. — RE VJE W.
Ex. 176. If two tetrahedrons have the faces including a trihedral of one similar to the faces including a trihedral of the other, each to each, and similarly placed, the tetrahedrons are similar.
Ex. 177. If two tetrahedrons have a dihedral angle of one equal to a dihedral angle of the other, and the faces including these dihedrals similar, each to each, and similarly placed, the tetrahedrons are similar.
Ex. 178. The perpendicular from the middle point of the diagonal of a rectangular parallelopiped upon a lateral edge bisects the edge, and is equal to one half the projection of the diagonal upon the base.
Ex. 179. In any polyhedron the number of edges increased by two is equal to the number of vertices increased by the number of faces. {Eider's Theorem. )
Ex. 180. The sum of the face angles of any polyhedron is equal to four right angles taken as many times, less two, as the polyhedron has vertices.
Ex. 181. If a plane is tangent to a circular cone, its intersection with the plane of the base is tangent to the base.
Ex. 182. If a plane is tangent to a circular cylinder, its intersection with the plane of the base is tangent to the base.
Ex. 183. What are the dimensions of a cylindrical measure whose alti- tude is half its diameter, if it holds a half bushel ?
Ex. 184. Find the weight of the water that will be contained in a vertical pipe 40 ft. high and 1 ft. in diameter. Also find the pressure per square inch on the base of the pipe.
Ex. 185. A rectangle revolves successively about two adjacent sides whose lengths are m and n respectively. Compare the volumes of the cylinders thus generated.
Ex. 186. A right triangle revolves successively about the perpendicular sides whose lengths are m and n respectively. Compare the volumes of the cones thus generated.
Ex. 187. If the sides including the right angle of a right triangle are m and n, what is the area of the surface generated by revolving the triangle about-its hypotenuse as an axis ?
Ex. 188 Find the altitude of a cylinder of revolution of radius r, if the cylinder is equivalent to a rectangular parallelopiped whose dimensions are I, m, and n.
Ex. 189. Find the altitude of a cone of revolution of radius r, equivalent to a rectangular parallelopiped whose dimensions are I, m, and n.
Ex. 190. The altitudes of two equivalent cylinders of revolution are in the ratio a :b. If the radius of one is r, what is the radius of the other ?
Ex. 191. Find the altitude of a regular quadrangular prism whose base edge is w, the prism being equivalent to a cylinder of revolution whose alti- tude is h and radius r.
GEOME TR Y. — RE VIE W. 381
Ex. 192. How must the dimensions of a cylinder of revolution be in- creased in order to form a similar cylinder whose total surface shall be n times that of the original cylinder ?
Ex. 193. How must the dimensions of a cylinder of revolution be increased in order to form a similar cylinder whose volume shall be n times that of the original cylinder ?
Ex. 194. What is the radius of the base of a circular cone whose altitude is A, the longest and the shortest elements being / and V respectively ?
Ex. 195. What is the slant height of a frustum of a cone of revolution whose lateral surface is 8 and whose lower and upper bases are B and h respectively ?
Ex. 196. A cone of revolution whose radius is r and altitude h is divided into two equivalent parts by a plane parallel to the base. What is the total area of the frustum thus formed ?
Ex. 197. The volume of a cylinder of revolution is equal to the area of its generating rectangle multiplied by the circumference generated by the point of intersection of the diagonals of the rectangle.
Ex. 198. On each base of the frustum of a cone of revolution there is a cone whose vertex is in the center of* the other base. If the radii of the lower and upper bases are r and r' respectively, what is the radius of the circle of intersection of the two cones ?
Ex. 199. A stone bridge 20 ft. wide has a circular arch of 140 ft. span at the water level. The crown of the arch is 140 (1 — ^ VS) ft. abave the sur- face of the water. How many square feet of surface must be gone over in cleaning so much of the under side of the arch as is above water ? {Harvard.)
Ex. 200. What part of the whole surface of a sphere is a spherical trian- gle whose angles are 57° 57', 75° 27', and 100° 36' ?
Ex. 201. What is the volume of a right cone whose altitude is 15 ft., inscribed in a sphere whose radius is 10 ft. ?
Ex. 202. How far from the base of a hemisphere must a plane be passed to divide the surface into two equivalent zones ?
Ex. 203. The volume of a spherical segment of one base is V and its altitude is h. What is the radius of the sphere ?
Ex. 204. Find an expression for the volume of a cube inscribed in a sphere whose radius is r.
Ex. 205. Two equal circles intersect in a diameter. If a plane is passed perpendicular to that diameter, prove that the fcfur points in which it inter- sects the circumferences lie in the circumference of a circle.
Ex. 206. The square on the diameter of a sphere and the square on the edge of an inscribed cube are to each other as 3 is to 1.
Ex. 207. Find an expression for the altitude of a zone of a sphere whose radius is r, the area of the zone being equal to that of a great circle of the sphere.
382 GEOMETRY. — REVIEW.
Ex. 208. Find an expression for tlie altitude of a zone whose area is A on a spliere whose volume is V.
Ex. 209. Assuming the atmosphere to extend to a height of 50 miles above the earth's surface and the earth to be a sphere whose radius is approxi- mately 4000 miles, what is the volume of the atmosphere ?
Ex. 210. Assuming the earth to be a sphere whose radius is approxi- mately 4000 miles, how far at sea is a lighthouse visible, if it is 80 ft. high ?
Ex. 211. A swimmer, whose eye is at the surface of the water, can just see the top of a buoy a mile distant. If the buoy is 8 in. out of the water, what is the radius of the earth ?
Ex. 212. How high above the surface of the earth must a man be in
order that he may see - of it ? n
Ex. 213. What is the area of the zone illuminated by a taper h deci- meters from the surface of a sphere whose radius is r decimeters ?
Ex. 214. In a cube whose edge is 1 ft. there are inscribed a cylinder, a cone, a sphere, and a square pyramid. What is the volume of each of these solids ?
Ex. 215. A cylindrical boiler with hemispherical ends has a total length of 12 ft. and its circumference is 10 ft. What is its surface ? What weight of water is required to fill it ?
Ex. 216. Find the diameter of a sphere which is circumscribed about a regular square pyramid whose base is 4 in. square and altitude 8 in.
Ex. 217. A sphere 8 in. in diameter has a 3-inch hole bored through its center. What is the remaining volume ?
Ex. 218. What is the volume of the portion of a sphere lying outside of an inscribed cylinder of revolution whose altitude is h and radius r ?
Ex. 219. Inscribe a circle in a given spherical triangle.
Ex. 220. Find the locus of the centers of the sections of a given sphere made by planes passing through a given straight line.
Ex. 221. Find the locus of the centers of the sections of a given sphere made by planes passing through a given point without the sphere.
Ex. 222. Having given the radius, construct a spherical surface to pass through three given points.
Ex. 223. Having given the radius, construct a spherical surface to pass through two given points and be tangent to a given plane or to a given sphere.
Ex. 224. Having given the radius, construct a spherical surface to pass through a given point and be tangent to two given planes.
Ex. 225. Having given the radius, construct a spherical surface to be tan- gent to three given planes.
GEOMETRY. — TABLES.
383
METRIC TABLES Measures of Length
10 Millimeters (m™) 10 Centimeters 10 Decimeters 10 Meters 10 Dekameters 10 Hektometers
= 1 Centimeter (c™) = 1 Decimeter (dm) = 1 Meter (") = 1 Dekameter (^m) = 1 Hektometer (Hm) = 1 Kilometer (J^)
Measures of Surface 100 Sq. Millimeters («i ""'n) = 1 Sq. Centimeter (^^ =»">
100 Sq. Centimeters 100 Sq. Decimeters 100 Sq. Meters 100 Sq. Dekameters 100 Sq. Hektometers
= 1 Sq. Decimeter (^i dm) = 1 Sq. Meter («q™) = 1 Sq. Dekameter ("Q Dm) = 1 Sq. Hektometer (sqHm) = 1 Sq. Kilometer*8'iKm)
A square hektometer is also called a hektare (h*>.
Measures of Volume 1000 Cu. Millimeters (c" mm) = 1 Cu. Centimeter («" cm) 1000 Cu. Centimeters = 1 Cu. Decimeter (cu dm)
1000 Cu. Decimeters = 1 Cu. Meter (^um)
Measures of Capacity
10 Milliliters (ml)
10 Centiliters
10 Deciliters
10 Liters
10 Dekaliters
10 Hektoliters
= 1 Centiliter (•!) = 1 Deciliter (di) = 1 Liter d) r= 1 Dekaliter (DD = 1 Hektoliter (Hi) = 1 Kiloliter (KD
The liter contains a volume equal to a cube whose edge is a decimeter.
384
GEOMETRY.-^ TABLES,
Measures of Weight 10 Milligrams ("s) = 1 Centigram (««)
10 Centigrams 10 Decigrams 10 Grams 10 Dekagrams 10 Hektograms
= 1 Decigram (<i8> = 1 Gram (s) = 1 Dekagram (I>k) = 1 Hektogram (««) = 1 Kilogram (Kg)
The weight of a gram is the weight of a cubic centimeter of distilled water t its greatest density.
Metric Equivalents
|
1 Meter |
= 39.37 in. = 1.0936 yd. |
1 Yard |
= .9144°^ |
|
1 Kilometer |
= .62138 Mile |
1 Mile |
= 1.6093Km |
|
1 Hektare |
= 2.471 Acres |
1 Acre |
= .4047Ha |
|
1 Liter |
_ f .908 qt. dry ~ 11.0567 qt. liquid |
1 qt. dry 1 qt. liq. |
= 1.1011 = .94631 |
|
1 Gram |
= 15.432 Grains |
1 Grain |
= .06488 |
|
1 Kilogram |
= 2.2046 lb. |
1 Pound |
= .4536Kg |
^ Approximate Metric Equivalents
1 Decimeter = 4 in. ^ ^^.^^^ ^ f t% qt- dry
1 Meter = 40 in. 1 1| qt. liquid
1 Kilometer = f Mile 1 Hektoliter = 2| bu.
1 Hektare = 2| Acres 1 Kilogram = 2^ lb.
Notes. — 1. The specific gravity of a substance (solid or liquid) is the ratio between the weight of any volume of the substance and the weight of a like volume of distilled water at its greatest density ; consequently, since a cubic centimeter of distilled water at its greatest density weighs one gram, the weight of any substance may be found if its specific gravity and volume are known.
2. A cubic foot of water weighs 62^ lb., or 1000 oz.
3. A bushel contains 2150.42 cu. in.
4. A gallon contains 231 cu. in.
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THE UNIVERSITY OF CALIFORNIA LIBRARY