Edward Bright
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ESSENTIALS OF GEOMETRY
BY
WEBSTER WELLS, S.B.
// . .
PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY
3>*:c
BOSTON, U.S.A. D. C. HEATH & CO., PUBLISHERS
1899
Copyright, 1898 and 1899, By WEBSTER WELLS.
Norfajootr i^ress
J. S. Gushing & Co. - Berwick & Smith Norwood Mass. U.S.A.
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PREFACE.
In the Essentials of Geometry, the author has endeavored to prepare a work suited to the needs of high schools and academies. It will also be found to answer as well the requirements of colleges and scientific schools.
In some of its features, the work is similar to the author's Revised Plane and Solid Geometry; but important improve- ments have been introduced, which are in line with the present requirements of many progressive teachers.
In a number of propositions, the figure is given, and a statement of what is to be proved ; the details of the proof being left to the pupil, usually with a hint as to the method of demonstration to be employed.
The propositions and corollaries left in this way for the pupil to demonstrate, in the Plane Geometry, will be found in the following sections : —
Book I., §§ 51, 75, 76, 78, 79, 96, 102, 110, 111, 112, 115, 117, 136.
Book II., §§ 158, 160, 165, 170, 172 (Case III.), 174, 178, 179, 193 (Case III.), 194, and 201.
Book III., §§ 251, 257, 261, 264, 268, 278, 282, 284, and 286.
Book IV., §§ 312 and 316.
Book v., §§ 346, 347, and 350.
iii
797944
iy PREFACE.
Book VI., §§ 405, 407, 412, 414, 415, 416, 417, 420, 421, 434, 437, 440, 442, and 444.
Book VII., §§ 491, 495, 507, 512, 513, 521, 528, 529, and 530.
Book VIII., §§ 554, 559, 578, 580, 581, 594, 595, 601, 603, 608, 613, 614, 625 (Case II.), 630, 631, 635, and 637.
Book IX., §§ 654, 656, 660, 673, and 679.
There are also Problems in Construction in which the construction or proof is left to the pupil.
Another important improvement consists in giving figures and suggestions for the exercises. In Book I., the pupil has a ligure for every non-numerical exercise; after that, they are only given with the more difficult ones.
In many of the exercises in construction, the pupil is expected to discuss the problem, or point out its limitations.
In Book I., and also in the first eighteen propositions of Book VI., the authority for each statement of a proof is given directly after the statement, in smaller type, enclosed in brackets. In the remaining portions of the work, the formal statement of the authority is omitted ; but the num- ber of the section where it is to be found is usually given.
In a number of cases, however, where the pupil is pre- sumed, from practice, to be so familiar with the authority as not to require reference to the section where it is to be found, there is given merely an interrogation-point.
In all these cases the pupil should be required to give the authority as carefully and accurately as if it were actu- ally printed on the page.
Another improvement consists in marking the parts of a demonstration by the words Given, To Pi'ove, and Proof, printed in heavy -faced type.
A similar system is followed in the Constructions, by the use of the words Given, Required, Construction, and Proof.
A minor improvement is the omission of the definite article in speaking of geometrical magnitudes; thus we speak of "angle ^," "triangle ABC,^^ etc., and not "the angle A,^^ "the triangle ABC,'' etc.
PREFACE.
Symbols and abbreviations have been freely used ; a list of these will be found on page 4.
Particular attention has been given to putting the propo- sitions in the first part of Book I. in a form adapted to the needs of a beginner.
The pages have been arranged in such a way as to avoid the necessity, while reading a proof, of turning the page for reference to the figure.
The Appendix to the Plane Geometry contains proposi- tions on Maxima and Minima of Plane Figures, and Sym- metrical Figures; also, additional exercises of somewhat greater difficulty than those previously given.
The Appendix to the Solid Geometry contains rigorous proofs of the limit statements made in §§ 639, 650, 667, and 674.
The author wishes to acknowledge, with thanks, the many suggestions which he has received from teachers in all parts of the country, which have added materially to the value of the work.
WEBSTER WELLS.
Massachusetts Institute of Technology,
Stereoscopic views of many of the figures in the Solid Geometry have been prepared. Full particulars may be obtained from the publishers.
Digitized by tine Internet Arciiive
in 2007 witii funding from
IVIicrosoft Corporation
littp://www.arcliive.org/details/essentialsofgeomOOwellricli
CONTENTS.
PAGE
Preliminary Definitions . . . . . . . i
PLANE GEOMETRY.
Book 1. Rectilinear Figures 5
Book IL The Circle 72
Book III. Theory of Proportion. — Similar Polygons 122
Book IV. Areas of Polygons 162
Book V. Regular Polygons. — Measurement of the
Circle . .188
APPENDIX TO PLANE GEOMETRY.
Maxima and Minima of Plane Figures . . . 211
Symmetrical Figures 217
Additional Exercises 220
SOLID GEOMETRY.
Book VI. Lines and Planes in Space. — Diedral
Angles. — Polyedral Angles . .' . 233
Book VII. Polyedrons 273
Book VIII. The Cylinder, Cone, and Sphere . . 319
Book IX. Measurement of the Cylinder, Cone, and
Sphere ' • • • 360
Appendix to Solid Geometry 386
GEOMETRY.
>>»i<
PRELIMINARY DEFINITIONS.
(^
A material body.
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A /\ |
/ |
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/ |
A geometrical solid.
1. A material body, such as a block of wood, occupies a limited or hounded portion of space.
The boundary which separates such a body from sur- rounding space is called the surface of the body.
2. If the material composing such a body could be con- ceived as taken away from it, without altering the form or shape of the hounding surface, there would remain a portion of space having the same bounding surface as the former material body ; this portion of space is called a geometrical solid, or simply a solid.
The surface which bounds it is called a geometrical sur- face, or simply a surface; it is also called the surface of the solid.
3. If two geometrical surfaces intersect each other, that which is common to both is called a geometriccd line, or simply a line.
Thus, if surfaces ^B and CD cut each other, their common intersection, EF, is a line.
1
GEOMETRY.
4'. If two ' geometrical lines intersect ^\ ^D
each other, that which is common to both is called a geometrical point, or simply a point. ^^ ^^
Thus, if lines AB and CD cut each other, their common intersection, 0, is a point.
5. A solid has extension in every direction; but this is not true of surfaces and lines.
A point has extension in no direction, but simply position in space.
6. A surface may be conceived as existing independently in space, without reference to the solid whose boundary it forms.
In like manner, we may conceive of lines and points as having an independent existence in space.
7. A straight line, or right line, is a line which has the same direction throughout its length ; as AB.
F G
A-
A curved line, or curve, is a line no portion of which is straight; as CD.
A broken line is a line which is composed of different successive straight lines ; as EFGH.
8. The word " line " will be used hereafter as signifying a straight line.
9. A plane surface, or plane, is a surface such that the straight line joining any two of its points jf lies entirely in the surface.
Thus, if P and Q are any two points in /^ surface MN, and the straight line joining P and Q lies entirely in the surface, then MN is a plane.
10. A curved surface is a surface no portion of which is plane.
PRELIMINARY DEFINITIONS. .3
11. We may conceive of a straight line as being of un- limited extent in regard to length; and in like manner we may conceive of a plane as being of unlimited extent in regard to length and breadth.
12. A geometrical figure is any combination of points, lines, surfaces, or solids.
A plane figure is a figure formed by points and lines all lying in the same plane.
A geometrical figure is called rectilinear, or nght-Uned, when it is composed of straight lines only.
13. Geometry treats of the properties, construction, and measurement of geometrical figures.
14. Plane Geometry treats of plane figures only.
Solid Geometry, also called Geometry of Space, or Geometry of Three Dimensions, treats of fi.gures which are not plane.
15. An Axiom is a truth which is assumed without proof as being self-evident.
A TJieorem is a truth which requires demonstration.
A Problem is a question proposed for solution.
A Proposition is a general term for a theorem or problem.
A Postulate assumes the possibility of solving a certain problem.
A Corollary is a secondary theorem, which is an imme- diate consequence of the proposition which it follows.
A Scholium is a remark or note.
An Hypothesis is a supposition made either in the state- ment or the demonstration of a proposition.
16. Postulates.
1. We assume that a straight line can be drawn between any two points.
2. We assume that a straight line can be produced (i.e., prolonged) indefinitely in either direction.
17. Axioms.
We assume the truth of the following:
4 GEOMETRY.
1. Things which are equal to the same thing, or to equals, are equal to each other.
2. If the same operation he performed upon equals, the results will be equal.
3. Bid one straight line can be drawn between two points.
4. A straight line is the shortest line between two points.
5. The whole is equal to the sum of all its parts.
6. The whole is greater than any of its parts.
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18. Since but one straight |
line can be drawn between two |
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points, a straight line is said to be determined by any two of |
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its points. |
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19. Symbols and Abbreviations. |
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The following symbols will be used in the work : |
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+, plus. |
A, triangle. |
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— , minus. |
A, triangles. |
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X , multiplied by. |
J_, perpendicular, is perpen- |
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=, equals. |
dicular to. |
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=c=, equivalent, is equivalent |
Js, perpendiculars. |
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to. |
II , parallel, is parallel to. |
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>, is greater than. |
lis, parallels. |
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<, is less than. |
O, parallelogram. |
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.-., therefore. |
m, parallelograms. |
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Z, angle. |
O, circle. |
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A, angles. |
CD, circles. |
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The following abbreviations will also be used : |
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Ax., Axiom. |
Sup., Supplementary. |
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Def., Definition. |
Alt., Alternate. |
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Hyp., Hypothesis. |
Int., Interior. |
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Cons., Construction. |
Ext., Exterior. |
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Rt., Right. |
Corresp., Corresponding. |
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Str., Straight. |
Rect., Rectangle, rec- |
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Adj., Adjacent. |
tangular. |
PLAICE GEOMETRY,
Book I.
RECTILINEAR FIGURES.
DEFINITIONS AND GENERAL PRINCIPLES.
20. An angle (Z) is the amount of diverg- ence of two straight lines which are drawn from the same point in different directions.
The point is called the vertex of the angle, and the straight lines are called its sides.
21. If there is but one angle at a given vertex, it may be designated by the letter at that vertex ; but if two or more angles have the same vertex, we avoid ambiguity by naming also a letter on each side, placing the letter at the vertex between the others.
Thus, we should call the angle of § 20 "angle 0"; but if there were other angles having the same vertex, we should read it either AOB or BOA.
Another way of designating an angle is by means of a letter placed between its sides; examples of this will be found in § 71.
22. Two geometrical figures are said to be equal when one can be applied to the other so that they shall coincide throughout.
To prove two angles equal, we do not consider the lengths of their sides.
5
6
PLANE GEOMETRY.— BOOK I.
Thus, if angle ABC can be applied to angle DEF in such a manner that point B shall fall on point E, and sides AB and BG on sides DE and EF, respec- tively, the angles are equal, even if sides AB and BC are not equal in length to sides DE and EF, respectively.
23. Two angles are said to be adjacent when they have the same vertex, and a common side between them; as AOB and
Boa 0^
PERPENDICULAR LINES.
24. If from a given point in a straight line a line be drawn meeting the given line in such a way as to make the adjacent angles equal, each of the equal angles is called a right angle, and the lines are said to be perpendicular (±) to each other. ^
Thus, if from point A in straight line CD line ABhe drawn in such a way as to make angles BAC and BAD equal, each of these angles is a right angle, and AB and CD are perpendicular to each other. A ~
Pkop. I. Theorem.
25. At a given point in a straight line, a perpendicular to the line can he drawn, and hut one.
A C B
Let C be the given point in straight line AB.
RECTILINEAR FIGURES. 7
To prove that a perpendicular can be drawn to AB at C, and but one.
Draw a straight line CD in such a position that angle BCD shall be less than angle ACD; and let line CD be turned about point C as a pivot towards the position CA.
Then, angle BCD will constantly increase; and angle ACD will constantly diminish, until it becomes less than angle BCD-, and it is evident that there is one position of CD, and only one, in which these angles are equal.
Let CE be this position ; then by the definition of § 24, CE is perpendicular to AB.
Hence, a perpendicular can be drawn to AB at C, and but one.
26. Cor. All right angles are equal.
Let ABC SindDEF be right angles.
To prove angles ABC and DEF equal. a
Let angle ABC be superposed (i.e., placed) upon angle DEF in such a way that point B shall fall upon point E, and line AB upon line DE.
Then, line BC will fall upon line EF; for otherwise we should have two lines perpendicular to DE at E, which is impossible.
[At a given point in a straight line, but one perpendicular to the line can be drawn. ] (§25)
Hence, angles ABC and DEF are equal (§ 22).
B D
DEFINITIONS.
27. An acute angle is an angle which is less than a right angle ; as ABC.
An obtuse angle is an angle which is greater than a right angle ; as DEF.
Acute and obtuse angles are called oblique angles; and intersecting lines which are not perpendicular, are said to be oblique to each other.
8 PLANE GEOMETRY. —BOOK I.
28. Two angles are said to be veHical, ^. or opposite, when the sides of one are the prolongations of the sides of the other ; as AEC and BED.
29. An angle is measured by finding how many times it contains another angle, adopted arbitrarily as the unit of measure.
The usual unit of measure is the degree, which is the ninetieth part of a right angle.
To express fractional parts of the unit, the degree is divided into sixty equal parts called minutes, and the min- ute into sixty equal parts, called seconds.
Degrees, minutes, and seconds are represented by the symbols, °, ', ", respectively.
Thus, 43° 22' 37" represents an angle of 43 degrees, 22 minutes, and 37 seconds.
30. If the sum of two angles is a right angle, or 90°, one is called the compleynent of the other; and if their sum is two right angles, or 180°, one is called the supplement of the other.
For example, the complement of an angle of 34° is 90° — 34°, or 56° ; and the supplement of an angle of 34° is 180° - 34°, or 146°.
Two angles which are complements of each other are called complementary; and two angles which are supple- ments of each other are called supplementary.
31. It is evident that
1. The complements of equal angles are equal.
2. The supplements of equal angles are equal.
EXERCISES.
1. How many degrees are there in the complement of 47° ? of 83° ? of 90°?
2. How many degrees are there in the supplement of 31° ? of 90° ? of 178° ?
RECTILINEAR FIGURES. 9
3. How many degrees are there in the complement, and in the supplement, of an angle equal to /j of a right angle ?
4. How many degrees are there in an angle whose supplement is equal to f f of its complement ?
5. Two angles are complementary, and the greater exceeds the less by 37°. How many degrees are there in each angle ?
Prop. II. Theorem.
32. If two adjacent angles have their exterior sides in the same straight line, their sum is equal to two right angles.
E
A C B
Let angles ACD and BCD have their sides AC and BC in the same straight line.
To prove the sum of angles ACD and BCD equal to two right angles.
Draw line CE perpendicular to AB at C.
[At a given point in a straight line, a perpendicular to the line can be drawn.] (§ 25)
Then, it is evident that the sum of angles ACD and BCD is equal to the sum of angles ACE and BCE.
But since CE is perpendicular to AB, angles ACE and BCE are right angles.
Hence, the sum of angles ACD and BCD is equal to two right angles.
33. Sch. Since angles ACD and BCD are supplementary (§ 30), the theorem may be stated as follows :
If two adjacent angles have their exterior sides in the same straight line, they are supplementary.
Such angles are called supplementary-adjacent.
10 PLANE GEOMETRY.— BOOK I.
34. Cor. I. TTie sum of all the angles on the same side of a straight line at a given point is equal to two right angles.
This is evident from § 32.
35. Cor. II. Tfie sum of all the angles about a point in a plane is equal to four right angles.
Let AOB, BOG, COD, and DOA be angles about the point 0.
To prove the sum of angles AOB, \ _B
BOG, GOD, and DOA equal to four \ ./^
right angles. -E -\^- A
Produce ^0 to ^. /
Then, the sura of angles AOB, BOG, I
and GOE is equal to two right angles. 'D
[The sum of all the angles on the same side of a straight line at a given point is equal to two right angles.] (§ 34)
In like manner, the sum of angles EOD and DOA is equal to two right angles.
Therefore, the sum of angles AOB, BOG, GOD, and DOA is equal to four right angles.
Ex. 6. If, in the figure of § 35, angles AOB, BOC, and COD are respectively 49°, 88°, and | of a right angle, how many degrees are there in angle AOD ?
36. Sch. The pupil will now observe that a demonstrar tion, in Geometry, consists of three parts :
1. The statement of what is given in the figure.
2. The statement of what is to be proved.
3. The proof
In the remaining propositions of the work, we shall mark clearly the three divisions of the demonstration by heavy- faced type, and employ the svmbols and abbreviations of § 20.
RECTILINEAR FIGURES. n
Prop. III. Theorem.
37. If the sum of two adjacent angles is equal to two right angles, their exterior sides lie in the same straight line.
.-"E
A C B
Given the sum of adj. A ACD and BCD equal to two
rt. A.
To Prove that AC and BC lie in the same str. line.
Proof. If AC and BC do not lie in the same str. line, let CE be in the same str. line with AC.
Then since ACE is a str. line, Z ECD is the supplement of A ACD.
[If two adj. A have their ext. sides in the same str. line, they are supplementary.] (§33)
But by hyp., A ACD + A BCD = two rt. A. Whence, A BCD is the supplement of A ACD. (§ 30) Then since both A ECD and A BCD are supplements of A ACD, A ECD = A BCD.
[The supplements of equal A are equal.] (§ 31)
Hence, EC coincides with BC, and AC and BC lie in the same str. line.
38. Sch. I. It will be observed that the enunciation of every theorem consists essentially of two parts ; the Hypoth- esis, and the Conclusion.
Thus, we may enunciate Prop. I as follows : Hypothesis. If a point be taken in a given straight line, Conclusion. A perpendicular to the line at the given point can be drawn, and but one.
12 PLANE GEOMETRY. —BOOK I.
39. Sch. II. We may enunciate Prop. II as follows : Hypothesis. If two adjacent angles have their exterior
sides in the same straight line,
Conclusion. Their sum is equal to two right angles.
Again, we may enunciate Prop. Ill :
Hypothesis. If the sum of two adjacent angles is equal to two right angles,
Conclusion. Their exterior sides lie in the same straight line.
One proposition is said to be the Converse of another when the hypothesis and conclusion of the first are, respectively, the conclusion and hypothesis of the second.
It is evident from the above considerations that Prop. Ill is the converse of Prop. II.
Prop. IV. Theorem.
40. If two straight lines intersect, the vertical angles are equal.
Given str. lines AB and CD intersecting at 0. To Prove ZAOO = ZBOD.
Proof. Since AAOC and AOD have their ext. sides in str. line CD, ZAOC is the supplement of Z AOD.
[If two adj. A have their ext. sides in the same str. line, they are supplementary.] (§ 33)
For the same reason, Z BOD is the supplement of Z AOD,
.-. ZAOC=ZBOD.
[The supplements of equal A are equal.] (§ 31)
In like manner, we may prove
ZAOD = ZBOa
RECTILINEAR FIGURES.
13
EXERCISES.
7. If, in the figure of Prop. IV., ZAOD = 137°, how many degrees are there in J50C? in^OC? in BOD?
8. Two angles are supplementary, and the greater is seven times the less. How many degrees are there in each angle ?
Prop. V. Theorem.
41. If a perpendicular be erected at the middle point of a straight line,
I. Any point in the perpendicular is equally distant from the extremities of the line.
II. Any point without the perpendicular is unequally dis- tant from the extremities of the line.
I. Given line CD ± to line AB at its middle point D, E any point in CD, and lines AE and BE.
To Prove AE = BE.
Proof. Superpose figure BDE upon figure ADE by fold- ing it over about line DE as an axis.
Now Z BDE = Z ADE.
[All rt. A are equal.] • (§ 26)
Then, line BD will fall upon line AD.
But by hyp., BD = AD.
Whence, point B will fall on point A.
Then line BE will coincide with line AE.
[But one str. line can be drawn between two points.] (Ax. 3)
.-. AE = BE.
14 PLANE GEOMETRY.— BOOK I.
A D B
II. Given line CD ± to line AB at its middle point D, F any point without CD, and lines AF and BF.
To Prove AF > BF,
Proof. Let AF intersect CD at E, and draw line J5^.
Now BE + EF>BF.
[A str. line is the shortest line between two points.] (Ax. 4)
But, BE = AE.
[If a _L be erected at the middle point of a str. line, any point in the ± is equally distant from the extremities of the line.] (§ 41, I)
Substituting for BE its equal AE, we have
AE -\- EF> BF, or AF > BF.
42 Cor. I. Every point which is equally distant from the extremities of a straight line, lies in the perpendicular erected at the middle point of the line.
43. Cor. II. Since a straight line is determined by any two of its points (§ 18), it follows from § 42 that
Two points, each equally distant from the extremities of a straight line, determine a perpendicular at its middle point.
44. Cor. III. When figure BDE is superposed upon figure ADE, in the proof of § 41, 1., ZEBD coincides with Z EAD, and Z BED with Z AED.
That is, Z EAD = Z EBD, and Z AED = Z BED. Then, if lines be dratvn to the extremities of a straight line from any point in the perpendicular erected at its middle pointy
1. They make equal ayigles with the line.
2. They make equal angles with the perpendicular.
RECTILINEAR FIGURES.
15
Prop. VI. Theorem.
45. From a given point without a straight line, a perpen- dicular can be drawn to the line, and hut one.
0 ^
H
Given point C without line AB.
To Prove that a _L can be drawn from C to AB, and but one.
Proof. Let line HKhQ ± to line FG at H.
[At a given point in a str. line, a _L to the line can be drawn.] (§ 25)
Apply line FO to line AB, and move it along until HK passes through C; let point // fall at D, and draw line CD.
Then, CD is _L AB.
If possible, let CE be another ± from C to AB.
Produce CD to C", making OD = CD, and draw line EC.
By cons., ED is ± to CC at its middle point D. .-. ZCED = ZC'ED.
[If lines be drawn to the extremities of a str. line from any point in the ± erected at its middle point, they make equal A with the ±.]
(§44)
But by hyp., Z CED is a rt. Z ; then, Z C'ED is a rt. Z. .-. Z CED + Z C'ED = two rt. A.
Then line CEC is a str. line.
[If the sum of two adj. A is equal to two rt. A, their ext. sides lie in the same str. line.] (§ 37)
But this is impossible, for, by cons., CDC is a str. line.
[But one str. line can be drawn between two points.] (Ax, 3)
Hence, CE cannot be _L AB, and CD is the only ± that can be drawn.
16
PLANE GEOMETRY.— BOOK I.
Prop. VII. Theorem.
46. The perpendicular is the shortest line that can be drawn from a point to a straight line.
Given CD the ± from point C to line AB, and CE any other str. line from C to AB.
To Prove CD < CE.
Proof. Produce CD to 6", making OD = CD, and draw line EC\
By cons., ED is ± to CC at its middle point D.
.'. CE^C'E.
[If a ± be erected at the middle point of a str. line, any point in the ± is equally distant from the extremities of the line.] (§ 41)
But CD + DC <CE + EC.
[A str. line is the shortest line between two points.] (Ax 4.)
Substituting for DC and EC their equals CD and CE, respectively, we have
2CD<2 CE.
.'. CD<CE.
47. Sch. The distance of a point from a line is understood to mean the length of the perpendicular from the point to the line.
Ex. 9. Find the number of degrees in the angle the sum of whose supplement and complement is 196°.
RECTILINEAR FIGURES. ^7
Prop. VIII. Theorem.
48. If two lines he drawn from a point to the extremities of a straight line, their sum is greater than the sum of two other lines similarly drawn, hut enveloped hy them.
Given lines AB and AC drawn from point A to the extremities of line BC', and DB and DC two other lines similarly drawn, but enveloped by AB and AC.
To Prove AB^AC>DB + DC
Proof. Produce BD to meet ^C at E.
Now AB->rAE> BE.
[A str. line is the shortest line between two points.] (Ax. 4)
Adding EC to both members of the inequality, BA + AC>BE + EC.
Again, DE + EC>DC
Adding BD to both members of the inequality, BE-\-EC>BD+DC.
Since BA + AC is greater than BE -f EC, which is itself greater than BD + DC, it follows that
AB-\-AC>DB-]-DC.
EXERCISES.
10. The straight line which bisects an angle bisects also its vertical angle.
(If 0^ bisects ZAOC, ZAOE = ZCOE; E and these A are equal to ABOF and DOF, respectively.)
18
PLANE GEOMETRY.— BOOK I.
11. The bisectors of a pair of vertical angles lie in the same straight line.
(Fig. of Ex. 10. To prove ^OFa str. line. Z COE= ZDOF, for they are the halves of equal A ; but ZDOE + Z GOE = 2 rt. A, and therefore ZDOE ■\-ADOF ='in. A.)
12. The bisectors of two supplementary ad- jacent angles are perpendicular to each other.
(We have ZACD + ZBCD = 2 rt. A; and ADCE and DCF are the halves of AACD and BCD, respectively.)
13. If the bisectors of two adjacent angles are perpendicular, the angles are supplementary.
(Fig. of Ex. 12. Sum of ADCE and DCF= 1 rt. Z, and ADCE and DCF are the halves of AACD and BCD, respectively.)
14. A line drawn through the vertex of an angle ^ pei*pendicular to its bisector makes equal angles with the sides of the given angle. o
(A AOD and BOE are complements of AAOC and BOC, respectively.) c
Prop. IX. Theorem.
49. If oblique lines be drawn from a point to a straight line,
I. Two oblique lines cutting off equal distances from the foot of the perpendicular from the point to the line are equal.
II. Of two oblique lines cutting off unequal distances from the foot of the perpendicular from the point to the line, the more remote is the greater.
C
A E
F B
I. Given CD the ± from point C to line AB; and CE and CF oblique lines from C to AB, cutting olf equal distances from the foot of CD.
RECTILINEAR FIGURES. ig
To Prove CE = CF.
Proof. By hyp., CD is ± to EF at its middle point D.
.-. CE=CF.
[If a _L be erected at the middle point of a str. line, any point in the ± is equally distant from the extremities of the line.] (§ 41)
II. Given CD the X from point C to line AB; and CE and CF oblique lines from C to AB, cutting off unequal distances from the foot of CD ; CF being the more remote.
To Prove CF>CE.
Proof. Produce CD to C", making CD = CD, and draw lines CE and CF.
By cons., AD is ± to CC" at its middle point D.
.-. CF= CF, and CE = CE.
[If a ± be erected at the middle point of a str. line, any point in the ± is equally distant from the extremities of the line.] (§ 41)
But CF + FC >CE + EC.
[If two lines be drawn from a point to the extremities of a str. line, their sum is < the sum of two other lines similarly drawn, but en- veloped by them.] (§ 48)
Substituting for FO and EC their equals CF and CE,
respectively, we have
2CF>2CE.
.'. CF>CE.
Note. The theorem holds equally if oblique line CE is on the opposite side of perpendicular CD from CF.
20 PLANE GEOMETRY.— BOOK I.
Prop. X. Theorem.
50. (Converse of Prop. IX., I.) If oblique lines he drawn from a point to a straight liiie, two equal oblique lines cut off equal distances from the foot of the perpendicular from the point to the line.
G
A E D F B
Given CD the ± from point C to line AB, and CE and CF equal oblique lines from C to AB.
To Prove DJE = DF.
Proof. We know that DE is either >, equal to, or < DF.
If we suppose DE > DF, CE would be > CF.
[If oblique lines be drawn from a point to a str. line, of two oblique lines cutting off unequal distances from the foot of the ± from the point to the line, the more remote is the greater.] (§ 49)
But this is contrary to the hypothesis that CE = CF.
Hence, DE cannot be > DF.
In like manner, if we suppose DE < DF, CE would be < CF, which is contrary to the hypothesis that CE = CF.
Hence, DE cannot be < DF.
Then, if DE can be neither > DF, nor < DF, we must have DE=DF.
Note. The method of proof exemplified in Prop. X is known as the " Indirect Method," or the " Beductio ad Ahsurdum.''''
The truth of a proposition is demonstrated by making every pos- sible supposition in regard to the matter, and sliowing that, in all cases except the one which we wish to prove, the supposition leads to something which is contrary to the hypothesis.
RECTILINEAR FIGURES. 21
51. Cor. (Converse of Prop. IX, II.) If two unequal oblique lines he drawn from a point to a straight line, the greater cuts off the greater distance from
the foot of the perpendicular from the point to the line.
Given CD the _L from point C to line AB ; and GE and CF unequal oblique lines from C to AB, CF being > CE.
To Prove DF>DE.
(Prove by Reductio ad Absurdum; by § 49, I, DE cannot equal DF, and by § 49, II, it cannot be > DF?)
PARALLEL LINES.
52. Def. Two straight lines are said to be parallel (II)
when they lie in the same plane, and ^ ^
cannot meet however far they may be
produced; as ^5 and CD. ^ ^
53. Ax. We assume that hut one straight line can he drawn through a given point parallel to a given straight line.
Prop. XI. Theorem.
54. Two perpendiculars to the same straight line are parallel. .
Given lines AB and CD A. to line AC.
To Prove AB II CD.
Proof. If AB and CD are not II, they will meet in some point if sufficiently produced (§ 52).
We should then have two Js from this point to AC, which is impossible.
[From a given point without a str. line, but one ± can be drawn to the line.] (§ ^5)
Therefore, AB and CD cannot meet, and are II.
22 PLANE GEOMETRY. — BOOK I.
Prop. XII. Theorem.
55. Two straight lines parallel to the same straight line are parallel to each other.
A B
E F
Given lines AB and CD II to line EF.
To Prove AB II CD.
Proof. If AB and CD are not II, they will meet in some point if sufficiently produced. (§ 52)
We should then have two lines drawn through this point II to EF, which is impossible.
[But one str. line can be drawn through a given point II to a given str. line.] (§ 53)
Therefore, AB and CD cannot meet, and are II.
Prop. XIII. Theorem.
56. A straight line perpendicular to one of two parallels is perpendicular to the other.
Given lines AB and CD II, and line AC ± AB. To Prove AC A. CD.
Proof. If CD is not ± AC, let line CE be ± AC. Then since AB and CE are ± AC, CE II AB. [Two Js to the same str. line are II.] (§ 54)
But by hyp., CD II AB.
Then, CE must coincide with CD.
[But one str. line can be drawn through a given point II to a given str. line.] (§53)
RECTILINEAR FIGURES. 23
But by hyp., AC ± CE.
Then since CE coincides with OD, we have AC ± CD.
TRIANGLES. DEFINITIONS.
57. A triangle (A) is a portion of a plane bounded by three straight lines ; as ABC.
The bounding lines, AB, BC, and CA, are called the sides of the triangle, and their points of intersection. A, B, and (7, the vertices.
The angles of the triangle are the angles GAB, ABC, and BCA, included between the adjacent sides.
An exterior angle of a triangle is the angle at any vertex between any side of the triangle and the adjacent side produced ; as ACD.
58. A triangle is called scalene when no two of its sides are equal ; isosceles when two of its sides are equal ; equi- lateral when all its sides are equal ; and equiangular when all its angles are equal.
Scalene. Isosceles. Equilateral.
59. A right triangle is a triangle which has a right angle; as ABC, which has a right ^A
angle at C.
The side AB opposite the right angle is called the hypotenuse, and the other sides, AC and BC, the legs.
24
PLANE GEOMETRY.— BOOK I.
60. If any side of a triangle be taken and called the base, the corresponding altitude is the perpendicular drawn from the opposite vertex to the base, produced if necessary.
In general, either side may be taken as the base ; but in an isosceles triangle, unless otherwise specified, the side which is not one of the equal sides is taken as the base.
When any side has been taken as the base, the opposite angle is called the ve?-- tical angle, and its vertex is called the vertex of the triangle.
Thus, in triangle ABC, BC is the base, AD the altitude, and BAG the vertical angle.
61. Since a straight line is the shortest line between two points (Ax. 4), it follows that
Any side of a triangle is less than the sum of the other two sides.
Prop. XIV. Theorem.
62. Any side of a triangle is greater than the difference of the other two sides.
Given AB, any side of A ABC; and side ^C> side AC. To Prove AB>BC- AC.
Proof. We have AB-}-AC>BC.
[A str. line is the shortest line between two points.] (Ax. 4)
Subtracting AC from both members of the inequality, AB>BC-Aa
RECTILINEAR FIGURES. 25
Prop. XV. Theorem.
63. Ti(}o triangles are equal when two sides and the in- cluded angle of one are equal respectively to two sides and the included angle of the other.
A B D
Given, in A ABC and DEF,
AB = DE, AC=DF, and ZA = ZD.
To Prove A ABC = A DEF.
Proof. Superpose A ABC upon A DEF in such a way that Z A shall coincide with its equal Z D ; side AB falling on side DE, and side AC on side DF.
Then since AB = DE and AC = DF, point B will fall on point E, and point C on point F.
Whence, side BC will coincide with side EF.
[But one str. line can be drawn between two points.] (Ax. 3)
Therefore, the A coincide throughout, and are equal.
64. Cor. Since ABC and DEF coincide throughout, we have ZB = ZE, ZC=ZF, 8^nd BC = EF.
65. Sch. I. In equal figures, lines or angles which are similarly placed are called homologous.
Thus, in the figure of Prop. XV, Z A is homologous to Z D ; AB is homologous to DE ; etc.
66. Sch. II. It follows from § 65 that
In equal figures, the homologous parts are equal.
67. Sch. III. In equal triangles, the equal angles lie opposite the equal sides.
26 PLANE GEOMETRY.— BOOK I.
Pkop. XVI. Theorem.
68. Two triangles are equal when a side and two adjacent angles of one are equal respectively to a side and two adjaxient angles of the other.
9. F
Given, in A ABC and DEF,
AB = DE, ZA = ZD, and ZB = ZE. To Prove A ABC = A DEF.
Proof. Superpose A ABC upon A DEF in such a way that side AB shall coincide with its equal DE-, point A falling on point D, and point B on point E.
Then since ZA = ZD, side AC will fall on side DF, and point C will fall somewhere on DF.
And since ZB = Z E, side BC will fall on side EF, and point C will fall somewhere on EF.
Then point C, falling at the same time on DF and EF, must fall at their intersection, F.
Therefore, the A coincide throughout, and are equal.
EXERCISES.
15. If, in the figure of Prop. XV., AB=EF, BC=DE, and ZB = ZE, whicli angle of triangle DEF is equal to J. ? which angle is equal to C ?
16. K, in the figure of Prop. XVI., AC = DF, ZA = ZF, and ZC= ZD, which side of triangle DEF is equal to AB ? which side is equal to BC?
17. If OD and OE are the bisectors of two complementary- adjacent angles, AOB and BOC, how many degrees are there in ZDOE?
RECTILINEAR FIGURES.
27
Prop. XVII. Theorem.
69. Two triangles are equal when the three sides of one are equal respectively to the three sides of the other.
Given, in A ABC and DEF,
AB = BE, BC = EF, and CA = FD.
To Prove A ABC = A DEF.
Proof. Place A DEF in the position ABF; side DE coinciding with its equal AB, and vertex F falling at F, on the opposite side of AB from C.
Draw line CF\
By hyp., AC = AF^ and BC = BF'.
Whence, AB is _L to CF' at its middle point.
[Two points, each equally distant from the extremities of a str. line, determine a ± at its middle point. ] (§ 43)
.-. ZBAC=ZBAF'.
[If lines be drawn to the extremities of a str. line from any point in the ± erected at its middle point, they make equal A with the ±.]
(§44)
Then since sides AB and AC and Z BAC of A ABC are
equal, respectively, to sides AB and AF' and Z BAF' of
AABF',
AABC=AABF'.
[Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.]
(§63) That is, A ABC = A DEF.
28
PLANE GEOMETRY. —BOOK I.
Prop. XVIII. Theorem.
70. Two right triangles are equal when the hypotenuse and an adjacent angle of one are equal respectively to the hypote- nuse and an adjacent angle of the other.
E
A CD
Given, in rt. A ABC and DEF,
hypotenuse AB = hypotenuse BE, and Z A = Z.D. To Prove A ABC = A DEF.
Proof. Superpose A ABC upon A DEF in such a way that hypotenuse AB shall coincide with its equal DE ; point A falling on point D, and point B on point E.
Then since Z.A = ZD, side AC will fall on side DF.
Therefore, side BC will fall on side EF.
[From a given point without a str. line, but one _L can be drawn to the line.] (§45)
Therefore, the A coincide throughout, and are equal.
71. Def . If two straight lines, AB and CD, are cut by a line EF, called a transversal, the angles are named as follows :
c, d, e, and / are called interior
and a, b.
and h exterior
angles, angles.
c and /, or d and e, are called alter- nate-interior angles.
a and h, or h and g, are called alternate-exterior angles.
a and e, b and /, c and g, or d and h, are called corre- sponding angles.
RECTILINEAR FIGURES. 29
Prop. XIX. Theorem.
72. If two parallels are cut by a transversal^ the alternate- interior angles are equal.
-B
Given lis AB and CD cut by transversal EF at points G and H, respectively.
To Prove Z AGH= Z OHD and Z BGH=Z OHG.
Proof. Through K, the middle point of GH, draw line LMl. AB', then, LM 1. CD.
[A str. line _L to one of two ||s is ± to the other.] (§ 56)
Now in rt. A GKL and HKM, by cons.,
hypotenuse GK = hypotenuse HK.
Also, Z GKL = Z HKM.
[If two str. lines intersect, the vertical A are equal.] (§ 40)
.-. A GKL = A HKM.
[Two rt. ^ are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.] (§ 70)
.-. ZKGL = ZKHM.
[In equal figures, the homologous parts are equal.] (§ QQ)
Again, Z KGL is the supplement of Z BGH, and Z KHM
the supplement of Z CHG.
[If two adj. A have their ext. sides in the same str. line, they are
supplementary.] (§ 33)
Then since Z KGL = Z KHM, we have ZBGH=ZCHG,
[The supplements of equal A are equal] (§ 31)
30 PLANE GEOMETRY.— BOOK I.
Prop. XX. Theorem.
73. (Converse of Prop. XIX.) If two straight lines are cut by a transversal, and the alternate-interior angles are equal, the tivo lines are parallel.
Given lines AB and CD cut by transversal EF at points G and H, respectively, and
Z AGH= Z GHD.
To Prove AB II CD.
Proof. If CD is not II AB, draw line KL through H II AB. Then since lis AB and KL are cut by transversal EF,
ZAGH=ZGHL.
[If two lis are cut by a transversal, the alt. int. A are equal.] (§ 72)
But by hyp., Z AGH = Z GHD.
.'. ZGHL = ZGHD.
[Things which are equal to the same thing are equal to each other.]
(Ax. 1) But this is impossible unless KL coincides with CD.
.: CDWAB.
In like manner, it may be proved that if AB and CD are cut by EF, and Z BGH= Z CHG, then AB II CD.
Ex. 18. If, in the figure of Prop. XIX., /LAGH=Q^°, how many degrees are there in BGW} in GHD ? in DHF'i
RECTILINEAR FIGURES. 3J^
Prop. XXI. Theorem.
74. If two parallels are cut by a transversal, the correspond- ing angles are equal.
Given lis AB and CD cut by transversal EF at points G and H, respectively.
To Prove Z AGE = Z CHG.
Proof. We have ZBGH=ZCHG.
[If two lis are cut by a transversal, the alt. int. A are equal.] (§ 72)
But, ZBGH=ZAGE.
[If two str. lines intersect, the vertical A are equal.] (§ 40)
.-. ZAGE = ZCHG.
[Things which are equal to the same thing are equal to each other.] (Ax. 1)
In like manner, we may prove
Z AGH=Z CHF, Z BGE=Z DHG, and Z BGH=Z DHF.
75. Cor. I. If two parallels are cut by a transversal, the alternate-exterior angles are equal.
(Fig. of Prop. XXI.)
Given lis AB and CD cut by transversal EF at points G and H, respectively.
To Prove Z AGE = Z DHF.
(Z BGH= Z CHG, and the theorem follows by § 40.)
What other two ext. A in the figure are equal ?
32 PLANE GEOMETRY.— BOOK I.
76. Cor. II. If two parallels are cut by a transversal, the sum of the interior angles on the same side of the transversal is equal to two right angles.
(Fig. of Prop. XXI.)
Given lis AB and CD cut by transversal EF at points G and H, respectively.
To Prove Z AGH + Z CHG = two rt. A.
(By § 32, /.AGH+ZAGE = t^o rt. A-, the theorem follows by § 74.)
What other two int. A in the figure have their sum equal to two rt. A ?
Prop. XXII. Theorem.
77. (Converse of Prop. XXI.) If two straight lines are cut by a transversal, and the corresponding angles are equal, the two lines are parallel.
'E
Given lines AB and CD cut by transversal EF at points G and H, respectively, and
Z AGE = Z CHG. To Prove AB II CD.
Proof. We have Z AGE = Z BGH. - ,
[If two str. lines intersect, the vertical A are equal.] (§ 40)
.-. ZBGH=^ZCHG.
[Things which are equal to the same thing are equal to each other.]
(Ax. 1)
RECTILINEAR FIGURES. 33
.-. ABWCD.
[If two str. lines are cut by a transversal, and the alt. int. A are equal, the two lines are II.] (-§ 73)
In like manner, it may be proved that if
Z AGH=Z CIIF, or Z BGE=ZDHG, or Z BGH=ZDHF,
then AB 11 CD.
78. Cor. I. (Converse of § 75.) If two straight lines are cut by a transversal, and the alternate-exterior angles are equal, the two lines are parallel.
(Fig. of Prop. XXII.)
Given lines AB and CD cut by transversal EF at points G and H, respectively, and
Z AGE = Z DHF.
To Prove AB 11 CD.
(Z AGE = Z BGH, and Z DHF= Z CEG; and the theo- rem follows by § 73.)
What other two ext. A are there in the figure such that, if they are equal, AB II CD ?
79. Cor. II. (Converse of § 76.) If two straight lines are cut by a transversal, and the sum of the interior angles on the same side of the transversal is equal to two right angles, the two lines are parallel.
(Fig. of Prop. XXII.)
Given lines AB and CD cut by transversal EF at points G and H, respectively, and
Z AGH+ Z CHG = two rt. A.
To Prove AB II CD.
(Z CHG is the supplement of Z AGH, and also of Z GHD ; then A AGH and GHD are equal by § 31, 2, and the theo- rem follows by § 73.)
What other two int. A are there in the figure such that, if their sum equals two rt. A, AB II CD ?
34 PLANE GEOMETRY.— BOOK I.
Prop. XXIII. Theorem. 80. Two parallel lines are everywhere equally distant. A E F B
C G H D
Given lis AB and CD, E and F any two points on AB, and Ea and FH lines ± CD.
To Prove EG = FH (§ 47).
Proof. Draw line FG.
We have EG±AB.
[A str. line i. to one of two ||s is ± to the other.] (§ 56)
Then, in rt. A EFG and FGH,
FG = FG.
And since lis AB and CD are cut by FG,
Z EFG = Z FGH.
[If two lis are cut by a transversal, the alt. int. A are equal.] (§ 72)
.-. A EFG = A FGH.
[Two rt. A are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.] (§ 70)
.-. EG = FH
[In equal figures, the homologous parts are equal.] (§ 66)
Prop. XXIV. Theorem.
81. Two angles whose sides are parallel, each to each, are equal if both pairs of parallel sides extend in the same direc- tion, or 171 opposite directions, from their vertices.
Note. The sides extend in the same direction if they are on the same side of a straight line joining the vertices, and in opposite direc- tions if they are on opposite sides of this line.
RECTILINEAR FIGURES. 35
fA /D
Given lines AB and BC \\ to lines DH and KF, respec- tively, intersecting at E.
I. To Prove that A ABC and DEF, whose sides AB and DE, and also BC and ^i^, extend in the same direction from their vertices, are equal.
Proof. Let BC and DH intersect at G. Since lis AB and DE are cut by BC,
ZABC=ZDGC
[If two lis are cut by a transversal, the corresp. A are equal.]
(§74) In like manner, since lis BC and EF are cut by DE,
ZDGC = ZDEF.
.-. ZABG = ZDEF. (1)
[Things which are equal to the same thing are equal to each other.]
(Ax. 1)
II. To Prove that A ABC and HEK, whose sides AB and EH, and also BC and EK, extend in opposite directions from their vertices, are equal.
Proof. From(l), Z ABC = Z DEF.
But, Z DEF = Z HEK.
[If two str. lines intersect, the vertical A are equal.] (§ 40)
.-. ZABC = ZHEK.
[Things which are equal to the same thing, are equal to each other.]
(Ax. 1)
36
PLANE GEOMETRY.— BOOK I.
82. Cor. Two angles whose sides are parallel, each to each, are supplementary if one pair of parallel sides extend in the same direction, and the other pair in opposite directions, from their vertices.
Given lines AB and BC \\ to lines DH and KF, respectively, intersecting at J57.
To Prove that A ABC and DEK, whose sides AB and DE extend in the same direction, and BC and EK in opposite directions, from their vertices, are supplementary.
Proof. We have Z ABC = Z DEF.
[Two A whose sides are ||, each to each, are equal if both pairs of || sides extend in the same direction from tlieir vertices.] (§ 81)
But Z DEF is the supplement of Z DEK. [If two adj. A have tlieir ext. sides in the same str. line, they are supplementary.] (§ 33)
Then its equal, Z ABC, is the supplement of Z DEK.
Prop. XXV. Theorem.
83. Two angles whose sides are perpejidicular, each to each, are either equal or supplementary.
Given lines AB and BC ± to lines DE and FG, respec- tively, intersecting at E.
To Prove Z ABC equal to Z DEF, and supplementary to Z DEC.
RECTILINEAR FIGURES. 37
Proof. Draw line EH± DE, and line EKl. EF. Then since EH and AB are ± DE,
EH II AB. [Two ±s to the same str, line are II.] (§ 54)
In like manner, since EK and BC are _L EF, EK II BO. .'. /.HEK=ZABC. [Two A whose sides are ||, each to each, are equal if both pairs of || sides extend in the same direction from their vertices.] (§ 81)
But since, by cons., A DEH and FEK are rfc. A, each of the A DEF and HEK is the complement of Z FEH. .'. A DEF = A HEK. [The complements of equal A are equal.] (§ 31)
.-. A ABC = A DEF. [Things which are equal to the same thing are equal to each other.]
(Ax. 1) Again, A DEF is the supplement of A DEG. [If two adj. A have their ext. sides in the same str. line, they are supplementary.] (§33)
Then, its equal, A ABC is the supplement of A DEG.
Note. The angles are equal if they are both acute or both obtuse ; and supplementary if one is acute and the other obtuse.
EXERCISES.
19. If, in the figure of Prop. XXIV., Z^BC'=59°, how many degrees are there in each of the angles formed about the point E ?
20. The line passing through the vertex of an angle perpendicular to its bisector bisects the supplementary adjacent angle.
(Fig. of Ex. 12. Let CE bisect ZACD, and suppose CFl. CE ; sum of AACD and BCD = 2 rt. A; then sum of ADCE and ^ J?CZ) = 1 rt. Z ; but sum ot ADCE and DCF is also 1 rt. Z; whence the theorem follows.)
21. Any side of a triangle is less than the half-sum of the sides of the triangle.
(Fig. of Prop. XIV. We have AB<BC-^ CA; then add AB to both members of the inequality.)
38 PLANE GEOMETRY.— BOOK I.
Prop. XXVI. Theorem.
84. Tlie sum of the angles of any triangle is equal to two right angles.
A CD
Given A ABC.
To Prove Z. A + /. B + AC =t^oit. A. Proof. Produce AC to D, and draw line GE II AB. Then, Z BCD + Z BCE + Z AGB = two rt. A. (1)
[The sum of all the A on the same side of a str. line at a given point is equal to two rt. A.'] (§ 34)
Now since lis AB and CE are cut by AD,
A ECI) = AA. [If two lis are cut by a transversal, the corresp. A are equal.] (§ 74) And since lis AB and CE are cut by BC,
A BCE = AB. [If two lis are cut by a transversal, the alt. int. A are equal.] (§ 72) Substituting in (1), we have
AA + AB + A ACB = two rt. A.
85. Cor. I. It follows from the above demonstration that Z BCD = A ECD + Z BCE = AA + AB', hence
1. An exterior angle of a triangle is equal to the sum of the two opposite interior angles.
2. An exterior angle of a triangle is greater than either of the opposite interior angles.
86. Cor. II. If two triangles have two angles of one equal respectively to two angles of the other, the third angle of the first is equal to the third angle of the second.
RECTILINEAR FIGURES.
39
87. Cor. III. A triangle cannot have two right angles, nor two obtuse angles.
88. Cor. rV. The sum of the acute angles of a right tri- angle is equal to one right angle.
89. Cor. V. Two right triangles are equal when a leg and an acute angle of one are equal respectively to a leg and the homologous acute angle of the other.
The theorem follows by §§ 86 and 68.
Prop. XXVII. Theorem.
90. Two right triangles are equal when the hypotenuse and a leg of one are equal respectively to the hypotenuse and a leg of the other.
A CD
Given, in rt. A ABC and DEF,
hypotenuse AB = hypotenuse DE, and BG = EF.
To Prove A ABC = A DEF.
Proof. Superpose A ABC upon A DEF in such a way that side BC shall coincide with its equal EF; point B falling on point E, and point C on point F.
We have ZC=ZF.
[All rt.^ are equal.] (§26)
Then, side AC will fall on side DF.
But the equal oblique lines AB and DE cut ofP upon DF equal distances from the foot of J_ EF.
[If oblique lines be drawn from a point to a str. line, two equal oblique lines cut off equal distances from the foot of the ± from the point to the line.] (§ ^0)
Therefore, point A falls on point D.
Hence, the A coincide throughout, and are equal.
40 PLANE GEOMETRY. — BOOK I.
Prop. XXVIII. Theorem.
91. If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, the third side of the first is greater than the third side of the second.
E
a
Given, in A ABC and DEF,
AB = DE, AO=DF, and ZBAC>ZD. To Prove BC>EF.
Proof. Place A DEF in the position ABO; side DE coinciding with its equal AB, and vertex F falling at G.
Draw line AH bisecting Z GAG, and meeting BG at H; also, draw line GH.
In A AGH and AGH, AH= AH.
Also, by hyp., AG = AG.
And by cons., Z GAH= A GAH.
.: A AGH = A AGH
[Two ^ are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (§ 63)
.-. GH=GH.
[In equal figures, the homologous parts are equal.] (§ 66)
But, BH+GH>BG.
[A str. line is the shortest line between two points.] (Ax. 4)
Substituting for GH its equal GH, we have
BH+GH>BG, or BG>EF.
RECTILINEAR FIGURES.
41
Prop. XXIX. Theorem.
92. (Converse of Prop. XXVIII.) // two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, the included angle of the first is greater than the included angle of the second.
Given, in A ABC and DBF,
AB = DE, AG=DF, and BC>EF.
To Prove /.A>^D.
Proof. We know that ZAis either <, equal to, or > Z Z>.
If we suppose ZA = ZD, A ABC would equal A DEF.
[Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (§ 63)
Then, BC would equal EF.
[In equal figures, the homologous parts are equal.] (§ 66)
Again, if we suppose ZA<ZD, BC would be < EF.
[If two i^ have two sides of one equal respectively to two sides of the other, but the included Z of the first > the included Z of the second, the third side of the first is > the third side of the second.]
(§91)
But each of these conclusions is contrary to the hypothe- sis that BC is > EF.
Then, ii ZA can be neither equal to Z D, nor < ZD,
ZA>ZD.
42 PLANE GEOMETRY. — BOOK I.
Prop. XXX. Tiikorem.
93. Tn an isosceles triangle, the angles opposite the equal sides are equal,
c
Given AC and BC the equal sides of isosceles A ABC* To Prove ZA^ZB,
Proof. Draw line CD ± AB, In It. A AOD iind BCD,
CD=CD, And by hyp., AC=BC
,-. AACD = ABCD. [Two rt. /S^ 9,re etjiial when the hypotenuse and a leg of one are equal reapeoUvely to the hypotenuse and a leg of the other,] (§ IH))
.'. Z .4 = Z 5.
[In equal figures, the homologous pai'ta are equal.] (§ GO)
94. Cor. I. From equal A ACD and BCD, we have
AD^^BDy and /LACD^ZBCDy hence,
1. The perpendicular from the vertex to the base of an iaoaoelea triangle bisects the base.
2. The perpendicular from the vertex to the base of an isosceles triangle bisects the vertical angle.
95. Cor. II. An eq^tilateml triangle is also equiangular.
Prop. XXXI, Theorem.
96. (Converse of Vvo\\ XXX.) Iftivo angles of a triangle
are equal, the sides opposite are equal.
RECTILINEAR FIGURES. 43
(Fig. of Prop. XXX.) Given, in A ABC, ZA = Z B.
To Prove AC = EC.
(Prove A ACD = A BCD by § 89.)
97. Cor. An equiangular triangle is also equilateral.
EXERCISES.
22. The angles A and B of a triangle ABC are 57° and 98° respec- tively ; how many degrees are there in the exterior angle at C ?
23. How many degrees are there in each angle of an equiangular triangle ?
Prop. XXXII. Theorem.
98. If two sides of a triangle are unequal, the angles oppo- site are unequal, and the greater angle lies opposite the greater side.
A
Given, in A ABC, AC > AB. To Prove ZABC>ZC.
Proof. Take AD = AB, and draw line BD. Then, in isosceles A ABD,
ZABD = ZADB. [In an isosceles A, the A opposite the equal sides are equal.] (§ 93) Now since Z ADB is an ext. Z of A BDG,
Z ADB >ZC. [An ext. Z of a A is > either of the opposite int. A.] (§ 86)
Therefore, its equal, Z ABD, is > ZC. Then, since Z ABC is > Z ABD, and Z ABD >ZC,
ZABOZa
44 PLANE GEOMETRY.— BOOK I.
Prop. XXXIII. Theorem.
99. (Converse of Prop. XXXII.) If two angles of a tri- angle are unequal, the sides opposite are unequal, and the greater side lies opposite the greater angle.
A
Given, in A ABC, Z ABC > Z C. To Prove AC > AB.
Proof. Draw line BD, making Z CBD = Z C, and meet- ing AC Sit D.
Then, in A BCD, BD = CD.
[If two A of a, A are equal, the sides opposite are equal,] (§ 96)
But, AD + BD>AB.
[A str. line is the shortest line between two points.] (Ax. 4)
Substituting for BD its equal CD, we have
AD+CD> AB, or AC>AB.
Prop. XXXIV. Theorem.
100. If straight lines he drawn from a point within a triangle to the extremities of any side, the angle included by them is greater than the angle included by the other two sides.
Given D, any point within A ABC, and lines BD and CD.
J
RECTILINEAR FIGURES.
To Prove ZBDC>ZA.
Proof. Produce BD to meet AC dJt E. Then, since Z BDC is an ext. Z of A CDE,
ZBDC>ZDEC.
[An ext. Z of a A is > either of the opposite int. zi.]
In like manner, since Z DEO is an ext. Z of A ABE,
ZDEOZA. Then, since Z ^Z>(7 is > Z DEC, and Z 2)^0 > Z ^
Zi?Z)0>ZA
45
(§85)
Prop. XXXV. Theorem.
101. Any point in the bisector of an angle is equally distant from the sides of the angle.
/A
Given P, any point in bisector BD of Z ABC, and lines PJf and FN± to AB and AC, respectively.
To Prove PM = PiV:
Proof. In rt. A BPM and BPN, BP=BP.
And by hyp., Z PBM= Z PBN.
.-. ABPM=ABPN.
[Two rt. A are equal when the hypotenuse and an adj. Z of one are equal respectively to the hypotenuse and an adj. Z of the other.]
(§70) .-. PM=PN.
[In equal figures, the homologous parts are equal]
(§ ^>«)
46
PLANE GEOMETRY. —BOOK I.
Prop. XXXVI. Theorem.
102. (Converse of Prop. XXXV.) Every point which is within an angle, and equally distant from its sides, lies in the bisector of the angle.
/A
Given point P within Z ABC, equally distant from sides AB and BC, and line BP.
To Prove
ZPBM=ZPBN.
(Prove ABPM = ABPNy by § 90; the theorem then follows by § QQ.)
EXERCISES.
24. The angle at the vertex of an isosceles triangle ABC is equal to five-thirds the sum of the equal angles B and C. How many degrees are there in each angle ?
25. If from a point 0 in a straight line AB lines OC and OD be drawn on opposite sides of AB, making Z.AOC = Z.BOD, prove that OC and OD lie in the same straight line,
(Fig. of Prop. IV. We have ZAOD + ZB0D = 2 rt. /i, and by hyp., ZBOD= ZAOC.)
26. If the bisectors of two adjacent angles make au angle of 45° with each other, the angles are com- plementary.
(Given OD and 0-E^ the bisectors of AAOB and BOC, respectively, and ZDOE = i^° ; to prove A AOB and BOC complementary.)
27. Prove Prop. XXX. by drawing CD to bisect Z ACB. (§ 63.)
28. Prove Prop. XXX. by drawing CD to the middle point of AB
29. Prove Prop. XXXI. by drawing CD to bisect ZACB. (§ 68.)
EECTILINEAR FIGURES. 47
QUADRILATERALS. DEFINITIONS.
103. A quadrilateral is a portion of a plane bounded by- four straight lines ; as ABCD. ^
The bounding lines are called the sides /\.
of the quadrilateral, and their points of / n.
intersection the vertices. / N.
The angles of the quadrilateral are the ^\™ ""/C
angles included between the adjacent ^\^^ /^
sides. D
A diagonal is a straight line joining two opposite vertices ; as AO.
104. A Trapezium is a quadrilateral no two of whose sides are parallel.
A Trapezoid is a quadrilateral two, and only two, of whose sides are parallel.
A Parallelogram (O) is a quadrilateral whose opposite sides are parallel.
TVapezium. Trapezoid. Parallelogram.
The bases of a trapezoid are its parallel sides; the alti- tude is the perpendicular distance between them.
If either pair of parallel sides of a parallelogram be taken and called the hases, the altitude corresponding to these bases is the perpendicular distance between them.
105. A Rhomboid is a parallelogram whose angles are not right angles, and whose adjacent sides are unequal.
A Rhombus is a parallelogram whose angles are not right angles, and whose adjacent sides are equal.
A Rectangle is a parallelogram whose angles are right angles.
48 PLANE GEOMETRY. —BOOK I.
A Square is a rectangle whose sides are equal.
Rhoniboid.
Rhombus.
Rectangle.
Square.
Prop. XXXVII. Theorem. 106. In any parallelogram, I. TJie opposite sides are equal. II. Tlie opposite angles are equal.
B,
Given 0^5(7Z>.
I. To Prove AB= CD and BC = AD. Proof. Draw diagonal AC.
In A ABC and ACD, AC=AC.
Again, since lis BC and AD are cut by AC,
Z BCA = Z CAD.
[K two lis. are cut by a transversal, the alt. int. A are equal.] (§ 72)
. In like manner, since lis AB and CD are cut by AC,
ZBAC=ZACD.
.-. A ABC = A ACD.
[Two A are equal when a side and two adj. A of one are equal respectively to a side and two adj. A of the other.] (§ 68)
.-. AB = CD and BC = AD.
[In equal figures, the homologous parts are equal.] (§ 66)
II. To Prove Z BAD = Z BCD Sind ZB = ZD. Proof. We have AB II CD, and AD 11 CB; and AB and
CD, and also AD and CB, extend in opposite directions from A and C.
RECTILINEAR FIGURES. 49
.-. Z BAD = Z BCD.
[Two A whose sides are ||, each to each, are equal if both pairs of II sides extend in opposite directions from their vertices.] (§ 81)
In like manner, ZB = Z D.
107. Cor. I. Parallel lines included between parallel lines are equal.
108. Cor. n. A diagonal of a parallelogram divides it into two equal triangles.
Prop. XXXVIII. Thj:orem.
109. (Converse of Prop. XXXVII, I.) If the opposite sides of a quadrilateral are equal, the figure is a parallelogram.
Given, in quadrilateral ABCD,
AB=CD SindBC = AD. To Prove ABCD a O. Proof. Draw diagonal AC. In A ABC and ACD, AC = AC. And by hyp., AB = CD and BC = AD.
.: A ABC = A ACD. [Two A are equal when the three sides of one are equal respec- tively to the three sides of the other.] (§ 69)
.-. ZBCA = ZCAD3indZBAC=ZACD.
[In equal figures, the homologous parts are equal.] (§ Q6)
Since Z BCA = Z CAD, BC II AD.
[If two str. lines are cut by a transversal, and the alt. int. A are equal, the two lines are jj.] (§ 73)
In like manner, since Z BAC = Z ACD, AB II CD. Then by del, ABCD is a O.
50 PLANE GEOMETRY.— BOOK I.
Ex. 30. If one angle of a parallelogram is 119°, how many degrees are there in each of the others ?
Prop. XXXIX. Theorem.
110. If two sides of a quadrilateral are equal and parallel, the Jigure is a parallelogram.
Given, in quadrilateral ABCD, BC equal and II to AD.
To Prove ABOD a O.
(Prove AABC=AACD, by §63; then, the other two sides of the quadrilateral are equal, and the theorem follows by § 109.)
Prop. XL. Theorem. HI. TJie diagonals of a parallelogram bisect each other.
Br^. ^ ^-^^C
Given diagonals AC and BD of O ABCD intersecting at ^.
To Prove AE = EC and BE = ED.
(Prove A AED = A BEC, by § 68.)
Note. The point E is called the centre of the parallelogram.
Prop. XLI. Theorem.
112. (Converse of Prop. XL.) If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram.
RECTILINEAR FIGURES. 51
(Fig. of Prop. XL.)
Given AG and BD, the diagonals of quadrilateral ABCD, bisecting each other at E.
To Prove ABCD a O.
(Frove A AED = A BEC, by § 63; then AD = BC] in like manner, AB= CD, and the theorem follows by § 109.)
Prop. XLII. Theorem.
113. Two parallelograms are equal when two adjacent sides and the included angle of one are equal respectively to two adjacent sides and the included angle of the other.
Given, in [U ABCD and EFGH,
AB = EF, AD = EH, and ZA = ZE.
To Prove CJABCD = CJ EFGH.
Proof. Superpose CJ ABCD upon CJ EFGH in such a way that Z A shall coincide with its equal Z E ; side AB falling on side EF, and side AD on side EH.
Then since AB = EF and AD = EH, point B will fall on point F, and point D on point H
Now since BC II AD and FG II EH, side BC will fall on side FG, and point G will fall somewhere on FG.
[But one str. line can be drawn through a given point jj to a given str. line.] (§ 53)
In like manner, side DC will fall on side HG, and point C will fall somewhere on HG.
Then point C, falling at the same time on FG and HG, must fall at their intersection G.
Hence, the CEJ coincide throughout, and are equal.
52
PLANE GEOMETRY.— BOOK I.
114. Cor. Tivo rectangles are equal if the base and alti- tude of one are equal respectively to the base and altitude of the other.
Prop. XLIII. Theorem.
115. The diagonals of a rectangle are equal. B^ ^C
Given AC and BD the diagonals of rect. ABCD, To Prove AC = BD.
(Prove rt. A ABD = rt. A ACD, by § 63.)
116. Cor. The diagonals of a square are equal.
Prop. XLIV. Theorem.
117. The diagonals of a rhombus bisect each other at right angles. ^
{AC and BD bisect each other at rt. z§ by § 43.)
EXERCISES.
31. The bisector of the vertical angle of an isosceles triangle bisects the base at right angles.
(Fig. of Prop. XXX. In equal A ACD and BCD, we have ZADC = ZBDC; then CD±AB by § 24.)
32. The line joining the vertex of an isosceles triangle to the middle point of the base, is perpendicular to the base, and bisects the vertical angle.
(Fig. of Prop. XXX. Prove CD ± AB as in Ex. 31.)
33. If one angle of a parallelogram is a right angle, the figure is a rectangle.
RECTILINEAR FIGURES.
63
POLYGONS.
DEFINITIONS.
118. A polygon is a portion of a plane bounded by three or more straight lines ; as ABCDE.
The bounding lines are called the sides of the polygon, and their sum is called the perimeter.
The angles of the polygon are the angles EAB, ABC, etc., included be- tween the adjacent sides ; and their vertices are called the vertices of the polygon.
A diagonal of a polygon is a straight line joining any two vertices which are not consecutive ; as AC.
119. Polygons are classified with reference to the number of their sides, as follows :
|
No. or Sides. |
Designation. |
No. OF Sides. |
Designation. |
|
3 4 5 6 7 |
Triangle. Quadrilateral. Pentagon. Hexagon. Heptagon. |
8 9 10 11 12 |
Octagon. Enneagon. Decagon. Hendecagon. Dodecagon. |
120. An equilateral polygon is a polygon all of whose sides are equal.
An equiangular polygon is a polygon all of whose angles are equal.
121. A polygon is called convex when no side, if produced, will enter the surface enclosed by the perimeter ; as ABCDE.
It is evident that, in such a case, each angle of the polygon is less than two right angles.
54
PLANE GEOMETRY.— BOOK I.
All polygons considered hereafter will be understood to be convex, unless the contrary is stated.
A polygon is called concave when at least two of its sides, if produced, will enter the surface enclosed G. ^
by the perimeter ; as FGHIK. / \n/^
It is evident that, in such a case, at least / \
one angle of the polygon is greater than I \
two right angles. -KT
Thus, in polygon FGHIK, the interior angle G HI is greater than two right angles.
SucK an angle is called re-entrant.
122. Two polygons are said to be mutually equilateral
when the sides of one are p/
C '
equal respectively to the ■ — V B,
sides of the other, when
taken in the same order. ^
Thus, polygons ABCD ^ ^ A
and A'B'C'D' are mutually equilateral if
AB = A'B', BC = B'C', GD=OD\ and DA = D'A\
Two polygons are said to be mutually equiangular when the angles of one are equal respectively to the angles of the other when taken in the F same order.
Thus, polygons EFGH and ^ E'F'G'H' are mutually equiangular if
ZE=ZE', ZF=ZF', AG = ZG', and ZH=ZH'.
123. In polygons which are mutually equilateral or mutually equiangular, sides or angles which are similarly placed are called homologous.
In mutually equiangular polygons, the sides included between equal angles are homologous.
124. If two triangles are mutually equilateral, they are also mutually- equiangular (§ 69).
I
RECTILINEAR FIGURES. 55
But with this exception, two polygons may be mutually equilateral without being mutually equiangular, or mutually equiangular without being mutually equilateral.
If two polygons are both mxUually equilateral and mutually equiangular, they are equal.
For they can evidently be applied one to the other so as to coincide throughout.
125. Two polygons are equal when they are composed of the same number of triangles, equal ea^h to each, and similarly placed.
For they can evidently be applied one to the other so as to coincide throughout.
Prop. XLV. Theorem.
126. The sum of the angles of any polygon is equal to two right angles taken as many times, less two, as the polygon has sides.
Given a polygon of n sides.
To Prove the sum of its A equal to n — 2 times two rt. A.
Proof. The polygon may be divided into n — 2 A by drawing diagonals from one of its vertices.
The sum of the A of the polygon is equal to the sum of the A of the A.
But the sum of the A of each A is two rt. A.
[The sum of the A of any A is equal to two rt. A.'\ (§ 84)
Hence, the sum of the A of the polygon is n — 2 times two rt. A.
56 PLANE GEOMETRY.— BOOK I.
127. Cor. I. The sum of the angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles.
For if B represents a rt. Z, and n the number of sides of a polygon, the sum of its A is (ri — 2)x2E, or 2 nR — ^B.
128. Cor. II. The sum of the angles of a quadrilateral is equal to four right angles; of a pentagon, six right angles; of a hexagon, eight right angles; etc.
Prop. XLVI. Theorem.
129. If the sides of any polygon be produced so as to make an exterior angle at each vertex, the sum of these exterior angles is equal to four right angles.
Given a polygon of n sides with its sides produced so as to make an ext. Z at each vertex.
To Prove the sum of these ext. A equal to 4 rt. A.
Proof. The sum of the ext. and int. A at any one vertex is two rt. A.
[If two adj. A have their ext. sides in the same str. line, their sum is equal to two rt. A.'] (§ 32)
Hence, the sum of all the ext. and int. A\^2n rt. A. But the sum of the int. A alone is 2 n rt. zi — 4 rt. A.
[The sum of the A of any polygon is equal to twice as many rt. A as the polygon has sides, less 4 rt. zi.] (§ 127)
Whence, the sum of the ext. z^ is 4 rt. A
RECTILINEAR FIGURES. 57
EXERCISES.
34. How many degrees are there in each angle of an equiangular hexagon ? of an equiangular octagon ? of an equiangular decagon ? of an equiangular dodecagon ?
35. How many degrees are there in the exterior angle at each vertex of an equiangular pentagon ?
36. If two angles of a quadrilateral are supplementary, the other two angles are supplementary.
37. If, in a triangle ABC, ZA=ZB, a line par- allel to AB makes equal angles with sides AC and BC.
(To prove Z CDE = Z CED.)
38. If the equal sides of an isosceles triangle be produced, the exterior angles made with the base are equal. (§31,2.) ^/ \b
d/ \e
39. If the perpendicular from the vertex to the base of a triangle bisects the base, the triangle is isosceles.
(Fig. of Prop. XXX. A ACD and BCD are equal by § 63.)
\C
40. The bisectors of the equal angles of an isosceles triangle form, with the base, another isosceles triangle.
41. If from any point in the base of an isosceles tri- angle perpendiculars to the equal sides be drawn, they make equal angles with the base. Ey
(ZADE = ZBDF, by § 31, 1.) A-
D
42. If the angles adjacent to one base of a b, -yC
trapezoid are equal, those adjacent to the other / \
base are also equal. /
(Given ZA = ZD; to prove ZB = ZC.) ^
43. Either exterior angle at the base of an isos- celes triangle is equal to the sum of a right angle and one-half the vertical angle.
U DAE is an ext. Z of A ACD.)
E
58
PLANE GEOMETRY. —BOOK I.
44. The straight lines bisecting the equal angles of an isosceles triangle, and terminating in the oppo- site sides, are equal.
45. Two isosceles triangles are equal when the base and vertical angle of one are equal respectively to the base and vertical angle of the other.
(Each of the remaining A of one A is equal to each of the remain- ing A of the other.)
46. If two parallels are cut by a transversal, ^^ e. the bisectors of the four interior angles form a rectangle. H^
(EHW FG, by § 73 ; in like manner, EF \\ GH; c then use Exs. 12 and 33.)
47. Prove Prop. XXVI. by drawing through B a line parallel to AC.
(Sum oiA&tB = 2 rt. A.)
MISCELLANEOUS THEOREMS.
Prop. XLVII. Theorem.
130. The line joining the middle points of two sides of a triangle is parallel to the third side, and equal to one-half of it.
Given line DE joining middle points of sides AB and AC, respectively, of A ABC.
To Prove DE 11 BC, and DE = ^BC.
Proof. Draw line BF II AC, meeting ED produced at F.
RECTILINEAR FIGURES. 59
In A ADE and BDF,
Z ADE = Z BDF.
[If two str. lines intersect, the vertical A are equal.] (§ 40)
Also, since lis AC and BF are cut by AB,
ZA = Z DBF.
[If two lis are cut by a transversal, the alt. int. A are equal.] (§ 72)
And by hyp., AD = BD.
.-. A ADE = A BDF.
[Two A are equal when a side and two adj. A of one are equal re- spectively to a side and two adj. A of the other.] (§ 68)
.-. DE = DF and AE = BF.
[In equal figures, the homologous parts are equal.] (§ 66)
Then since, by hyp., AE — EC, BF is equal and II to CE.
Whence, BCEF is a O.
[If two sides of a quadrilateral are equal and ||, the figure is a O.]
(§ 110) .-. DEWBC.
Again, since DE = DF,
DE = \FE = \BC.
[In any O, the opposite sides are equal.] (§ 106)
131. Cor. The line which bisects one side of a triangle, and is parallel to another- side, bisects also the third side.
Given, in A ABC, D the middle point of side AB, and line DE II BC.
To Prove that DE bisects AC.
Proof. A line joining D to the middle point of AC will be 11 BC.
[The line joining the middle points of two sides of a A is || to the third side.] (§ 130)
Then this line will coincide with DE.
[But one str. line can be drawn through a given point || to a given str. line.] (§ 53)
Therefore, DE bisects AC.
60 PLANE GEOMETRY. —BOOK I.
Prop. XLVIII. Theorem.
132. TJie line joining the middle points of the non-parallel sides of a trapezoid is parallel to the bases, and equal to one- half their sum.
Given line EF joining middle points of non-ll sides AB and CD, respectively, of trapezoid ABCD.
To Prove EF II to AD and EC, and EF = \ {AD + BC).
Proof. If EF is not II to AD and BC, draw line EK II to AD and BC, meeting CD at K-, and draw line BD in- tersecting EF at G, and EK at H.
In A ABD, EH is II AD and bisects AB ; then it bisects BD.
[The line which bisects one side of a A, and is |I to another side, bisects also the third side.] (§ 131)
In like manner, in ABCD, HK is II BC and bisects BD; then it bisects CD.
But this is impossible unless EK coincides with EF.
[But one str. line can be drawn between two points.] (Ax. 3)
Hence, EF is II to AD and BC.
Again, since EG coincides with EH, and EH bisects AB
^nd BD, EG = \AD. (1)
[The line joining the middle points of two sides of a A is equal to one-half the third side.] (§ 130)
In like manner, since GF bisects BD and CD,
GF=\BC. (2)
Adding (1) and (2),
EG+GF=^AD + ^BG. Or, EF=\{AD+BC).
RECTILINEAR FIGURES. 61
133. Cor. The line which is parallel to the bases of a traj)- ezoid, and bisects one of the non-parallel sides, bisects the other also.
Prop. XLIX. Theorem.
134. TJie bisectors of the angles of a triangle intersect at a common point.
C
Given lines AD, BE, and CF bisecting A A, B, and 0, respectively, of A ABC.
To Prove that AD, BE, and CF intersect at a common point.
Proof. Let AD and BE intersect at 0. Since 0 is in bisector AD, it is equally distant from sides AB and AC.
[Any point in the bisector of an Z is equally distant from the sides of the Z.] (§ 101)
In like manner, since 0 is in bisector BE, it is equally distant from sides AB and BC.
Then 0 is equally distant from sides AC and BC, and therefore lies in bisector CF.
[Every point which is within an Z, and equally distant from its sides, lies in the bisector of the Z.] (§ 102)
Hence, AD, BE, and CF intersect at the common point 0.
135. Cor. TJie point of intersection of the bisectors of the angles of a triangle is equally distant from the sides of the triangle.
62
PLANE GEOMETRY. — BOOK I.
Prop. L. Theorem.
136. The perpendiculars erected at the middle points of the sides of a triangle intersect at a common point.
Given DG, EH, and FK the Js erected at middle points D, E, and F, of sides BC, GA, and AB, respectively, of A ABC.
To Prove that DG, EH, and FK intersect at a common point.
(Let DG and EH intersect at 0 ; by § 41, 0 is equally distant from B and C; it is also equally distant from A and (7; the theorem follows by § 42.)
137. Cor. The point of intersection of the perpendiculars erected at the middle poirits of the sides of a triangle, is equally distant from the vertices of the triangle.
EXERCISES.
48. If the diagonals of a parallelogram are equal, the figure is a rectangle.
(Fig. of Prop. XLIII. A ABD and ACD are equal, and therefore ZBAJ)=/.ADC ; also, these A are supplementary.)
49. If two adjacent sides of a quadrilateral are equal, and the diagonal bisects their included angle, the other two sides are equal,
(Given AB = AD, and AC bisecting Z BAD; to prove BC= CD.)
I
50. The bisectors of the interior angles of a
parallelogram form a rectangle,
(By Ex. 46, each Z of EFGH is a rt. Z.) /^ h
RECTILINEAR FIGURES.
63
Prop. LI. Theorem.
138. The perpendiculars from the vertices of a triangle to the opposite sides intersect at a common point.
K^
Given AD, BE, and Ci^the Js from the vertices of A ABC to the opposite sides.
To Prove that AD, BE, and CF intersect at a common point.
Proof. Through A, B, and C, draw lines HK, KG, and GH II to BC, CA, and AB, respectively, forming A GHK Then AD, being i. BC, is also ± HK.
[A str. line ± to one of two ||s is ± to the other.] (§ 56)
Now since, by cons., ABCH a,nd ACBK&ve UJ,
AH=BC and AK=Ba [In any O, the opposite sides are equal.] (§ 106)
.-. AH=AK
[Things which are equal to the same thing, are equal to each other.]
(Ax. 1)
Then AD is i. fiT/f at the middle point of HK
In like manner, BE and CF are _L to KG and GH, respec- tively, at their middle points.
Then, AD, BE, and CF being i. to the sides of A GHK at their middle points, intersect at a common point.
[The ± erected at the middle points of the sides of a A intersect at a common point.] (§1^6)
64 PLANE GEOMETRY. — BOOK I.
139. Def. A median of a triangle is a line drawn from any vertex to the middle point of the opposite side.
Prop. LII. Theorem.
140. The medians of a triangle intersect at a common pointy which lies two-thirds the way from each vertex to the middle point of the opposite side.
, C
A F B
Given AD, BE, and CF the medians of A ABC.
To Prove that AD, BE, and CF intersect at a common point, which lies two-thirds the way from each vertex to the middle point of the opposite side.
Proof. Let AD and BE intersect at 0.
Let G and H be the middle points of OA and OB, respec- tively, and draw lines ED, OH, EG, and DH.
Since ED bisects AC and BC,
ED\\AB2.n& =\AB.
[The line joining the middle points of two sides of a A is || to the third side, and equal to one-half of it.] (§ 130)
In like manner, since GH bisects OA and OB,
GH 11 AB and = i AB. Then ED and GH are equal and II.
[Things which are equal to the same thing, are equal to each other.]
(Ax. 1) [Two str. lines 11 to the same str. line are || to each other.] (§ 55) Therefore, EDHG is a O.
[If two sides of a quadrilateral are equal and ||, the figure is a O.]
(§ 110)
RECTILINEAR FIGURES. 65
Then GD and EH bisect each other at 0.
[The diagonals of a O bisect each other.] (§ HI)
But by hyp., O is the middle point of OA, and H of OB. .-. AG=OG= OD, and BH= 0H= OE.
That is, AD and BE intersect at a point 0 which lies two-thirds the way from A to D, and from B to ^.
In like manner, AD and CF intersect at a point which lies two-thirds the way from A to D, and from C to F.
Hence, AD, BE, and CF intersect at the common point 0, which lies two-thirds the way from each vertex to the middle point of the opposite side.
LOCI.
141. Def. If a series of points, all of which satisfy a certain condition, lie in a certain line, and every point in this line satisfies the given condition, the line is said to be the locus of the points.
For example, every point which satisfies the condition of being equally distant from the extremities of a straight line, lies in the perpendicular erected at the middle point of the line (§ 42).
Also, every point in the perpendicular erected at the middle point of a line satisfies the condition of being equally distant from the extremities of the line (§ 41).
Hence, the perpeyidicular erected at the middle point of a straight line is the Locus of points which are equally distant from the extremities of the line.
Again, every point which satisfies the condition of being within an angle, and equally distant from its sides, lies in the bisector of the angle (§ 102).
Also, every point in the bisector of an angle satisfies the condition of being equally distant from its sides (§ 101).
Hence, the bisector of an angle is the locus of points which are within the angle, and equally distant from its sides.
66 PLANE GEOMETRY.— BOOK I.
EXERCISES.
51. Two straight lines are parallel if any two points of either are equally distant from the other.
(Prove by Beductio ad Absurdum.)
52. What is the locus of points at a given distance from a given straight line ? (Ex. 51.)
53. What is the locus of points equally distant from a pair of intersecting straight lines ?
54. What is the locus of points equally distant from a pair of parallel straight lines ?
D n
55. The bisectors of the interior angles of a trapezoid form a quadrilateral, two of whose angles are right angles. (Ex. 46.)
56. If the angles at the base of a trapezoid B^ n
are equal, the non-parallel sides are also equal. /-X \
(Given ZA = ZD; to prove AB = CD. Draw / \ \
BEWCD.) ^^ i ^D
57. If the non-parallel sides of a trapezoid are equal, the angles which they make with the bases are equal.
(Fig. of Ex. 56. Given AB= CD; to prove Z^ = ZD, and also ZABC=ZC. DrsiW BE \\ CD.)
58. The perpendiculars from the extremities of the base of an isosceles triangle to the opposite sides are equal.
59. If the perpendiculars from the extremities of the base of a triangle to the opposite sides are equal, the triangle is isosceles.
(Converse of Ex. 58. Prove AACD = ABCE.)
60. The angle between the bisectors of the equal angles of an isosceles triangle is equal to the exte- rior angle at the base of the triangle.
(Z ADB = 180° - (Z BAD + Z ABD).)
61. If a line joining two parallels be bisected, any line drawn through the point of bisection and included between the parallels will be bisected at the point.
(To prove that GH is bisected at 0.)
RECTILINEAR FIGURES.
67
62. If through a point midway between two parallels two transversals be drawn, they inter- cept equal portions of the parallels.
(Draw 0K± AB, and produce KO to meet CD at L. Then A OGK = A OIIL.)
63. If perpendiculars BE and DF be drawn from vertices B and D of parallelogram ABCD to the diagonal AC, prove BE = DF. (§ 70.)
64. The lines joining the middle points of the sides of a triangle divide it into four equal trian- gles. (§130.)
65. If from any point in the base of an isosceles triangle parallels to the equal sides be drawn, the perimeter of the parallelogram formed is equal to the sum of the equal sides of the triangle. (§ 96.)
66. The bisector of the exterior angle at the ver- tex of an isosceles triangle is parallel to the base. (§ 85, 1.)
67. The medians drawn from the extremities of the base of an isosceles triangle are equal.
68. If from the vertex of one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, it makes with the base an angle equal to one-half the vertical angle of the triangle.
(To prove Z BAD = ^ZC.)
69. If the exterior angles at the vertices A and B of triangle ABC are bisected by lines vsrhich meet at D, prove
Z ADB = 90° - 1 C.
(Z ADB = 180° - (Z BAD + Z ABD) .)
68
PLANE GEOMETRY.— BOOK I.
70. The diagonals of a rhombus bisect its angles. (Fig. of Prop. XLIV.)
71. If from any point in the bisector of an angle a parallel to one of the sides be drawn, the bisector, the parallel, and the remaining side form an isosceles triangle.
72. If the bisectors of the equal angles of an isos- celes triangle meet the equal sides at B and E, prove DE parallel to the base of the triangle.
(Prove A CED isosceles.)
73. If at any point D in one of the equal sides AB of isosceles triangle ABC, DE be drawn per- pendicular to base BC meeting CA produced at E, prove triangle ADE isosceles.
74. From C, one of the extremities of the base ^C of isosceles triangle ABC, a line is drawn meet- ing BA produced at D, making AD = AB. Prove CD perpendicular to BC. (§ 84.)
(A ACD is isosceles.)
75. If the non-parallel sides of a trapezoid are equal, its diagonals are also equal. (Ex. 57.)
76. If ADC is a re-entrant angle of quadrilateral ABCD, prove that angle ADC, exterior to the fig- ure, is equal to the sum of interior angles A, B, and C. (§ 128.) B
77. If a diagonal of a quadrilateral bisects two of its angles, it is perpendicular to the other diagonal, a
(Prove AC±DB, by § 43.)
78. In a quadrilateral ABCD, angles ABD and CAD are equal to ACD and BDA, respec- tively ; prove BC parallel to AD.
(Prove AB= CD; then prove BE = CF.)
79. State and prove the converse of Prop. XLIV. (J 41, I.)
RECTILINEAR FIGURES,
69
80. State and prove the converse of Ex.66, p. 67. (§ 96.)
81. The bisectors of the exterior angles at two vertices of a triangle, and the bisector of the inte- rior angle at the third vertex meet at a common point.
(Prove as in § 134.)
82. A BCD is a trapezoid whose parallel sides AD and BG are perpendicular to CD. If E is the middle point of AB, prove EC= ED. (§ 41 , T.)
(Draw^jPMZ).) A
83. The middle point of the hypotenuse of a right triangle is equally distant from the vertices of the triangle.
(To prove AD=BD= CD. Draw DE\\BC.)
84. The bisectors of the angles of a rectangle B form a square.
(By Ex. 50, EFGH is a rectangle. Now prove AF = BH and AE = BE.) A
85. If D is the n)iddle point of side BC of triangle ABC, and BE and CF are perpendiculars from B and C to AD, produced if necessary, prove BE = CF. B
86. The angle at the vertex of isosceles triangle ABC is equal to twice the sum of the equal angles B and C. If CD be drawn perpendicular to BC, meeting BA produced at D, prove triangle ACD equilateral.
(Prove each Z of A ACD equal to 60°.)
87. If angle B of triangle ABC is greater than angle C, and BD be drawn to ^ C making AD = AB, prove
ZADB=l(B+ C), and ZCBD = l(B- C). (Fig. of Prop. XXXII.)
88. How many sides are there in the polygon the sum of whose interior angles exceeds the sum of its exterior angles by 540° ?
70
PLANE GEOMETRY.— BOOK I.
89. The sum of the lines drawn from any point within a triangle to the vertices is greater than the half-sum of the three sides.
(Apply § 61 to each of the ^ABD, ACD, and BCD.)
90. The sum of the lines drawn from any point within a triangle to the vertices is less than the sum of the three sides. (§ 48. )
(Fig. of Ex. 89.)
91. If D, E, and F are points on the sides A B, BC, and CA, respectively, of equilateral triangle ABC, such that AD = BE = CF, prove DEF an equilateral triangle.
(Prove AADF, BDE, and C^i?' equal.)
92. If E, F, G, and H are points on the sides AB, BC, CD, and DA, respectively, of parallelogram ABCD, such that AE = CG and BF = DH, prove EFGH a parallelogram.
93. If E, F, G, and H are points on sides AB, BC, CD, and DA, respectively, of square ABCD, such that AE= BF= CG = DH, prove EFGH a square.
(First prove EFGH equilateral. Then prove ^FEH=^\)
94. If on the diagonal BD of square ABCD a distance BE be taken equal to AB, and EFhe drawn perpendicular to BD, meeting AD at F, prove that AF=EF= ED.
95. Prove the theorem of § 127 by drawing lines from any point within the polygon to the vertices. (§35.)
96. If CD is the perpendicular from the ver- tex of the right angle to the hypotenuse of right triangle ABC, and CE the bisector of angle C, meeting AB at E, prove ZDCE equal to one-half the difference of angles A and B. B
(To prove Z DCE = ^ (ZA - Z B).)
97. State and prove the converse of Ex. 70, p. 68. (Fig. of Prop. XLIV. Prove the sides all equal,)
RECTILINEAR FIGURES.
71
98. State and prove the converse of Ex. 75, p. 68. (Fig. of Ex. 78. Prove AACF and BDE equal.)
99. D is any point in base BC oi isosceles triangle ABC. The side AG \^ produced from C to E, so that CE = CD, and DE is drawn meeting AB at F. Prove Z AFE = 3 Z AEF.
(Z AFE is an ext. Z of A BFD.)
100. If ABC and ABD are two triangles on the same base and on the same side of it, such that AC = BD and AD = BC, and AD and BC intersect at 0, prove triangle OAB isosceles.
101. If D is the middle point of side ^C of equi lateral triangle ABC, and DE be drawn perpen dicular to BC, prove EC = ^ BC.
(Draw DF to the middle point of BC.)
102. If in parallelogram A BCD, E and F are the middle points of sides BC and AD, re- spectively, prove that lines AE and CF trisect diagonal BD. A
(By § 131, AE bisects BH, and CF bisects DG.)
103. If CD is the perpendicular from C to the hypotenuse of right triangle ABC, and E is the middle point of AB, prove ZDCE equal to the difference of angles A and 5. (Ex. 83.)
104. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter leg.
(Fig. of Ex. 86. Draw CA to middle point of BD.)
105. If AC be drawn from the vertex of the right angle to the hypotenuse of right triangle BCD so as to make ZACD = ZD, it bisects the hypotenuse.
(Fig. of Ex. 74. Prove A ABC isosceles.)
106. If D is the middle point of side BC of triangle ABC, prove AD>^iAB -\- AC - BC).
Note. For additional exercises on Book I., see p. 220.
Book II.
THE CIRCLE. DEFINITIONS.
142. A circle (O) is a portion of a plane bounded by a curve called a circiimference, all points
of which are equally distant from a point within, called the centre; as ABGD.
An arc is any portion of the circum- ference ; as AB.
A radius is a straight line drawn from the centre to the circumference ; as OA.
A diameter is a straight line drawn through the centre, having its extremities in the circumference ; as AG.
143. It follows from the definition of § 142 that
All radii of a circle are equal.
Also, all its diameters are equal, since each is the sum of two radii.
144. Two circles are equal when their radii are equal.
For they can evidently be applied one to the other so that their circumferences shall coincide throughout.
145. Conversely, the radii of equal circles are equal.
146. A semi-circumference is an arc equal to one-half the circumference.
A qiiadrant is an arc equal to one-fourth the circumference.
Concentric circles are circles having the same centre. 72
THE CIRCLE
73
147. A chord is a straight line joining the extremities of an arc ; as AB.
The arc is said to be subtended by its chord.
Every chord subtends two arcs; thus chord AB subtends arcs AMB and ACDB.
When the arc subtended by a chord is spoken of, that arc which is less than a semi-circumference is understood, unless the contrary is specified.
A segment of a circle is the portion included between an arc and its chord ; as AMBN.
A semicircle is a segment equal to one-half the circle.
A sector of a circle is the portion included between an arc and the radii drawn to its extremities ; as OCD.
148. A central angle is an angle whose vertex is at the centre, and whose sides are radii ; as AOC.
An inscribed angle is an angle whose ver- tex is on the circumference, and whose sides are chords ; as ABC.
An angle is said to be inscribed in a segment when its vertex is on the arc of the segment, and its sides pass through the extremities of the subtending chord.
Thus, angle B is inscribed in segment ABC.
149. A straight line is said to be tangent to, or touch, a circle when it has but one point in com- mon with the circumference ; as AB.
In such a case, the circle is said to be tangent to the straight line.
The common point is called the poi7it of contact, or poiyit of tangency.
A secant is a straight line which intersects the circumference in two points ; as CD.
74
PLANE GEOMETRY. —BOOK 11.
150. Two circles are said to be tangent to each other when they are both tangent to the same straight line at the same point.
They are said to be tangent internally or externally accord- ing as one circle lies entirely within or entirely without the other.
A common tangent to two circles is a straight line which is tangent to both of them.
151. A polygon is said to be inscribed in a circle when all its vertices lie on the circumference ; as ABCD.
In such a case, the circle is said to be circumscribed about the polygon.
A polygon is said to be inscriptible when it can be inscribed in a circle.
A polygon is said to be circumscribed about a circle when all its sides are tan- gent to the circle ; as EFGH.
In such a case, the circle is said to be itiscribed in the polygon.
Prop. I. Theorem.
152. Every diameter bisects the circle and its circumference.
B
Given AC a diameter of O ABCD.
To Prove that AC bisects the O, and its circumference.
THE CIRCLE. 75
Proof. Superpose segment ABC upon segment ADC, by folding it over about AC as an axis.
Then, arc ABC will coincide with arc ADC; for other- wise there would be points of the circumference unequally distant from the centre.
Hence, segments ABC and ADC coincide throughout, and are equal.
Therefore, AC bisects the O, and its circumference.
Prop. II. Theorem.
153. A straight line cannot intersect a circumference at more than two points.
0
/\
/ \\ / \ \
MA B C N
Given 0 the centre of a O, and MJ^ any str. line.
To Prove that MN cannot intersect the circumference at more than two points.
Proof. If possible, let JOT intersect the circumference at three points, A, B, and C; draw radii OA, OB, and OC.
Then, OA=OB= OC (§ 143)
We should then have three equal str. lines drawn from a point to a str. line.
But this is impossible ; for it follows from § 49 that not more than two equal str. lines can be drawn from a point to a str. line.
Hence, MN cannot intersect the circumference at more than two points.
Ex. 1. What is the locus of points at a given distance from a given point?
76
PLANE GEOMETRY.— BOOK II.
Prop. III. Theorem.
154. In equal circles, or in the same circle, equal central angles intercept equal arcs on the circumference.
Given ACB and A'C'B' equal central A of equal (D AMB and A'M'B', respectively.
To Prove arc AB = arc A'B'.
Proof. Superpose sector ABC upon sector A^B'O in such a way that Z C shall coincide with its equal Z C.
Now, AC = A' a and BC = B'C. (§ 145)
Whence, point A will fall at A', and point B at B'.
Then, arc AB will coincide with arc A'B' ; for all points of either are equally distant from the centre. .'. arc AB = arc A'B'.
Prop. IV. Theorem.
155. (Converse of Prop. III.) In equal circles, or in the same circle, equal arcs are intercepted by equal central angles.
Given ACB and A'C'B' central A of equal © AMB and A'M'B', respectively, and arc AB = arc A'B'.
THE CIRCLE.
77
To Prove AC = ^C'.
Proof. Since the (D are equal, we may superpose OAMB upon O A'M'B' in such a way that point A shall fall at A', and centre C at O',
Then since arc AB = arc A'B', point ^ will fall at B'.
Whence, radii AC and BC will coincide with radii A'C and 5'C", respectively. (Ax. 3)
Hence, Z G will coincide with Z C. .-. ZC'=ZC".
156. Sch. In equal circles, or in the same circle,
1. The greater of two central angles intercepts the greater arc on the circumference.
2. TJie greater of two arcs is intercepted by the greater cen- tral angle.
Prop. V. Theorem.
157. In equal circles, or in the same circle, equal chords subtend equal arcs.
M M'
Given, in equal (D AMB and A'M'B',
chord AB = chord A'B'.
To Prove arc AB = arc A'B'.
Proof. Draw radii AC, BC, A'C, and B'C.
Then in A ABC and A' B'C, by hyp., AB = A'B'.
Also, AC = A'C and BC = B'C.
.-. AABC=AA'B'C'.
.'. zc=zc.
.'. SLicAB = 2iVcA'B'.
(?)
(?)
(?)
(§ 154)
78
PLANE GEOMETRY. — BOOK II.
Prop. VI. Theorem.
158. (Converse of Prop. V.) In equal circles, or in the same circle, equal arcs are subtended by equal chords.
(Fig. of Prop. V.)
Given, in equal © AMB and AM'B\ arc AB = arc AB' ; and chords AB and A'B\
To Prove chord AB = chord A'B'.
(Prove A ABC = A A'B'C, by § 63.)
Ex. 2. If two drcumferences intersect each other, the distance between their centres is greater than the difference of their radii.
(§ 62.)
Prop. VII. Theorem.
159. In equal circles, or in the same circle, the greater of two arcs is subtended by the greater chord; each arc being less than a semircircumfereyice.
Given, in equal (D AMB and A'M^B', arc AB > arc A^B\ each arc being < a semi-circumference, and chords AB and
a:b'.
To Prove chord AB > chord A'B'.
Proof. Draw radii AC, BC, A'C, and B'C. Then in A ABC and A'B'C,
AC = A'C and BG = B'C. (?)
And since, by hyp., arc AB > arc A'B', we have
ZC>ZC. (§156,2)
.-. chord AB > chord A'B'. (§ 91)
THE CIRCLE.
79
Prop. VIII. Theorem.
160. (Converse of Prop. VII.) In equal circles, or in the same circle, the greater of two chords subtends the greater arc; each arc being less than a semi-circumference.
(Fig. of Prop. VII.)
(ZC>Z C, by § 92 ; the theorem follows by § 156, 1.)
161. Sch. If each arc is greater than a semi-circumfer- ence, the greater arc is subtended by the less chord ; and conversely the greater chord subtends the less arc.
Prop. IX. Theorem.
162. TJie diameter perpendicular to a chord bisects the chord and its subtended arcs.
Given, in O ABD, diameter CD _L chord AB.
To Prove that CD bisects chord AB, and arcs ACB and ADB.
Proof. Let 0 be the centre of the O, and draw radii OA and OB.
Then, OA = OB. (?)
Hence, A OAB is isosceles.
Therefore, CD bisects AB, and Z AOB. (§ 94)
Then since Z AOC= Z BOC, we have
Sivc. AC = arc. BC. (§ 154)
Again, Z AOD = ZBOD. (§ 31, 2)
.-. SiVGAD=avGBD. (?)
Hence, CD bisects AB, and arcs ACB and ADB.
80 PLANE GEOMETRY.— BOOK II.
163. Cor. The perpendicular erected at the middle point of a chord passes through the centime of the circle, and bisects the arcs subtended by the chord.
EXERCISES.
3. The diameter which bisects a chord is perpendicular to it and bisects its subtended arcs. (§43.)
(Fig. of Prop. IX. Given diameter CD bisecting chord AB.)
4. The straight line which bisects a chord and its subtended arc is perpendicular to the chord. a,
(By § 158, chord AC = chord BC.)
Prop. X. Theorem.
164. In the same circle, or in equal circles, equal chords are equally distant from the centre.
Given AB and CD equal chords of O ABC, whose centre is 0, and lines OE and OF ± to AB and CD, respectively. To Prove OE = OF. (§47)
Proof. Draw radii OA and 0(7. Then in rt. A OAE and OCF,
OA = OC (?)
Now, E is the middle point of AB, and F of CD. (§ 162) .-. AE=CF, being halves of equal chords AB and CD, respectively.
.-. A OAE = A OCF. (?)
.-. OE=OF. (?)
THE CIRCLE. gj
Prop. XI. Theorem.
165. (Converse of Prop. X.) In the same circle, or in equal circles, chords equally distant from the centre are equal.
(Fig. of Prop. X.)
Given 0 the centre of O ABC, and AB and CD chords equally distant from 0.
To Prove chord AB = chord CD.
(Rt. A OAE = rt. A OCF, and AE=CF, E is the middle point of AB, and F of CD.)
Prop. XII. Theorem.
166. In the same circle, or in equal circles, the less of two chords is at the greater distance from the centre.
B
Given, in O ABC, chord AB < chord CD, and Js OF and OG drawn from centre 0 to AB and CD, respectively. To Prove OF>OQ.
Proof. Since chord AB < chord CD, we have
arc^S<arcCZ>. (§ 160)
Lay off arc CE = arc AB, and draw line CE.
.: chord CE = chord AB. (§ 158)
Draw line OHl. CE, intersecting CD at K.
.-. 0H= OF. (§ 164)
But, OH>OK.
And, 0K>00. ' (?)
Whence, OH, or its equal OF, is > OG.
82 PLANE GEOMETRY.— BOOK 11.
Prop. XIII. Theorem.
167. (Converse of Prop. XII.) In the same circle, or in equal circles, if two chords are unequally distant from the centre, the more remote is the less.
Given 0 the centre of (d ABC, and chord AB more re- mote from 0 than chord CD.
To Prove chord AB < chord CD.
Proof. Draw lines 00. and QUI. to AB and CD respec- tively, and on OC lay off 0K= OH. Through K draw chord EFl. OK.
.'. chord EF= chord CD. (§ 165)
Now, chord AB II chord EF. (§ 54)
Then it is evident that arc AB is < arc EF, for it is only a portion of arc EF.
.: chord AB < chord EF. (§ 159)
.-. chord ^5 < chord (7i>.
168. Cor. A diameter of a circle is greater than any other chord; for a chord which passes through the centre is greater than any chord which does not. (§ 167)
EXERCISES.
5. The diameter which bisects an arc bisects its chord at right angles.
6. The perpendiculars to the sides of an inscribed quadrilateral at their middle points meet in a common point. (§ 163.)
THE CIRCLE.
83
Prop. XIV. Theorem.
169. A straight line perpendicular to a radius of a circle at its extremity is tangent to the circle.
ADC B
Given line AB± to radius OC of O EC at G. To Prove AB tangent to the O.
Proof. Let D be any point of AB except C, and draw line on.
.'. oD>oa (?)
Therefore, point D lies without the O. Then, every point of AB except C lies without the O, and AB is tangent to the O. (§ 149)
Prop. XV. Theorem.
170. (Converse of Prop. XIV.) A tangent to a circle is perpendicular to the radius drawn to the point of contact.
A C B
Given line AB tangent to O EC at C, and radius OC. To Prove OC±AB.
(OC is the shortest line that can be drawn from 0 to AB.)
171. Cor. A line perpendicular to a tangent at its point of contact passes through the centre of the circle.
84
PLANE GEOMETRY.— BOOK II.
Prop. XVI. Theorem.
172. Two parallels intercept equal arcs on a circumference. Case I. When one line is a tangent and the other a secant.
^ B
Given AB a tangent to O CED at E, and CD a secant II AB, intersecting the circumference at G and D. To Prove arc CE = arc DE.
Proof. Draw diameter EF.
.'. EF LAB.
.-. EF LCD.
.'. arc CE = arc DE.
(§ 170)
(?) (§ 162)
Case H. When both lines are secants.
Given, in O ABC, AB and CD II secants, intersecting the circumference at A and B, and C and D, respectively.
To Prove sltg AC= arc BD.
Proof. Draw tangent EF II AB, touching the O at G.
.'. EF II CD. (?)
Now, arc AG = arc BG,
and arc CG = arc DG. (§ 172, Case I)
THE CIRCLE.
Subtracting, we have
arc AG — arc CG = arc BG — arc DG.
.'. arc AC = arc BD.
Case III. When both lines are tangents.
E
85
Given, in O EGF, AB and CD II tangents, touching the O at E and F, respectively. To Prove arc EGF= arc EHF.
(Draw secant GH II AB.)
173. Cor. The straight line joining the points of contact of two imrallel tangents is a diameter.
Prop. XVII. Theorem.
174. Tlie tangents to a circle from an outside point are equal.
(Rt. A GAB = rt. A OAC, by § 90 ; then AB = AC.)
175. Cor. From equal A GAB and GAC,
Z GAB = Z GAC and Z AGB = Z AGC
Then, the line joining the centre of a circle to the point of intersection of two tangents makes equal angles with the tan- gerits, and also with the radii drawn to the points of contact.
86
PLANE GEOMETRY.— BOOK II.
Prop. XVIII. Theorem.
176. Through three jjoints, not in the same straight line, a drcumference can he draivn, and but one.
Given points A, B, and C, not in the same straight line.
To Prove that a circumference can be drawn through A, B, and C, and but one.
Proof. Draw lines AB and BC, and lines Z>i^ and EG ± to AB and BC, respectively, at their middle points, meeting at 0.
Then 0 is equally distant from A, B, and C. (§ 137)
Hence, a circumference described with 0 as a centre and OA as a radius will pass through A, B, and C.
Again, the centre of any circumference drawn through A, B, and C must be in each of the Js DF and EG. (§ 42)
Then as DF and EG intersect in but one point, only one circumference can be drawn through A, B, and C.
177. Cor. Two cii-cumferences can intersect in hut two 'points; for if they had three common points, they would have the same centre, and coincide throughout.
Prop. XIX. Theorem.
178. If two circumferences intersect, the straight line joining their centres bisects their common chord at right angles.
THE CIRCLE.
87
Given 0 and 0' the centres of two (D, whose circumfer- ences intersect at A and B, and lines 00' and AB. To Prove that 00' bisects AB at rt. A. (The proposition follows by § 43.)
Prop. XX. Theorem.
179. If two circles are tangent to each other, the straight line joining their centres passes through their point of contact.
B
Given 0 and 0' the centres of two CD, which are tangent to line AB at A.
To Prove that str. line joining 0 and 0' passes through A.
(Draw radii OA and O'A ; since these lines are ± AB, OAO' is a str. line by § 37 ; the proposition follows by Ax. 3.)
EXERCISES.
7. The straight line which bisects the arcs sub- tended by a chord bisects the chord at right angles.
8. The tangents to a circle at the extremities of a diameter are parallel. D
9. If two circles are concentric, any two chords of the greater which are tangent to the less are C equal. (§ 165.)
10. The straight line drawn from the centre of a circle to the point of intersection of two tangents bisects at right angles the chord joining their points of contact. (§ 174.)
88 PLANE GEOMETKY.— BOOK II.
ON MEASUREMENT.
180. The ratio of a magnitude to another of the same kind is the quotient of the first divided by the second.
Thus, if a and h are quantities of the same kind, the ratio
of a to 6 is -; it may also be expressed a : b. h
A magnitude is measured by finding its ratio to another magnitude of the same kind, called the uyiit of measure.
The quotient, if it can be obtained exactly as an integer or fraction, is called the numerical measure of the magnitude.
181. Two magnitudes of the same kind are said to be commensurable when a unit of measure, called a common measure, is contained an integral number of times in each.
Thus, two lines whose lengths are 2| and 3f inches are commensu- rable ; for the common measure i^ inch is contained an integral num- ber of times in each ; i.e., 55 times in the first line, and 76 times in the second.
Two magnitudes of the same kind are said to be incommen- surable when no magnitude of the same kind can be found which is contained an integral number of times in each.
For example, let AB and CD be two lines such that
CD
As V2 can only be obtained approximately, no line, however small, can be found which is contained an integral number of times in each line, and AB and CD are incommensurable.
182. A magnitude which is incommensurable with respect to the unit has, strictly speaking, no numerical measure (§ 180); still if CD is the unit of measure, and ^= V2, we shall speak of V2 as the numerical measure of AB.
183. It is evident from the above that the ratio of two magnitudes of the same kind, whether commensurable or incommensurable, is equal to the ratio of their numerical measures when referred to a common unit.
THE CIRCLE.
THE METHOD OF LIMITS.
184. A variable quantity, or simply a variable, is a quan- tity which may assume, under the conditions imposed upon it, an indefinitely great number of different values.
185. A constant is a quantity which remains unchanged throughout the same discussion.
186. A limit of a variable is a constant quantity, the dif- ference between which and the variable may be made less than any assigned quantity, however small, but cannot be made equal to zero.
In other words, a limit of a variable is a fixed quantity to which the variable approaches indefinitely near, but never actually reaches.
187. Suppose, for example, that a point moves from A towards B under the condition that it ^ C D E B
shall move, during successive equal in- I 1 1 — I — I
tervals of time, first from A to C, half-way between A and B ; then to D, half-way between C and B ; then to E, half- way between D and B] and so on indefinitely.
In this case, the distance between the moving point and B can be made less than any assigned distance, however small, but cannot be made equal to 0.
Hence, the distance from A to the moving point is a variable which approaches the constant distance AB as a limit.
Again, the distance from the moving point to 5 is a variable which approaches the limit 0.
As another illustration, consider the series 1 11 1 _i_ ...
-*-' 2"' 4' "8? 16' J
where each term after the first is one-half the preceding.
In this case, by taking terms enough, the last term may be made less than any assigned number, however small, but cannot be made actually equal to 0.
90 PLANE GEOMETKY.— BOOK 11.
Then, the last term of the series is a variable which ap- proaches the limit 0 when the number of terms is indej&- nitely increased.
Again, the sum of the first two terms is 1^ ; the sum of the first three terms is If ; the sum of the first four terms is 1 J ; etc.
In this case, by taking terms enough, the sum of the terms may be made to differ from 2 by less than any as- signed number, however small, but cannot be made actually equal to 2.
Then, the sum of the terms of the series is a variable which approaches the limit 2 when the number of terms is indefinitely increased.
188. The Theorem of Limits. If ttvo variables are always equal, and each approaches a limit, the limits are equal.
AM C B
\ \ I I
A' M' B'
I \ I
Given AM and A'M^ two variables, which are always equal, and approach the limits AB and A'B', respectively.
To Prove AB = A'B'.
Proof. If possible, let AB be > A'B'; and lay off, on AB,AO=A'B'.
Then, variable AM may have values > AC, while vari- able A'M' is restricted to values < AC; which is con- trary to the hypothesis that the variables are always equal.
Hence, AB cannot be > A'B'.
In like manner, it may be proved that AB cannot be <A'B'.
Therefore, since AB can be neither >, nor <,A'B', we have
AB = A'B'.
THE CIRCLE. gj
MEASUREMENT OF ANGLES.
Prop. XXI. Theorem.
189. In the same circle, or in equal circles, two central angles are in the same ratio as their intercepted arcs.
Case I. When the arcs are commensurable (*§ 181).
Given, in QABC, AOB and BOG central A intercepting commensurable arcs AB and BO, respectively.
To Prove ZAOB^^tcAB^
Z BOC arc BC
Proof. Since, by hyp., arcs AB and BG are commensur- able, let arc AD be a common measure of arcs AB and BC-, and suppose it to be contained 4 times in arc AB, and 3 times in arc BG.
SiTcAB _ 4 /-. s
'' arc 50 "3* ^ ^
Drawing radii to the several points of division of arc AC, Z.AOB will be divided into 4 A, and A BOG into 3 A, all of which A are equal. (§ 155)
A AOB 4
* A BOG 3
From (1) and (2), we have
AAOB^^x^AB A BOG slvcBG'
(2) (?)
92 PLANE GEOMETRY.— BOOK II.
Case n. When the arcs are incommensurable (§ 181).
B
Given, in Q ABG,AOB and BOC central A intercepting incommensurable arcs AB and BC, respectively.
-, - ZAOB SiTcAB
To Prove . ^^^, = ^77-
/.BOC arcJ3C
Proof. Let arc AB be divided into any number of equal arcs, and let one of these arcs be applied to arc^C as a unit of measure.
Since arcs AB and BC are incommensurable, a certain number of the equal arcs will extend from B to O, leaving a remainder C'C less than one of the equal arcs.
Draw radius OC.
Then, since by const., arcs AB and BC are commensurable,
ZbOC'^^^^W (§ 189, Case I.)
Now let the number of subdivisions of arc AB be indefi- nitely increased.
Then the unit of measure will be indefinitely diminished ; and the remainder C'C, being always less than the unit, will approach the limit 0.
Then Z BOC will approach the limit Z BOC, and arc BC will approach the limit arc BC.
Hence, ^lA2^ will approach the limit ^^^^, ' ZBOC ^^ ZBOC
and ^tgAB ^.j^ approach the limit ^:E^A^.
THE CIRCLE. 93
^ow, and — — are variables which are always
Z.B0O SLTcBC ZAOB An
equal, and approach the limits — — ^ and ^^'^ ^^ respec- tively. ^^^^ ^^^^^
By the Theorem of Limits, these limits are equal. (§ 188) . Z AOB ^ arc AB ' ' Z BOG arc BC'
190. Sch. The usual unit of measure for arcs is the degree, which is the ninetieth part of a quadrant (§ 146).
The degree of arc is divided into sixty equal parts, called minutes, and the minute into sixty equal parts, called seconds.
If the sum of two arcs is a quadrant, or 90°, one is called the complement of the other ; if their sum is a semi-circum- ference, or 180°, one is called the supplement of the other.
191. Cor. I. By § 154, equal central A, in the same O, intercept equal arcs on the circumference.
Hence, if the angular magnitude about the centre of a O be divided into four equal A, each Z will intercept an arc equal to one-fourth of the circumference.
That is, a right central angle intercepts a quadrant on the circumference. (§ 35)
192. Cor. II. By § 189, a central Z of n degrees bears the same ratio to a rt. central Z that its intercepted arc bears to a quadrant.
But a central Z of n degrees is — of a rt. central Z.
Hence, its intercepted arc is — of a quadrant, or an arc of n degrees.
The above principle is usually expressed as follows :
A ceiitral angle is measured by its intercepted arc.
This means simply that the number of angular degrees in a central angle is equal to the number of degrees of arc in its intercepted arc.
PLANE GEOMETRY. — BOOK II.
Prop. XXII. Theorem.
193. An inscribed angle is measured by one-half its inter- cepted arc.
Case I. When one side of the angle is a diameter.
A
Oiven AC a diameter, and AB a chord, of O ABO. To Prove that Z BAG is measured by ^ arc BG. Proof. Draw radius OB ; then, OA = OB. Then A OAB is isosceles, and Z B = ZA. But since BOG is an ext. Z of A OAB, Z BOG = ZA-\-ZB. .'. ZB0G=2ZA, or ZA = iZBOG. But, Z BOG is measured by arc BG. Whence, ZAis measured by ^ arc BG.
Case 11. When the centre is within the angle. A
(?) (?)
(§ 85, 1) (§ 192)
Given AB and AG chords of O ABC, and the centre of the O within Z BAG.
To Prove that Z BAG is measured by J arc BG.
THE CIRCLE.
95
Proof. Draw diameter AD. Then, Z BAD is measured by | arc BD, and Z CAD is measured by ^ arc CD. (§ 193, Case I)
.-. Z BAD + Z C^Z) is measured by | arc BD + ^ arc 0/>. .-. Z.^^O is measured by ^ arc ^0. Case III. When the centre is without the angle.
A
(The proof is left to the pupil.)
194. Cor. I. Angles inscribed in the same segment are equal.
Given A, B, and (7 zi inscribed in seg- ment ADE of O ABC.
To Prove ZA=Z.B = ZG. (The proposition follows by § 193.)
195. Cor. II. An angle inscribed in a semicircle is a right angle.
Given BC a diameter, and AB and AC B chords, of O ABD.
To Prove Z BAC a rt. Z.
Proof. ZBAC is measured by ^ of 180°, or 90°. (§ 193)
196. Cor. III. Tlie opposite angles of an inscribed quadrilateral are supplement- ary.
For their sum is measured by ^ of 360°, or 180°. (?)
96
PLANE GEOMETRY. — BOOK II.
Prop. XXIII. Theorem.
197. The angle between a tangent and a chord is measured by one-half its intercepted arc.
B
Given AE a tangent to O BCD at B, and BQ a chord. To Prove that Z ABC is measured by i arc BO. Proof. Draw diameter BD ; then, BD A. AE. (?)
Now a rt. Z is measured by one-half a semi-circumference.
.-. Z ABD is measured by ^ arc BCD. Also, Z (7jBZ) is measured by \ arc CD. (§ 193)
.-. Z ABD - Z (75i> is measured by | arc 5(7i) -| arc Oi>.
.-. Z ABC is measured by ^ arc BG. Similarly, Z ^JB(7 is measured by ^ arc BDG.
Prop. XXIV. Theorem.
198. The angle between two chords, intersecting within the circumference, is measured by one-half the sum of its inter- cepted arc, and the arc intercepted by its vertical angle.
-4.
Given, in 0 ABC, chords AB and CD intersecting within the circumference at E.
THE CIRCLE.
97
To Prove that
Z. AEG is measured by \ (arc ^(7 + arc BTf), Proof. Draw chord BG. Then, since AEG is an ext. Z of A BGE^ ,
ZAEG = ZB + ZG. (?)
But, Z 5 is measured by \ arc AG,
and Z C is measured by \ arc 5Z). (?)
.*. Z AE;(7 is measured by ^ (arc ^C-f- arc ^D).
Prop. XXV. Theorem.
199. The angle between tivo secants, intersecting without the circumference, is measured by one-half the difference of the intercepted arcs.
Given, in (DABG, secants AE and GE intersecting with- out the circumference at E, and intersecting the circumfer- ence at A and B, and G and D, respectively.
To Prove that Z ^ is measured by ^ (arc J. (7— arc BD).
Proof. Draw chord BG.
Then since ABG is an ext. Z of A BGE,
ZABG=ZE + ZG. (?)
.-. ZE = ZABG-ZG. But, Z ABG is measured by ^ arc AG, and ZG is measured by ^ arc BD. (?)
.-. Z ^ is measured by J (arc AG — sue BD).
98
PLANE GEOMETRY.— BOOK II.
200. Cor. (Converse of § 196.) If the opposite angles of a quadrilateral are supple- mentary, the quadrilateral can he inscribed in a circle.
Given, in quadrilateral ABCD, Z.A sup. Ji to Z C, and ZB to Z Z> ; also, a circumfer- ence drawn through A, B, and G. (§ 176)
To Prove that D lies on the circumference.
Proof. Since ZD is sup. to Z B, it is measured by ^ arc ABO. (§ 193)
Then, D must lie on the circumference; for if it were within the O, ZD would be measured by ^ an arc > ABC', and if it were without the O, ZD would be measured by 1 an arc < ABC. (§§ 198, 199)
Pkop. XXVI. Theorem.
201. The angle between a secant and a tangent, or ttvo tangents, is measured by one-half the difference of the inter- cepted arcs.
Fig. 1.
Fig. 2.
1. Given AE a tangent to 0 BDC at B, and EC a secant intersecting the circumference at C and D. (Fig. 1.)
To Prove that Z^ is measured by i (arc BFC— arc BD). (We have ZE = Z ABC - Z C.)
2. (In Mg. 2, Z ^ = Z ABD - Z BDE ; then use § 197.)
THE CIRCLE. 99
202. Cor. Since a circumference is an arc of 360°, we have i (arc BFD - arc BGD)
= i (360° - arc BGD - arc BGD) = i (360° - 2 arc BGD) = 180° - arc BGD. Then, Z E is measured by 180° — arc BGD. Hence, the angle between two tangents is measured by the supplement of the smaller of the two intercepted arcs.
EXERCISES.
11. If, in figure of § 197, arc BC = 107°, how many degrees are there in angles ABC and EBC?
12. If, in figure of § 198, arc AC = 'I^% and ZAEC = 5r, how many degrees are there in arc BD ?
13. If, in figure of § 199, arc ^C = 117°, and ZC= 14°, how many degrees are there in angle E?
14. If, in figure of § 199, ^C is a quadrant, and ZE = 39°, how many degrees are there in arc BD ?
15. If, in Fig. 1 of § 201, arc BFC = 197°, and arc CD = 75°, how many degrees are there in angle E ?
16. If, in Fig. 1 of § 201, Z J5: = 53°, and arc BD is one-fifth of the circumference, how many degrees are there in arc BFC ?
17. If, in Fig. 2 of § 201, arc BFD is thirteen-sixteenths of the circumference, how many degrees are there in angle E ?
18. Three consecutive sides of an inscribed quadrilateral subtend arcs of 82°, 99°, and 67° respectively. Find each angle of the quad- rilateral in degrees, and the angle between its diagonals.
19. Prove Prop. XXIV. by drawing through B a chord parallel to CD. (§ 172.)
20. Prove Prop. XXV. by drawing through B a chord parallel to CD.
21. Prove Prop. XXVI. for Fig. 1 by drawing through D a chord parallel to AE.
22. An angle inscribed in a segment greater than a semicircle is acute ; and an angle inscribed in a segment less than a semicircle is obtuse. (§ 193.)
100
PLANE GEOMETRY. — BOOK II.
23. In an inscribed trapezoid the non-parallel sides are equal, and also the diagonals.
(The non-parallel sides, and also the diagonals, subtend equal arcs.)
24. If the inscribed and circumscribed circles of a triangle are con- centric, prove the triangle equilateral. (§ 165.)
25. If AB and AC are the tangents from point A to the circle whose centre is 0, prove Z BAG = 2 Z OBC. (Ex. 10, p. 87.)
c
26. If two chords intersect at right angles within the circumference of a circle, the sum of the oppo- / \ \ B site intercepted arcs is equal to a semi-circumfer- ence.
27. Two intersecting chords which make equal angles with the diameter passing through their point of intersection are equal. (§ 165.)
(Prove that 01f = 0ir.)
28. Prove Prop. XXIII. by drawing a radius perpendicular to BO. (§ 162.)
29. If AB and AC are two chords of a circle making equal angles with the tangent at J., prove AB = AC.
30. From a given point within a circle and not coincident with the centre, not more than two equal straight lines can be drawn to the circumference.
(If possible, let AB, AC, and AD be three equal straight lines from point A to circumference BCD; then, by § 163, A must coincide with the centre.)
31. The sum of two opposite sides of a circum- scribed quadrilateral is equal to the sum of the other two sides. (§ 174.)
(To prove AB + CD = AD + BC.)
THE CIRCLE. . IQl-
32. Prove Prop, VI. by superposition. , > J . , ^ ; ' - , •> '. '*, i
33. In a circumscribed trapezoid, the straight line joining the middle points of the non-parallel sides is equal to one-fourth the perimeter of the trapezoid. (§ 132.)
34. If the opposite sides of a circumscribed quadrilateral are paral- lel, the figure is a rhombus or a square. (Ex. 31.)
(Prove the sides all equal.)
35. If tangents be drawn to a circle at the extremities of any pair of diameters which are not perpendicular to each other, the figure formed is a rhombus. (Ex. 34.)
36. If the angles of a circumscribed quadrilateral are right angles, the figure is a square.
37. If two circles are tangent to each other at point A, the tangents to them from any point in the common tangent which passes through A are equal. (§ 174.)
38. If two circles are tangent to each other externally at point A, the common tangent which passes through A bisects the other two common tangents. (Ex. 37.)
(To prove that FG bisects BC and DE.)
39. The bisector of the angle between two tangents to a circle passes through the centre.
(The bisector of the Z between the tangents bisects at rt. A the chord joining their points of contact.)
40. The bisectors of the angles of a circumscribed quadrilateral pass through a common point.
41. If AB is one of the non-parallel sides of a trapezoid circum- scribed about a circle whose centre is 0, prove AOB a right angle. (§ 175.)
B
42. If two circles are tangent to each other internally, the distance between their centres is equal to the difference of their radii.
43. Prove the theorem of § 168 by drawing radii to the extremities of the chord. (Ax. 4.)
44. Prove the theorem of § 202 by drawing radii to the points of contact of the tangents. (§ 192.)
45. If any number of angles are inscribed in the same segment, their bisectors pass through a common point. (§ 193.)
102
PLANE GEOMETRY. — BOOK II.
.' ' 46. Prove- Prap. XIII. by Beductio ad Absurdum. (§§ 164, 166.)
47. Two chords perpendicular to a third chord at its extremities are equal. (§ 158.)
48. If two opposite sides of an inscribed quadrilat- eral are equal and parallel, the figure is a rectangle. (Arc BCD is a semi-circumference.)
49. If the diagonals of an inscribed quadrilateral intersect at the centre of the circle, the figure is a rectangle. (§ 195.)
50. The circle described on one of the equal sides of an isosceles triangle as a diameter, bisects the base. (§ 195.)
51. If a tangent be drawn to a circle at the ex- tremity of a chord, the middle point of the sub- tended arc is equally distant from the chord and from the tangent.
(JB2> bisects Z^^C.)
52. If sides AB, BC, and CD of an inscribed quadrilateral sub- tend arcs of 99°, 106°, and 78°, respectively, and sides BA and CD produced meet at E, and sides AD and BC Sit F, find the number of degrees in angles AED and AFB.
53. If 0 is the centre of the circumscribed circle of triangle ABC, and OD be drawn perpendicular to BC, prove
ZBOD = ZA. (§192.)
54. If Z>, E, and F are the points of contact of sides AB, BC, and CA respectively of a triangle circumscribed about a circle, prove
Z DEF = 90° - I A. (§ 202.)
55. If sides AB and BC of an inscribed quadrilateral ABCD sub- tend arcs of 69° and 112°, respectively, and angle AED between the diagonals is 87°, how many degrees are there in each angle of the quadrilateral ?
56. If any number of parallel chords be drawn in a circle, their middle points lie in the same straight line. (§ 162.)
57. What is the locus of the middle points of a system of parallel chords in a circle ?
THE CIRCLE.
103
58. What is the locus of the middle points of a system of chords of given length in a circle ?
59. If two circles are tangent to each other, any straight line drawn through their point of contact subtends arcs of the same number of degrees on their circumferences. (§ 197.)
(Let the pupil draw the figure for the case when the (D are tangent internally.)
60. If a straight line be drawn through the point of contact of two circles which are tan- gent to each other externally, terminating in their circumferences, the radii drawn to its extremities are parallel. (§ 73.)
(Let the pupil state the corresponding theo- rem for the case when the (D are tangent internally.)
61. If a straight line be drawn through the point of contact of two circles which are tangent to each other externally, terminating in their circumferences, the tangents at its extremities are parallel. (§ 73.)
(Let the pupil state the corresponding theorem for the case when the © are tangent internally.)
62. If sides AB and DC of inscribed quadrilateral ABCD be pro- duced to meet at E, prove that triangles ACE and BDE^ and also triangles ADE and BCE, are mutually equiangular.
(For second part, see § 196.)
63. The sum of the angles subtended at the centre of a circle by two opposite sides of a circum- scribed quadrilateral is equal to two right angles. (§ 175.)
(To prove Z AOB + Z COD = 180°.)
64. If a circle is inscribed in a right triangle, the sum of its diameter and the hypotenuse is equal to the sum of the legs. (§ 174.)
65. If a circle be described on the radius of another circle as a diameter, any chord of the greater passing through the point of con- tact of the circles is bisected by the circumference of the smaller. (§ 196.)
104
PLANE GEOMETRY. —BOOK II.
66. If sides AB and CD of inscribed quad- rilateral ABCD make equal angles with the diameter passing through their point of inter- section, prove AB = CD. (§ 165.)
67. If angles^, B, and G of circumscribed quadrilateral ylBC2> are 128°, 67°, and 112°, respectively, and sides AB, BC, CD, and DA are tangent to the circle at points E, F, 6?, and H, respectively, find the number of degrees in each angle of quadrilateral EFGH.
68. The chord drawn through a given point within a circle, perpendicular to the diameter passing through the point, is the least chord which can be drawn through the given point. (§ 165.)
(Given chords AB and CD drawn through P, and AB ± OP. To prove AB < CD.)
69. If ADB, BEC, and CFA are angles inscribed in segments ABD, BCE, and ACF, respectively, exterior to inscribed triangle ABC, prove their sum equal to four right angles. (§ 196.)
Note. For additional exercises on Book II. , see p. 222.
CONSTRUCTIONS.
Prop. XXVII. Problem.
203. At a given point in a straight line to erect a perpen- dicular to that line.
First Method.
F,,-
A D C E B
Given C any point in line AB.
THE CIRCLE. 105
Required to draw a line ± to AB at C.
Construction. Take points D and E on AB equally dis- tant from 0.
With D and E as centres, and with equal radii, describe arcs intersecting at F^ and draw line CF.
Then, CF is J_ to AB at C.
Proof. By cons., C and F are each equally distant from D and E.
Whence, CF is ± to i)^ at its middle point. (?)
Second Method.
E I
Given (7 any point in line AB.
Required to draw a line ± to ^B at C.
Construction. With any point 0 without line AB as a centre, and distance OC as a radius, describe a circumfer- ence intersecting AB at C and B.
Draw diameter Z>^, and line CE.
Then, O^ is ± to ^B at C.
Proof. Z i)(7^, being inscribed in a semicircle, is a rt. Z. (§ 195)
.-. CE ± CZ).
Note. The second method of construction is preferable when the given point is near the end of the line.
EXERCISES.
70. Given the base and altitude of an isosceles triangle, to con- struct the triangle.
71. Given an acute angle, to construct its complement.
106 PLANE GEOMETRY.— BOOK II.
Prop. XXVIII. Problem.
204. From a given point without a straight line to draw a perpendicular to that line.
C /
f
Given C any point without line AB.
Required to draw from C a line ± to AB.
Construction. With (7 as a centre, and any convenient radius, describe an arc intersecting AB 2X D and E.
With D and E as centres, and with equal radii, describe arcs intersecting at F.
Draw line CF.
Then, CF ± AB.
Proof. Since, by cons., C and F are each equally distant from D and F, CF is ± to BF at its middle point. (?)
Prop. XXIX. Problem. 205. To bisect a given straight line.
"M
\E
-B
%
Given line AB.
THE CIRCLE. IQ7
Required to bisect AB.
Construction. With A and B as centres, and with equal radii, describe arcs intersecting at C and D. Draw line CD intersecting AB at E. Then, E is the middle point of AB. (The proof is left to the pupil.)
Prop. XXX. Problem. 206. To bisect a given arc.
Given arc AB. Required to bisect arc AB.
Construction. With A and B as centres, and with equal radii, describe arcs intersecting at C and D. Draw line CD intersecting arc AB at E. Then E is the middle point of arc AB. Proof. Draw chord AB.
Then, CD is ± to chord AB at its middle point. (?)
Whence, CD bisects arc AB. (§ 163)
EXERCISES.
72. Given an angle, to construct its supplement. • 73. Given a side of an equilateral triangle, to construct the tri-
74. To construct an angle of 60^ (Ex. 73); of 30° (Ex. 71).
75. To construct an angle of 120° (Ex. 72); of 160°.
108 PLANE GEOMETRY.— BOOK n.
Prop. XXXI. Problem. 207. To bisect a given angle.
Given Z AOB.
Required to bisect Z. AOB.
Construction. With 0 as a centre, and any convenient radius, describe an arc intersecting OA at C, and OB at D.
With C and D as centres, and with the same radius as before, describe arcs intersecting at E, and draw line OE.
Then, OE bisects Z AOB.
Proof. Let OE intersect arc CD at F.
By cons., 0 and E are each equally distant from C and D.
Whence, OE bisects arc CD at F (§ 206).
.-. ZCOF=ZDOF. (?)
That is, OE bisects Z AOB.
Prop. XXXII. Problem.
208. With a given vertex and a given side, to construct an angle equal to a given angle.
C ^ A
Given 0 the vertex, and OA a side, of an Z, and Z 0',
THE CIRCLE. IQg
Required to construct, with 0 as the vertex and OA as a side, an Z equal to Z 0'.
Construction. With 0' as a centre, and any convenient radius, describe an arc intersecting the sides of Z 0' at 0 and Z> ; and draw chord CD.
With 0 as a centre, and with the same radius as before, describe the indefinite arc AE.
With J. as a centre and CD as a radius, describe an arc intersecting arc AE at B, and draw line OB.
Then, ZAOB = ZO'.
(The chords of arcs AB and CD are equal, and the propo- sition follows by §§ 157 and 155.)
Prop. XXXIII. Problem.
209. Through a given point without a given straight li7ie, to draw a parallel to the line.
/F
/c
-D
/E
Given C any point without line AB. Required to draw through C a line || to AB. Construction. Through C draw any line EF, meeting AB at E, and construct Z FCD = Z CEB. (§ 208)
Then, CD || AB. (?)
EXERCISES.
76. To construct an angle of 45° ; of 135° ; of 22^° ; of 67^°.
77. Through a given point without a straight line to draw a line making a given angle with that line.
(Draw through the given point a II to the given line.)
110
PLANE GEOMETRY. — BOOK U.
Prop. XXXIV. Problem. 210. Given two angles of a triangle, to find the third.
<F
-B
Given A and B two A of a, A. Required to construct the third Z.
Construction. At any point E of the indefinite line CD, construct Z DEF= Z A. (§ 208)
Also, Z FEG adjacent to Z DEF, and equal to Z B. Then, Z CEG is the required Z. (The proof is left to the pupil.)
Prop. XXXV. Problem.
211. Given tvjo sides and the included angle of a triangle, to construct the triangle.
ca
m /
ny
Given m and n two sides of a A, and A^ their included Z.
Required to construct the A.
Construction. Draw line AB == m.
Construct Z BAD = Z A'. (§ 208)
On AD take AO = n, and draw line BC.
Then, ABC is the required A.
212. Sch. The problem is possible for any values of the given parts.
THE CIRCLE.
Ill
Prop. XXXVI. Problem.
213. Given a side and two adjacent angles of a triangle^ to construct the triangle.
D
Given a side m, and the adj. A A' and B' of a A. (The construction is left to the pupil.)
214. Sch. I. If a side and any two angles of a triangle are given, the third angle may be found by § 210, and the triangle may then be constructed as in § 213.
215. Sch. n. The problem is impossible when the sum of the given angles is equal to, or greater than, two right angles. (§ 84)
Prop. XXXVII. Problem.
216. Given the three sides of a triangle, to construct the triangle.
m
Given m, n, and p the three sides of a A.
Required to construct the A.
Construction. Draw line AB = m.
With ^ as a centre and n as a radius, desc^ribe an arc ; with 5 as a centre and j9 as a radius, describe an arc inter- secting the former arc at C, and draw lines AC and BC.
Then, ABG is the required A.
112 PLANE GEOMETRY. — BOOK II.
217. Sch. The problem is impossible when one of the given sides is equal to, or greater than, the sum of the other two. (§ 61)
Prop. XXXVIII. Problem.
218. Given two sides of a triangle, and the angle opposite to one of them, to construct the triangle.
Given m and n two sides of a A, and A' the Z opposite to n.
Required to construct the A.
Construction. Construct Z DAE = ZA' (§ 208), and on AE take AB = m.
With ^ as a centre and n as a radius, describe an arc.
Case I. When A' is acute, and m> n.
There may be three cases :
1. The arc may intersect AD in two points.
Let Ci and Cg be the points in which the arc intersects AD, and draw lines BCi and BG2'
Then, either ABCi or ABC2 is the required A.
Note. This is called the ambiguous case.
2. The arc may be tangent to AD. In this case there is but one A.
And since a tangent to a O is JL to the radius drawn to the point of contact (§ 170), the A is a right A.
3. The arc may not intersect AD at all. In this case the problem is impossible. Case II. When A^ is acute, and m — n.
THE CIRCLE. 113
In this case, the arc intersects AD in two points, one of which is A.
Then there is but one A ; an isosceles A.
Case m. WJien A' is acute, and m<n.
m
In this case, the arc intersects AD in two points.
Let Ci and C2 be the points in which the arc intersects AD, and draw lines BCi and BC^.
Now A ABCi cloes not satisfy the conditions of the prob- lem, since it does not contain the given Z A'.
Then there is but one A ; A ABC2.
Case rV. When A' is right or obtuse, and m<n. In each of these cases, the arc intersects AD in two points on opposite sides of A. Then there is but one A.
219. Sch. If A' is right or obtuse, and m = n or m>n, the problem is impossible ; for the side opposite the right or obtuse angle in a triangle must be the greatest side of the triangle. (§ 99)
The pupil should construct the triangle corresponding to each case of § 218.
EXERCISES.
78. Given one of the equal sides and the altitude of an isosceles triangle, to construct the triangle.
What restriction is there on the values of the given lines ?
79. Given two diagonals of a parallelogram and their included angle, to construct the parallelogram. (§ 111.)
114 PLANE GEOMETRY. — BOOK II.
Prop. XXXIX. Problem.
220. Oiven two sides and the included angle of a parallelo- gram, to construct the parallelogram.
m
Given m and n two sides, and A' the included Z, of a O. (The construction and proof are left to the pupil.)
Prop. XL. Problem. 221. To inscnhe a circle in a given triangle.
^^ ^-^ '^g
Given A ^5(7.
Required to inscribe a O in A ABC. Construction. Draw lines AD and BE bisecting A A and B, respectively (§ 207).
From their intersection 0, draw line OM A. AB (§ 204). With 0 as a centre and OM as a radius, describe a O. This O will be tangent to AB, BC, and CA. (The proof is left to the pupil ; see § 135.)
Ex. 80. To construct a right triangle, having given the hypotenuse and an acute angle.
(The other acute Z is the complement of the given Z.)
THE CIRCLE. II5
Prop. XLI. Problem. 222. To circumscribe a circle about a given triangle.
C
Given A ABC.
Required to circumscribe a O about A ABC.
Construction. Draw lines DF and EG ± to AB and AC,
respectively, at their middle points (§ 205). Let DF and EO intersect at 0.
With 0 as a centre, and OA as a radius, describe a O. The circumference will pass through A, B, and G. (The proof is left to the pupil ; see § 137.)
223. Sch. The above construction serves to describe a circumference through three given points not in the same straight line, or to find the centre of a given circumference or arc.
EXERCISES.
81. To construct a right triangle, having given a leg and the oppo- site acute angle.
(Construct the complement of the given Z.)
82. Given the base and the vertical angle of an isosceles triangle, to construct the triangle,
(Each of the equal A is the complement of one-half the vertical Z.)
83. Given the altitude and one of the equal angles of an isosceles triangle, to construct the triangle.
(One-half the vertical Z is the complement of each of the equal A.)
84. To circumscribe a circle about a given rectangle^ (Draw the diagonals.)
116
PLANE GEOMETRY. — BOOK II.
Prop. XLII. Problem.
224. To draw a tangent to a circle through a given point on the circumference.
BAG
Given A any point on the circumference of O AD. Required to draw through A a tangent to O AD. Construction. Draw radius OA. Through A draw line BC ± OA (§ 203). Then, BG will be tangent to O AD.
(?)
Prop. XLIII. Problem.
225. To draw a tangent to a circle through a given point without the circle.
Given A any point without O BC.
Required to draw through A a tangent to O BG.
Construction. Let 0 be the centre of Q BG, and draw line OA.
On OA as a diameter, describe a circumference, cutting the given circumference at B and G.
Draw lines AB and AG.
THE CIRCLE.
117
Then, AB and AC are tangents to O BC.
Proof. Draw line OB.
Z ABO, being inscribed in a semicircle, is a rt. Z.
Therefore, AB is tangent to O BC.
In like manner, ^C is tangent to O BC.
(?) (?)
Prop. XLIV. Problem.
226. Upon a given straight line, to descHbe a segment which shall contain a given angle.
a
|
•■■■^ |
> |
(F |
|
v. J |
r |
B
Given line AB, and Z A'.
Required to describe upon AB a segment such that every Z inscribed in the segment shall equal Z A'.
Construction. Construct Z BAC = Z A'. (§ 208)
Draw line DE ± to AB at its middle point. (§ 205)
Draw line AFl. AC, intersecting DE at 0. With 0 as a centre and OA as a radius, describe O AMBN. Then, AMB will be the required segment.
Proof, Let AOB be any Z inscribed in segment AMB.
Then, Z AGB is measured by \ arc ANB. But, by cons., AC _L OA. Whence, ^C is tangent to O AMB. Therefore, Z BAC is measured by ^ arc ANB.
.'. ZAGB = ZBAC = ZA'.
(?)
(?) (§ 197)
(?)
Hence, every Z inscribed in segment AMB equals Z A'.
(§ 194)
118
PLANE GEOMETRY. — BOOK II.
EXERCISES.
85. Given the middle point of a chord of a circle, to construct the chord.
(To draw through C a chord which is bisected at C.)
86. To draw a line tangent to a given circle and parallel to a given straight line. (To draw a tangent || AB.)
87. To draw a line tangent to a given circle and perpendicular to a given straight line.
88. To draw a straight line through a given point within a given acute Z, forming with the sides of A the angle an isosceles triangle.
89. Given the base, an adjacent angle, and the altitude of a tri- angle, to construct the triangle.
(Draw a II to the base at a distance equal to the altitude,)
90. Given the base, an adjacent side, and c^-— the altitude of a triangle, to construct the ^,--''^S^iD triangle. n\p
Discuss the problem for the following cases :
A
1. n>p.
3. n<p.
91. To construct a rhombus, having given its base and altitude. (Draw a II to the base at a distance equal to the altitude.) What restriction is there on the values of the given lines ?
92. Given the altitude and the sides includ- ing the vertical angle of a triangle, to construct the triangle.
What restriction is there on the values of the given lines ?
Discuss the problem for the following cases : 1. m < w or > w. 2. m — n.
THE CIRCLE. 219
93. Given the altitude of a triangle, and the angles at the extremi- ties of the base, to construct the triangle.
(The Z between the altitude and an adjacent side is the complement of the Z at the extremity of the base, if acute, or of its supplement, if obtuse.)
94. To construct an isosceles triangle, having given the base and the radius of the circumscribed circle.
What restriction is there on the values of the given lines ?
95. To construct a square, having given one of its diagonals. (§ 195.)
96. To construct a right triangle, having
given the hypotenuse and the length of the d...^ <■" "^n
perpendicular drawn to it from the vertex of i //^...^ "\ the right angle. i// ^^""-^^ \
What restriction is there on the values of A*^ ^ ^^B
the given lines ?
97. To construct a right triangle, having given the hypotenuse and a leg.
What restriction is there on the values of the given lines ?
98. Given the base of a triangle and the ^ E, perpendiculars from its extremities to the other ~"'*>s^, c,-f'^ sides, to construct the triangle. (§ 225.)
What restriction is there on the values of the given lines ?
99. To describe a circle of given radius tangent to two given intersecting lines.
(Draw a II to one of the lines at a distance equal to the radius.)
100. To describe a circle tangent to a given straight line, having its centre at a given point without the line.
101. To construct a circle having its centre in a given line, and passing through two given points without the line. (§ 163.)
What restriction is there on the positions of the given points ?
102. In a given straight line to find a point equally distant from two given intersecting lines. (§ 101.)
103. Given a side and the diagonals of a parallelogram, to con- struct the parallelogram.
What restriction is there on the values of the given lines ?
104. Through a given point without a given straight line, to describe a circle tangent to the given line at a given point. (§ 163.)
120 PLANE GEOMETRY.— BOOK II.
105. Through a given point within a circle to draw a chord equal to a given chord. (§ 164.)
What restriction is there on the position of the given point?
106. Through a given point to describe a circle of given radius tangent to a given straight line.
(Draw a || to the given line at a distance equal to the radius.)
107. To describe a circle of given radius tangent to two given circles.
(To describe a O of radius m tangent to two given (D whose radii are n and p, respec- tively.)
What restriction is there on the value of m ?
108. To describe a circle tangent to two given parallels, and pass- ing through a given point.
What restriction is there on the position of the given point ?
109. To describe a circle of given radius, tangent to a given line and a given circle.
(Draw a II to the given line at a distance equal to the given radius.)
110. To construct a parallelogram, having given a side, an angle, and the diagonal drawn from the vertex of the angle.
111. In a given triangle to inscribe a rhombus, having one of its angles coincident with an angle of the triangle.
(Bisect the Z which is common to the A and the rhombus.)
112. To describe a circle touching two given intersecting lines, one of them at a given point. (§ 169.)
113. In a given sector to inscribe a circle. (The problem is the same as inscribing a O in
A O'CD.)
114. In a given right triangle to inscribe a square, having one of its angles coincident with the right angle of the triangle.
115. Through a vertex of a triangle to draw a straight line equally distant from the other vertices.
THE CIRCLE.
121
116. Given the base, the altitude, and the vertical angle of a tri- angle, to construct the triangle. (§ 226. )
(Construct on the given base as a chord a segment which shall contain the given Z. )
117. Given the base of a triangle, its vertical angle, and the median drawn to the base, to construct the triangle.
118. To construct a triangle, having given the middle points of its sides.
119. Given two sides of a triangle, and the median drawn to the third side, to construct the triangle.
(Construct AABD with its sides equal to g m, w, and 2p, respectively.)
What restriction is there on the values of the given lines ?
120. Given the base, the altitude, and the radius of the circum- scribed circle of a triangle, to construct the triangle.
(The centre of the circumscribed O lies at a distance from each vertex equal to the radius of the~0.)
121. To draw common tangents to two given circles which do not intersect.
(To draw exterior common tangents, describe O AA' with its radius equal to the difference of the radii of the given (D.
To draw interior common tangents, describe O AA' with its radius equal to the sum of the radii of the given (D.)
Note. For additional exercises on Book II., see p. 224.
Book III.
THEORY OP PROPORTION. -SIMILAR POLYGONS.
DEFINITIONS.
227. A Proportion is a statement that two ratios are equal.
228. The statement that the ratio of a to 5 is equal to the ratio of c to d, may be written in either of the forms
a : 0 = c : a, or - = — b d
229. The first and fourth terms of a proportion are called the extremes, and the second and third terms the means.
The first and third terms are called the antecedents, and the second and fourth terms the consequents.
Thus, in the proportion a'.h = c:d, a and d are the extremes, h and c the means, a and c the antecedents, and h and d the consequents.
230. If the means of a proportion are equal, either mean is called a mean proportional between the first and last terms, and the last term is called a third proportional to the first and second terms.
Thus, in the proportion a:h — h:c,h is a mean propor- tional between a and c, and c a third proportional to a and h.
231. A fourth proportional to three quantities is the fourth term of a proportion, whose first three terms are the three quantities taken in their order.
Thus, in the proportion a:h = c: d, d is a fourth propor- tional to a, b, and c.
122
THEORY OF PROPORTION. 123
Prop. I. Theorem.
232. In any proportion, the product of the extremes is equal to the product of the means.
Given the proportion a:b = c:d.
To Prove ad = be.
Proof. By §228, - = -^.
b d
Multiplying both members of this equation by bd,
ad = be.
233. Cor. The mean proportional between two quantities is equal to the square root of their product.
Given the proportion a:b = b:c. (1)
To Prove b = Vac:
Proof. From (1), b' = ac. (§ 232)
.-. b =^/ac.
Prop. II. Theorem.
234. (Converse of Prop. I.) If the product of two quan- tities is equal to the product of two others, one pair may be made the extremes, and the other pair the means, of a proportion.
Given ad — be. (1)
To Prove a:b = c.d.
Proof. Dividing both members of (1) by bd,
ad _bc
bd~bd
|
Or, |
b~~d |
|
Then by § 228, |
a:b = c:d. |
|
In like manner, |
a:c = b : d. |
|
b:a = d: c, etc. |
124 PLANE GEOMETRY.— BOOK III.
Prop. III. Theorem.
235. In any proportion, the terms are in proportion by Alternation ; that is, the first term is to the third as the
|
second term |
is to the fourth. |
|||
|
Given the |
proportion |
a:b: |
= c: |
d. |
|
To Prove |
a: c: |
= b: |
d. |
|
|
Proof, rrom (1), |
ad: |
= bc |
'. |
|
|
••• |
a: c: |
= b: |
d. |
|
|
Prop. |
IV. |
Theorem. |
(1)
(§ 232) (§ 234)
236. In any proportion, the terms are in proportion by Inversion; that is, the second term is to the first as the fourth term is to the third.
Given the proportion a:b = c:d. (1) To Prove b:a = d: c.
Proof. From (1), ad = be. (?)
.-. b:a = d:c. (?)
Prop. V. Theorem.
237. In any proportion, the terms are in proportion by Composition ; that is, the sum of the first two terms is to the first term as the sum of the last two terms is to the third term.
Given the proportion a:b — c:d. (1)
To Prove a-\-b:a = c-[-d:c.
Proof. From (1), ad = be. (?)
Adding both members of the equation to ac,
ac-\- ad = ac + be. Factoring, a(c ■i-d) = c{a + b).
.-. a + b'.a = c-\-d'.c. (§ 234)
In like manner, a -\-b:b — c-\- d:d.
THEORY OF PROPORTION. 125
Prop. VI. Theorem.
238. In any proportion, the terms are in proportion by Division ; that is, the difference of the first two terms is to the first term as the difference of the last two terms is to the third term.
Given the proportion a : 6 = c : c?, (1)
in which a>h and c>d. To Prove a — h:a=c — d:c.
Proof. From (1), ad = he. (?)
Subtracting both members of the equation from acj
ac — ad = ac — be. Factoring, a(c — d) = c(a — b).
.: a — b : a = c ~ d : c. (?)
In like manner, a — b: b = c — did.
Prop. VII. Theorem.
239. In any proportion, the terms are in proportion by Composition and Division; that is, the sum of the first two terms is to their difference as the sum of the last two terms is to their difference.
Given the proportion a:b = c:d, (1)
in which a>b and c>d. To Prove a-\-b:a—b = c + d:c — d.
Proof. From(l), a±b^c_±d^ ^g 237)
a c
and ^L^^^Jzl. (§238)
a c
Dividing the first equation by the second,
a-^b _c +_d
a — b c — d
.'. a + b:a — b = c-\-d:c — d.
126 PLANE GEOMETRY. -BOOK III.
Prop. VIII. Theorem.
240. In a series of equal ratios, the sum of all the antece- dents is to the sum of all the consequents as any antecedent is to its consequent.
Given a:b = c:d = e:f (1)
To Prove a-\- c-^e-.b + d +/= a : b.
Proof. We have ba = ab.
Also, from (1), be = ad,
and be = af (?)
Adding, ba + be + be= ab + ad -\- af.
.'. b{a^c + e) = a{b-\-d+f). .'. a-{-c-\-e:b-j-d -f /= a : b. (?)
Prop. IX. Theorem.
241. In any proportion, like powers or like roots of the terms are in proportion.
Given the proportion a:b = c:d. (1)
To Prove a" : 6" = c" : d*".
Proof. From(l), ^ = |
Raising both members to the nth power,
b^ ~~ d^ .'. a^ : 5» = c« : 6?**. In like manner, Va : V6 = ->/c : -^d.
Note. The ratio of two magnitudes of the same kind is equal to the ratio of their numerical measures when referred to a common unit (§ 183) ; hence, in any proportion involving the ratio of two magnitudes of the same kind, we may regard the ratio of the mag- nitudes as replaced by the ratio of their numerical measures when referred to a common unit.
Thus, let AB, CD, EF, and GH be four lines such that AB:CD = EF: GH.
Then, ABx GH=CDx EF. (§ 232)
THEORY OF PROPORTION. 127
This means that the product of the numerical measures of AB and GH is equal to the product of the numerical measures of CD and EF.
An interpretation of this nature must be given to all applications of §§ 232, 233 and 241.
EXERCISES.
1. Find a fourth proportional to 65, 80, and 91.
2. Find a mean proportional between 28 and 63.
3. Find a third proportional to | and f .
4. What is the second term of a proportion whose first, third, and fourth terms are 45, 160, and 224, respectively ?
PROPORTIONAL LINES. Prop. X. Theorem.
242. If a series of parallels, cutting two straight lines, intercept equal distances on one of these lines, they also inter- cept equal distances on the other. A c/ \C'
\F"
is'
Given lines AB and A'B' cut by lis CC, DD', EE\ and FF^ at points G, D, E, F, and O, D', E', F', respectively, so that CD = DE = EF.
To Prove OD' = D'E'= E'F'.
Proof. In trapezoid CC'E'E, by hyp., DD' is II to the bases, and bisects CE ; it therefore bisects C'E'. (§ 133) .-. C'D' = D'E'. . (1)
In like manner, in trapezoid DD'F'F, EE' is II to the bases, and bisects DF.
.'. D'E' = E'F'. (2)
From (1) and (2), CD' = D'E' = E'F'. (?)
128 PLANE GEOMETRY. — BOOK III.
243. Def. Two straight lines are said to be divided pro- portionally when their corresponding segments are in the same ratio as the lines themselves. ^ ^ ^ ^
Thus, lines AB and CD are divided ! ^ 1
proportionally if C F D
AE^BE^AB CF DF CD
Prop. XI. Theorem.
244. A parallel to one side of a triangle divides the other two sides proportionally.
Case I. Wlien the segments of each side are commensvr rdble.
A
fA\
Given, in A ABC, segments AD and BD of side AB commensurable, and line DE II BC, meeting AC at E.
To Prove ^ = ^.
BD CE
Proof. Let AF be a common measure of AD and BD ;
and let it be contained 4 times in AD, and 3 times in BD.
...^ = i (1)
BD 3 ^ ^
Drawing lis to BC through the several points of division of AB, AE will be divided into 4 parts, and CE into 3 parts, all of which, parts are equal. (§ 242)
.-. ^ = i (2)
From (1) and (2), |§ = g. (?)
THEORY OF PROPORTION. 129
Case n. When the segments of each side are incommenr surahle.
A
Given, in A ABC, segments AD and BD of side AB incommensurable, and line DE II BC, meeting AG at E.
To Prove bD=CE'
Proof. Let AD be divided into any number of equal parts, and let one of these parts be applied to BD as a unit of measure.
Since AD and BD are incommensurable, a certain num- ber of the equal parts will extend from D to B', leaving a remainder BB' < one of the equal parts.
Draw line B'C II BO, meeting ^C at C.
Then, since AD and B'D are commensurable,
|g = g. (§244, Case I.)
Now let the number of subdivisions of AD be indefinitely increased.
Then the unit of measure will be indefinitely diminished, and the remainder BB' will approach the limit 0.
Then, — — - will approach the limit —-,
B'D ^^ BD
and will approach the limit — — •
C'E ^^ CE
By the Theorem of Limits, these limits are equal. (?)
AD^AE
' ' BD CE
130 PLANE GEOMETRY.— BOOK III.
245. Cor. I. We may write the result of § 244,
AD:BD = AE:CE. (1)
.-. AD + BD'.AD = AE+ CE : AE. (§ 237)
.-. AB:AD=^AC'.AE. (2)
In like manner, AB:BD = AC: CE. (3)
246. Cor. n. From (2), §245,.
AB.AC = AD:AE,
and from (3), AB : AC = BD : CE. (§ 235)
mi, u A i AB AD BD x.x
Then, by Ax. 1, _ = _ = _. (4)
247. Sch. The proportions (1), (2), (3), and (4), of § § 245 and 246, are all included in the general statement,
A parallel to one side of a triangle divides the other two sides proportionally.
Prop. XII. Theorem.
248. (Converse of Prop. XI.) A line which divides two sides of a triangle proportionally is parallel to the third side.
A
B C
Given, in A ABC, line DE meeting AB and AC at D and
E respectively, so that
AB^AC
AD AE
To Prove DE II BG.
Proof. If DE is not II BC, draw line DF II BC, meeting
AC at F.
,.,4^ = A£. (§247)
AD AF ^\ ^
(?)
THEORY OF PROPORTION. 131
But by hyp., 4^ = A^,
^ ^^' AD AE
AC^AG
' ' AE AF .-. AE = AF.
Then, DF coincides with DE, and DE II BC. (Ax. 3)
Prop. XIII. Theorem.
249. In any ttiangle, the bisector of an angle divides the opposite side into segments proportional to the adjacent sides. E
B D
Given line AD bisecting Z J. of A ABC, meeting BC at D.
Proof. Draw line BE II AD, meeting CA produced at E. Then, since lis AD and BE are cut by AB,
ZABE = ZBAD. (?)
And since lis AD and BE are cut by CE,
Z AEB = Z CAD. (?)
But by hyp., ZBAD = Z CAD.
.-. ZABE = ZAEB. (?)
.-. AB = AE. (?)
Now since AD is II to side BE of A BCE,
M = 4R. (§247)
DC AC ^ ^
Putting for AE its equal AB, we have
DB^AB
DC AG
132 PLANE GEOMETRY. — BOOK III.
250. Def. The segments of a line by a point are the dis- tances from the point to the extremities of the line, whether the point is in the line itself, or in the line produced.
Prop. XIV. Theorem.
251. In any triangle the bisector of an exterior angle divides the opposite side externally into segments proportional to the adjacerit sides.
Note. The theorem does not hold for the exterior angle at the vertex of an isosceles triangle. E
D B
Given line AD bisecting ext. Z BAE of A ABC, meeting GB produced at D.
To Prove ^ = ^.
DC AC
(Draw BF II AD ; then Z ABF=Z AFB, and AF=AB;
BF is II to side AD of A ACD.)
SIMILAR POLYGONS.
252. Def. Two polygons are said to be similar if they are mutually equiangular (§ 122), and have their homolo- gous sides (§ 123) proportional. B
E D E' D
Thus, polygons ABCDE and A'B'C'D'F' are similar if
ZA=ZA', ZB = ZB',etG., , AB BC CD .
""^' A^'^WC'-^Cr^'"'''
SIMILAR POLYGONS. 133
253. Sch. The following are given for reference :
1. In similar polygons, the homologous angles are equal.
2. In similar polygons, the homologous sides are propor- tioned.
Prop. XV. Theorem.
254. Two triangles are similar when they are mutually equiangular.
A
B ■ U
Given, in A ABC and A'B'O,
ZA = ZA', ZB = ZB',3in([ZC=ZC'. To Prove A ABC and A'B'O similar. Proof. Place AA'B'C in the position ADE; Z A' coin- ciding with its equal Z A, vertices B' and C falling at D and E, respectively, and side B'C at DE.
Since, by hyp., Z ADE = Z B, DE II BC. (?)
.-.—=—. (§247)
AD AE ^^ ^
That IS, A^' = ^C'' (^>
In like manner, by placing A A' B'C so that Z B' shall
coincide with its equal Z B, vertices A' and C falling on
sides AB and BC, respectively, we may prove
AB ^ BC
A'B' B'C'
(2)
From (1) and (2), -^, = -^, = j^,- (?)
Then, A ABC and A' B'C have their homologous sides proportional, and are similar. (§ 252)
134
PLANE GEOMETRY. — BOOK III.
255. Cor. I. Tico triangles are similar when two angles of one are equal respectively to two angles of the other.
For their remaining A are equal each to each. (§ 86)
256. Cor. II. Two nght triangles are similar ivhen an acute angle of one is equal to an acute aiigle of the other.
257. Cor. III. If a line he drawn between two sides of a triangle iiavallel to the third side, the tri- angle formed is similar to the given triangle.
Given line DE II to side BG of A ABC, meeting AB and AC at D and E, respectively.
To Prove A ADE similar to A ABC.
(The A are mutually equiangular.)
258. Sch. I7i similar triangles, the homologous sides lie opposite the equal angles.
Prop. XVI. Theorem.
259. Two triangles are siynilar when their homologous sides are proportional.
A
B C
Given, in A ABC and A'B'C, AB AC
BC B'O'
A'B' A'C To Prove A ABC and A'B'C similar. Proof. On AB and AC, take AD = A'B' and AE = A'C Praw line DJEJ ; then, from the given proportion, AB^AC AD AE
SIMILAR POLYGONS. I35
.-. DE II BO. (§ 248)
Then, A ADE and ABC are similar. (§ 257)
. AB BG ^^ AB BO ,. ^kq ox
''AD='DE^'''A^'^nE' ^^^'^'^^
1. .V. T. AB BC
But by hyp., 'aB^^B^^'
.-. DE = B'a. .'. A ADE = A A'B'O. (§ 69)
But, A ADE has been proved similar to A ABO. Hence, AA'B'O is similar to A ^50.
260. Sch. To prove that two polygons in general are similar, it must be shown that they are mutually equiangu- lar, and have their homologous sides proportional (§ 252) ; but in the case of two triangles, each of these conditions involves the other (§§ 254, 259), so that it is only neces- sary to show that one of the tests of similarity is satisfied.
Prop. XVII. Theorem.
261. Two triangles are similar when they have an angle of one equal to an angle of the other, and the sides including these angles proportional.
B C
Given, in A ABO and A'B' O',
To Prove A ABO and A'B'O' similar. (Place A A'B'O' in the position ADE ; by § 248, DE II BO', the theorem follows by § 257.)
136
PLANE GEOMETRY. — BOOK III.
Prop. XVIII. Theorem.
262. Two triangles are similar when their sides are paral- lel each to each, or perpendicular each to each.
Fig. 1.
Fig. S.
Given sides AB, AC, and BC, of A ABC, II respectively to sides A'B', A'C, and B'C of AA'B'C in Pig. 2, and ± respectively to sides A'B', A'C, and B'C of AA'B'C in Fig. 3.
To Prove A ABC and A' B'C similar.
Proof. Since the sides of A A and A' are II each to each, or _L each to each, A A and A' are either equal or supple- mentary. " (§§ 81, 82, 83)
In like manner, A B and B', and A C and C, are either equal or supplementary.
We may then make the following hypotheses with regard to the A of the A :
1. ^ + ^' = 2rt. A, B + B' = 2Tt. A, C+C' = 2rt A.
2. A-\-A' = 2 rt. A, B + B' = 2 rt. A, C= C.
3. A-\-A' = 2vt A, B = B', (7+C" = 2rt.A
4. A = A', B + B' = 2Tt A, C+C = 2Tt.A.
5. A = A', B=B', whence C=C. (§86)
The first four hypotheses are impossible; for, in either case, the sum of the six A of the two A would be > 4 rt. A.
(§ 84) We can then have only A = A',B = B', and C = C. Therefore, A ABC and A'B'C are similar. (§ 254)
SIMILAR POLYGONS.
137
263. Sch. 1. In similar triangles whose sides are parallel each to each, the parallel sides are homologous.
2. In similar tnangles whose sides are perpendicular each to each, the perpendicular sides are homologous.
Prop. XIX. Theorem.
264. The homologous altitudes of two similar triangles are in the same ratio as any two homologous sides.
A
B ^D C B' ly c
Given AD and A'D' homologous altitudes of similar A ABC and A'B'C.
AD ^ AB ^AO ^ BO B'C'
To Prove
A'D' A'B' A'C (Rt. AABD and A'B'D' are similar by § 256.)
265. Sch. In two similar triangles, any two homologous lines are in the same ratio as any two homologous sides.
EXERCISES.
5. The sides of a triangle are AB = S, BC=6^ and CA=7 ; find the segments into which each side is divided by the bisector of the opposite angle.
6. The sides of a triangle are AB = 5, BC—7, and CA=8; find the segments into which each side is divided by the bisector of the exterior angle at the opposite vertex.
7. The sides of a triangle are 5, 7, and 9. The shortest side of a similar triangle is 14. What are the other two sides ?
8. Two isosceles triangles are similar when their vertical angles are equal. (§ 255.)
9. The base and altitude of a triangle are 5 ft. 10 in. and 3 ft.
6 in., respectively. If the homologous base of a similar triangle is
7 ft. 6 in. , find its homologous altitude.
138 PLANE GEOMETRY. — BOOK III.
Prop. XX. Theorem.
266. Two polygons are similar when they are composed of the same number of triangles, similar each to each, and similarly placed.
Given, in polygons AC and A'C, A ABE similar to AA'B'E', ABCE to AB'C'E', and A CDE to A C'D'E'. To Prove polygons AC and A'C similar. Proof. Since A ABE and A'B'E' are similar,
ZA = ZA'. (?)
Also, * Z ABE = Z A'B'E'.
And since A BCE and B'C'E' are similar, ZEBC=ZE'B'C'. . : Z ABE + Z EBC = Z A'B'E' + Z E'B'O. Or, ZABC = ZA'B'C'.
In like manner, Z BCD = Z B'C'D', etc. Then, AC and A'C are mutually equiangular. Again, since A ABE is similar to A A'B'E', and A BCE to AB'C'E',
^^ ^^ and ^^ = ^. (?)
TDI TTll -Dim V /
A'B' B'E' B'E' B'C
AB ^ BC A'B' B'C'
In like manner, ^ = ^^ = -^, etc.
A'B' B'C CD''
(?)
Then, AC and A'C have their homologous sides propor- tional.
Therefore, AC and A'C are similar. (§ 252)
SIMILAR POLYGONS. I39
Prop. XXI. Theorem.
267. (Converse of Prop. XX.) Two similar polygons may be decomposed into the same number of triangles, similar each to each, and sim.i-^^ ly placed.
Given E and E^ homologous vertices of similar polygons AC and A'C, and lines EB, EC, E'B', and E'C.
To Prove A ABE similar to A A'B'E', A BGE to A B'C'E', and A CDE to A C'D'E'.
Proof. Since polygons AG and A'C are similar,
Z^ = Z^' and 4^ = ^. (?)
A'E' A'B' ^ '
Then, A ABE and A!B'E' are similar. (§ 261)
Again, since the polygons are similar,
Z.ABC=/.AB^a.
And since A ABE and A'B'E' are similar,
Z ABE = Z A'B'E'.
.'. Z ABC - Z ABE = Z A'B'C - Z A'B'E'.
Or, ZEBC=ZE'B'C'.
AB BC
Also, since the polygons are similar.
And since A ABE and A'B'E' are similar.
A'B' B'C
AB BE
A'B' B'E BC BE
B'C B'E'
(?)
Then, since Z EBC = Z E'B'C, and J^ = ^, A BCE
and B'C'E' are similar. (?)
In like manner, we may prove A CDE and C'D'E' similar.
140
PLANE GEOMETRY— BOOK III.
Prop. XXII. Theorem.
268, Tlie perimeters of tivo similar polygons are in the same ratio as any two homologous sides.
Given AB and A'B', EC and S'C, CD and C'D\ etc., homologous sides of similar polygons J.(7and AC. To Prove
AB+BC+CD + Qta. AB BO CD
A'B' + B'C'-\-G'D'-{-etG. A'B' B'C CD' (Apply § 240 to the equal ratios of § 252.)
-, etc.
Prop. XXIII. Theorem.
269. If a perpendicular be drawn from the vertex of the right angle to the hypotenuse of a right triangle,
I. The triangles formed are similar to the whole triangle, and to each other.
11. The perpendicular is a mean proportional between the segments of the hypotenuse.
III. Either leg is a mean proportional between the whole hypotenuse and the adjacent segment.
A D B
^iven line GD J. hypotenuse AB of rt, A ABC.
SIMILAR POLYGONS. 141
I. To Prove A ACD and BCD similar to A ABC, and to each other.
Proof. In rt. A ACD and ABC,
/.A = ZA. Then, A ACD is similar to A ABC. (§ 256)
In like manner, A BCD is similar to A ABC. Then, A ACD and BCD are similar to each other, for each is similar to A ABC.
II. To Prove AD: CD = CD: BD. Proof. Since A ACD and BCD are similar,
Z ACD = ZB3indZA = Z BCD. (§ 253, 1)
In A ACD and BCD, AD and CD are homologous sides,
for they lie opposite the equal A ACD and B, respectively ;
also, CD and BD are homologous sides, for they lie opposite
the equal A A and BCD, respectively. (§ 258)
/. AD:CD=CD:BD. (?)
III. To Prove AB : AC == AC : AD. Proof. Since A ABC and ACD are similar,
Z ACB = Z ADC 3i,nd ZB = Z ACD. (?)
In A ABC and ACD, AB and AC are homologous sides, for they lie opposite the equal A ACB and ADC, respec- tively; also, AC and AD are homologous sides, for they lie opposite the equal A B and ACD, respectively. . (?)
.-. AB:AC=AC:AD. (?)
In like manner, AB:BC = BC: BD.
270. Cor. I. Since an angle inscribed in a semicircle is a right angle (§ 195), it follows that :
If a perpendicular be drawn from any j)oint in the circumference of a circle to a diameter,
1. The perpendicular is a mean proportional between the segments of the diameter.
142 PLANE GEOMETRY. — BOOK III.
2. The chord joining the point to either extremity of the diameter is a mean proportional between the whole diameter and the adjacent segment.
271. Cor. II. The three proportions of § 269 give
Cff=AD xBD, AO^=ABxAD, and W=ABxBD. (?)
Hence, if a perpendicular be drawn from the vertex of the right angle to the hypotenuse of a right triangle,
1. The square of the perpendicular is equal to the product of the segments of the hypoteiiuse.
2. The square of either leg is equal to the product of the whole hypotenuse and the adjacent segment.
As stated in Note, p. 126, these equations mean that the square of the numerical measure of CD is (equal to the product of the numerical measures of AD and BD, etc.
Prop. XXIV. Theorem.
272. In any right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.
A D B
Given AB the hypotenuse of rt. A ABO. To Prove AB' = AC' + BC'.
Proof. Draw line CD ± AB. Then, AC' = ABxAD,
and BC^ = AB x BD. (§ 271, 2)
Adding, AC^ + BC" = AB x (AD + BD) = AB x AB.
.'. ab^ = ac^ + bg\
SIMILAR POLYGONS. 143
273. Cor. I. It follows from § 272 that
AC' = AB'-BC\ and BC' = AB'-Aa'.
That is, in any right tnangle, the square of either leg is equal to the sq^iare of the hypotenuse, minus the square of the other leg.
274. Cor. II. If ^(7 is a diagonal of square ABCD, ACP = AB' + BC^ = AB' + AB'. (§272)
.-. AC' = 2AB'. Dividing both members by AB^,
^ = 2,or^=V2.
Hence, the diagonal of a square is incommensurable with Us side (§ 181).
EXERCISES.
10. The perimeters of two similar polygons are 119 and 68 ; if a side of the first is 21, what is the homologous side of the second?
11. What is the length of the tangent to a circle whose diameter is 16, from a point whose distance from the centre is 17 ?
12. What is the length of the longest straight line which can be drawn on a floor 33 ft. 4 in. long, and 16 ft. 3 in. wide ?
13. A ladder 32 ft. 6 in. long is placed so that it just reaches a window 26 ft. above the street ; and when turned about its foot, just reaches a window 16 ft. 6 in. above the street on the other side. Find the width of the street.
14. The altitude of an equilateral triangle is 5 ; what is its side ?
15. Find the length of the diagonal of a square whose side is 1 ft. 3 in.
16. One of the non-parallel sides of a trapezoid is perpendicular to the bases. If the length of this side is 40, and of the parallel sides 31 and 22, respectively, what is the length of the other side ?
17. The length of the tangent to a circle, whose diameter is 80, from a point without the circumference, is 42. What is the distance of the point from the centre ?
18. If the length of the common chord of two intersecting circles is 16, and their radii are 10 and 17, what is the distance between their centres ? (§ 178.)
144
PLANE GEOMETRY. — BOOK III.
DEFINITIONS.
275. The projection of a point upon a straight line of indefinite length, is the foot of the per- pendicular drawn from the point to the line.
Thus, if line AA^ be perpendicular to line CD, the projection of point A on line CD is point A\
276. The projection of a finite straight line upon a straight line of indefinite length, is that portion of the second line included between the projections of the extremities of the first.
Thus, if lines AA and BB^ be perpendicular to line CD, the projection of line AB upon line CD is line AB\
Prop. XXV. Theorem
277. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides, minus tivice the product of one of these sides and^ the projection of the other side upon it.
Fig. ^.
Given C an acute Z of A ABC, and CD the projection of side AC upon side CB, produced if necessary. (§ 276)
To Prove AB^ ^BC" -\- AC' -2BCx CD. Proof. Draw line AD ; then, AD± CD. (§ 276)
There will be two cases according as D falls on CB (Fig. 1), or on CB produced (Fig. 2).
SIMILAK POLYGONS.
145
In Fig. 1, BD = BC- CD.
In Fig. 2, BD=CD- EG.
Squaring both members of the equation, we have by the algebraic rule for the square of the difference of two num- bers, in either case,
Bff=^BG'+CD^-2BCxCD.
Adding AD" to both members,
Iff +BD' ^W + Aff + W -2 BG X CD, But in rt. A ABD and ACD,
Aff + Bff = A&,
and Aff + Cff = AG^. (§ 272)
Substituting these values, we have
AB'=BG^ + AC^-2BCxGD.
Prop. XXVI. Theorem.
278. In any triangle having an obtuse angle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the other two sides, plus twice the product of one of these sides and the projection of the other side upon it.
Given C an obtuse Z of A ABG, and GD the projection of side AG upon side BG produced.
To Prove AB' = BG' -{- AG' + 2 BG x GD.
(We have BD = BG -I- GD; square both members, using the algebraic rule for the square of the sum of two num- bers, and then add AD' to both members.)
146
PLANE GEOMETRY.— BOOK III.
Prop. XXVII. Theorem.
279. In any triangle^ if a median he drawn from the vertex to the base,
I. The sum of the sqiiares of the other two sides is equal to twice the square of half the base, plus twice the square of the median.
II. The difference of the squares of the other two sides is equal to twice the product of the base and the projection of the median upon the base.
G
Given DE the projection of median CD upon base AB of A ABC', s^nd AC >BC.
To Prove
I. AC -\-BC=2AD' + 2CD.
II. AC -BC =2ABxDE.
Proof. Since AC > BC, E falls between B and D. Then, Z ADC is obtuse, and Z BDC acute. Hence, in A ADC and BDC,
AG^ = AD' + Gff + 2ADx DE, and BC^ = BD" + Cff-2BDx DE.
But by hyp., BD = AD.
.'. AG"" = Aff +CD'-\-ABx DE, and W = ad' +Gff-ABx DE.
Adding (1) and (2), we have
AG' + BC' = 2AD'-^2GD\ Subtracting (2) from (1), we have
AG' -BG' = 2ABx DE.
(§ 278) (§ 277)
(1) (2)
SIMILAR POLYGONS. I47
Prop. XXVIII. Theorem.
280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one chord is equal to the product of the segments of the other.
Given AB and A'B' any two chords passing through fixed point P within O AA'B. To Prove AP x BP= A'P x B'P.
Proof. Draw lines AA' and BB'. Then, in A AA'P and BB'P,
ZA=ZB', for each is measured by ^ arc A'B. (?)
In like manner, Z A' = Z B.
Then, A AA'P and BB'P are similar. (?)
In similar A AA'P and BB'P, sides AP and B'P are
homologous, as also are sides A'P and BP. (§ 258)
.-. AP:A'P=B'P:BP. (?)
.-. APxBP= A'P X B'P. (?)
281. Sch. The proportion of § 280 may be written AP_B^ AP_ 1 A'P~BP' ^^ AJP~BP' B'P
If two magnitudes, such as the segments of a chord pass- ing through a fixed point, are so related that the ratio of any two values of one is equal to the reciprocal of the ratio of the corresponding values of the other, they are said to be reciprocally proportional.
148 PLANE GEOMETRY. — BOOK III.
Then, the theorem may be written,
If any two chords be drawn through a fixed point within a cirdej their segments are reciprocally proportional.
Prop. XXIX. Theorem.
282. If through a fixed poirit without a circle a secant and a tangent be drawn, the product of the ivhole secant and its external segment is equal to the square of the tangent.
Given AP a secant, and CP a tangent, passing through fixed point P without O ABC.
To Prove APx BP= UP.
{/. A = Z. BOP, for each is measured by ^ arc BC (?) ; then A AOP and BCP are similar, and their homologous sides are proportional.)
283. Cor. I. If through a fixed point without a circle a secant and a tangent be drawn, the tangent is a mean pro- portional between the whole secant and its external segment.
284. Cor. n. If any two secants ^ — -^ be drawn through a fixed point without / ^^^^^^^^ a circle, the product of one and its external segment is equal to the prod- uct of the other and its external seg- ment. j^~
Given P any point without OABC, and AP and A'P secants intersecting the circumference at A and B, and A' and B', respectively.
SIMILAR POLYGONS. 149
To Prove APxBP= A'P x B'P.
(We have AP x BP and A'P x B'P each equal to CP.)
285. Cor. III. If any two secants be drawn through a fixed point without a circle, the whole secants and their ex- ternal segments are reciprocally proportional (§ 281).
EXERCISES.
19. Find the length of the common tangent to two circles whose radii are 11 and 18, if the distance between their centres is 25,
20. AB is the hypotenuse of right triangle ABC. If perpendicu- lars be drawn to AB at A and JB, meeting AC produced at Z), and BC produced at E, prove triangles ACE and BCD similar.
Pbop. XXX. Theorem.
286. In any triangle, the product of any two sides is equal to the diameter of the circumscribed circle, multiplied by the perpendicular drawn to the third side from the vertex of the opposite angle.
Given AD a diameter of the circumscribed QAOD of A ABC, and line AE ± BC.
To Prove AB x AC = AD x AE.
(In rt. AABD and ACE, ZD = ZC; then, the A are similar, and their homologous sides are proportional.)
287. Cor. In any triangle, the diameter of the circum- scribed circle is equal to the product of any two sides divided by the perpendicular drawn to the third side from the vertex of the opposite angle.
150
PLANE GEOMETRY. — BOOK III.
Prop. XXXI. Theorem.
288. In any triangle, the product of any two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector.
Tic
— t.— - E
Given, in A ABC, line AD bisecting /.A, meeting BQ 't D.
To Prove AB x AC =BD x DC -\- Ajf. Proof. Circumscribe a O about A ABC\ produce AD to meet the circumference at E, and draw line CE. Then in A ABD and ACE, by hyp.,
Z BAD = Z CAE. Also, ZB = ZE,
since each is measured by i arc AC. (?)
Then, A ABD and ACE are similar. (?)
Tn A ABD and ACE, sides AB ?,nd AE are homologous,
(§ 258)
as also are sides AD and AC.
.'. AB:AD = AE: AC. .-. ABxAC=ADxAE
= ADx (DE -h AD) = AD X DE + Alf. But ADx DE = BD X DC.
.'. AB X AC=BD X DC + Aff.
(?)
(§ 280)
SIMILAR POLYGONS. 151
EXERCISES.
21. The square of the altitude of an equilateral triangle is equal to three-fourths the square of the side.
22. If AD is the perpendicular from A to side BC of triangle ABC, prove _
AB^-AC^= BD^- CD^.
23. If one leg of a right triangle is double the other, the perpen- dicular from the vertex of the right angle to the hypotenuse divides it into segments which are to each other as 1 to 4. (§ 271.)
24. If two parallels to side BC of triangle ABC meet sides AB and ^C at D and F, and E and G, respectively, prove
BD^BF=DF^ (§247.) CE CG EG ^^ ^
25. C and D are respectively the middle points of a chord AB and its subtended arc. If AD = 12 and CD = 8, what is the diame- ter of the circle ? (§ 27L)
26. If AD and BE are the perpendiculars from vertices A and B of triangle ABC to the opposite sides, prove
AC : DC = BC : EC. (Prove AACD and BCE similar.)
27. If D is the middle point of side BC of triangle ABCj right- angled at C, prove AE^ - A& = 3 CD^.
28. The diameters of two concentric circles are 14 and 50 units, respectively. Find the length of a chord of the greater circle which is tangent to the smaller. (§ 273.)
29. The length of a tangent to a circle from a point 8 units dis- tant from the nearest point of the circumference, is 12 units. What is the diameter of the circle ?
(Let X represent the radius. )
30. The non-parallel sides AD and BC ot trapezoid ABCD inter- sect at 0. If AB = 15, CD = 24, and the altitude of the trapezoid is 8, what is the altitude of triangle GAB ? (§ 2G4.)
(Draw CEWAD.)
31. If the equal sides of an isosceles right triangle are each 18 units in length, what is the length of the median drawn from the vertex of the right angle ?
32. The non-parallel sides of a trapezoid are each 53 units in length, and one of the parallel sides is 50 units longer than the other. Find the altitude of the trapezoid.
152 PLANE GEOMETRY. — BOOK III.
33. AB is a chord of a circle, and CE is any chord drawn through the middle point C of arc AB, cutting chord AB at Z>. Prove AG a mean proportional between CD and CE.
(Prove AACD and ACE similar.)
34. Two secants are drawn to a circle from an outside point. If their external segments are 12 and 9, respectively, while the internal segment of the former is 8, what is the internal segment of the latter ? (§284.)
35. If, in triangle ABC, ZC= 120°, prove
AB^ = BG^ + AG^ -{-ACx BC. (Fig. of Prop. XXVI. A AC D is one-half an equilateral A.)
36. BC is the base of an isosceles triangle ABC inscribed in a circle. If a chord AD be drawn cutting BC 2it E, prove
AD:AB = AB: AE. (Prove A ABD and ABE similar.)
37. Two parallel chords on opposite sides of the centre of a circle are 48 units and 14 units long, respectively, and the distance between their middle points is 31 units. What is the diameter of the circle ?
(Let X represent the distance from the centre to the middle point of one chord, and 31 — « the distance from the centre to the middle point of the other. Then the square of the radius may be expressed in two ways in terms of x. )
38. ABC is a triangle inscribed in a circle. Another circle is drawn tangent to the first- externally at G, and AG and BC are pro- duced to meet its circumference at D and E, respectively. Prove tri- angles ABC and CDE similar. (§ 197.)
(Draw a common tangent to the (D at G. Then BC and CE are arcs of the same number of degrees. )
39. ABC and A'BG are triangles whose vertices A and A' lie in a parallel to their com- mon base BC. If a parallel to BC cuts AB and AG Sit D and E, and A'B and A'C at D' and E', respectively, prove DE = D'E'.
(prove ^=^^.) \ BC BG I
40. A line parallel to the bases of a trapezoid, passing through the intersection of the diagonals, and terminating in the non-parallel sides, is bisected by the diagonals. (Ex. 39.)
41. If the sides of triangle ABC are AB = 10, BG=U, and CA = 16, find the lengths of the three medians. (§ 279, I.)
SIMILAR POLYGONS. I53
42. If the sides of a triangle are AB = i, AC=5, and BC = 6, find the length of the bisector of angle A. (§§ 249, 288 )
P
43. The tangents to two intersecting circles from any point in their common chord produced are equal. (§ 282.)
44. If two circles intersect, their common chord produced bisects their common tangents.
45. AB and A C are the tangents to a circle from point A. If CD be drawn perpendicular to radius OB at D, prove
AB:OB = BD: CD.
(Prove A OAB and BCD similar by § 262.)
46. ABC is a triangle inscribed in a circle, A line AD is drawn from A to any point of BC, and a chord BE is drawn, making ZABE=ZADC. Prove
ABxAC = ADxAE.
(Prove AB : AE =z AD : AC.)
47. The radius of a circle is 22^ units. Find the length of a chord which joins the points of contact of two tangents, each 30 units in length, drawn to the circle from a point without the circumference.
(By § 271, 2, the radius is a mean proportional between the dis- tances from the centre to the chord and to the point without the cir- cumference ; in this way the distance from the centre to the chord can be found.)
48. If, in right triangle ABC, acute angle B is double acute angle A, prove AC^ = 3 BC\ (Ex. 104, p. 71.)
49. Find the product of the segments of any chord drawn through a point 9 units from the centre of a circle whose diameter is 24 units.
50. The hypotenuse of a right triangle is 5, and the perpendicular to it from the opposite vertex is 2f. Find the legs, and the segments into which the perpendicular divides the hypotenuse. (§ 271.)
(Let X represent one of the segments of the hypotenuse.)
51. State and prove the converse of Prop. XIII.
(Fig. of Prop. XIII. To prove Z BAD = Z CAD. Produce CA to E, making AE = AB.)
52. State and prove the converse of Prop. XIV. (Fig. of Prop. XIV. Lay off AF = AB.)
154 PLANE GEOMETEY. — BOOK III.
53. If D is the middle point of hypotenuse AB of right triangle ABC^ prove
(JD^ = \ {AB^ + BC^ + CA^) . (Ex. 83, p. 69. )
54. If a line be drawn from vertex C of isosceles triangle ABC^ meeting base AB produced at Z>, prove
CD^ - CB^ = ADx BD. (§ 278.)
55. If AB is the base of isosceles triangle ABC^ and AD be drawn perpendicular to BC^ prove
3 AL^ + Blf + 2CD^ = AB^ + BC^ + (Ja\ (We have 3 AB^ = AD^ + 2 AD^. )
56. The middle points of two chords are distant 5 and 9 units, respectively, from the middle points of their subtended arcs. If the length of the first chord is 20 units, find the length of the second.
(Find the diameter by aid of § 270, 1.)
57. The sides AB and AC, of triangle ABC, are 16 and 9, respec- tively, and the length of the median drawn from C is 11. Find side BC. (§279,1.)
58. The diameter which bisects a chord whose length is 33| units, is 35 units in length. Find the distance from either extremity of the chord to the extremities of the diameter.
(Let X represent one segment of the diameter made by the chord.)
59. The equal angles of an isosceles triangle are each 30°, and the equal sides are each 8 units in length. What is the length of the base ? (Ex. 104, p. 71.)
60. The diagonals of a trapezoid, whose bases are AD and BC, intersect at U. If AE = 9, EC = 3, and BD = 16, find BE and ED.
(A AED and BEC are similar. Find BE by § 237. )
61. Prove the theorem of § 284 by drawing A'B and AB'.
62. The parallel sides, AD and BC, of a circumscribed trapezoid are 18 and 6, respectively, and the other two sides are equal to each other. Find the diameter of the circle.
(Find AB by Ex. 31, p. 100. Draw through 5 a || to CD.)
63. An angle of a triangle is acute, right, or obtuse according as the square of the opposite side is less than, equal to, or greater than, the sum of the squares of the other two sides.
(Prove by Beductio ad Ahsurdum. )
64. Is the greatest angle of a triangle whose sides are 3, 5, and 6, acute, right, or obtuse ?
SIMILAR POLYGONS. I55
65. Is the greatest angle of a triangle whose sides are 8, 9, and 12, acute, right, or obtuse ?
66. Is the greatest angle of a triangle whose sides are 12, 35, and 37, acute, right, or obtase ?
67. If two adjacent sides and one of the diagonals of a parallelo- gram are 7, 9, and 8, respectively, find the other diagonal.
(One-half of either diagonal is a median of the A whose sides are, respectively, the given sides and the other diagonal of the O.)
68. If D is the intersection of the perpendiculars from the vertices of triangle ABC io the opposite sides, prove
AB^ - AG^ = Bjf - CD^. (§ 272.)
69. If a parallel to hypotenuse AB of right triangle ABC meets AC and BC a.t D and E, respectively, prove
AE^ + B& = AB^ + DE"^.
70. The diameters of two circles are 12 and 28, respectively, and the distance between their centres is 29. Find the length of the common tangent which cuts the straight line joining the centres.
(Find the ± drawn from the centre of the smaller O to the radius of the greater O produced through the point of contact.)
71. State and prove the converse of Prop. XXIII. , III. (Fig. of Prop. XXIII. A ABC and ACD are similar.)
72. State and prove the converse of Prop. XXIIL, II.
73. The sum of the squares of the distances of any point in the circumference of a circle from the vertices of an inscribed square, is equal to twice the square of the diameter of the circle. (§195.) ______ _, •
(To prove P^^+PB^-fP(7^+PD^=2^C^.) m ^^ ^D
74. The sides AB, BC, and GA, of triangle ABC, are 13, 14, and 15, respectively. Find the segments into which AB and BC are di- vided by perpendiculars drawn from C and A, respectively.
{A BAC and A CB are acute by § 98. Find the segments by § 277.)
75. In right triangle ABC is inscribed a square DEFGr, having its vertices D and- C in hypotenuse BC, and its vertices E and F in sides AB and AC, respectively. Prove BD : DE = DE : CQ.
(Prove A BDE and CFG similar.)
Note. For additional exercises on Book III., see p. 226.
156 PLANE GEOMETKY.— BOOK III.
CONSTRUCTIONS.
Prop. XXXII. Problem.
289. To divide a given straight line into any number of equal parts.
.^
_ 5< \ \
A H K L B
Given line AB. Required to divide AB into four equal parts.
Construction. On the indefinite line AC, take any con- venient length AD., on DC take DE — AD\ on EG take EF=AD', on FC take FG = AD ; and draw line BG.
Draw lines DH, EK, and FL II BG, meeting AB at H, Kf and L, respectively.
.'. AH= HK= KL = LB. (§ 242)
Prop. XXXIII. Problem.
290. To construct a fourth proportional (§ 231) to three given straight lines.
^^-■'
my A -"---■
P F Q
Given lines m, n, and p.
Required to construct a fourth proportional to m, n, and p. Construction. Draw the indefinite lines AB and AC, making any convenient /. with each other.
SIMILAR POLYGONS. X67
On AB take AD = m ; on DB take DE = n; on AC take AF = p.
Draw line DF; also, line EG II DF, meeting ^Oat G. Then, FG is a fourth proportional to ?ai, 71, and 2?- Proof. Since DF is 11 to side EG of A ^^G^,
AD:DE = AF:FG. (?)
That is, m:n=p: FG.
291. Cor. If we take AF = n, the proportion becomes
m:n = n: FG. In this case, FG is a ^/uVd proportional (§ 230) to m and n.
Prop. XXXIV. Problem.
292. To construct a mean proportional (§ 230) between two given straight lines.
P
A rn B n G E
Given lines m and n.
Required to construct a mean proportional between m and 71.
Construction. On the indefinite line AE, take AB = m ; on BE take BC = n.
With J.(7 as a diameter, describe the semi-circumference ADC.
Draw line BD1.AC, meeting the arc at D.
Then, BD is a mean proportional between m and n.
(The proof is left to the pupil ; see § 270.)
293. Sch. By aid of § 292, a line may be constructed equal to Va, where a is any number whatever.
Thus, to construct a line equal to V3, we take AB equal to 3 units, and BC equal to 1 unit.
Then, BD = VABxBC (§ 232) = V3 x 1 = V3.
158 PLANE GEOMETRY.— BOOK IH.
Prop. XXXV. Problem.
294. To divide a given straight line into parts proportional to any number of given lines.
F -C
^-"' / / /
/
G H Given line AB, and lines m, n, and p.
Required to divide AB into parts proportional to m, n, and p.
Construction. On the indefinite line AG, take AD = m ; on DC take DE = n; on EG take EF = p\ and draw line BF.
Draw lines DG and EH II to ^l^", meeting AB at G^ and IT, respectively.
Then, AB is divided into parts AG, GH, and HB propor- tional to m, ti, and p, respectively.
Proof. Since DGi^W to side ^iT of A AEH,
AH^AG^GH .^.
AE AD DE' ^'^
rri-i , • AH AG GH /^x
That IS, — — - = = (1)
' AE m n ^ ^
And since EH is II to side BF of A ABF,
AH^HB^HB ,2^
AE EF p ' ^ ^
From (1) and (2), 4^ = ^=^. (?)
m n p
Ex. 76. Construct a Ime equal to V2 ; to V5 ; to \/6.
SIMILAR POLYGONS.
Prop. XXXVI. Problem.
15^
295. Upon a given side, homologous to a given side of a given polygon, to construct a polygon similar to the given polygon.
D
A B A' B'
Given polygon ABODE, and line AB\
Required to construct upon side A'B', homologous to AB, a polygon similar to ABODE.
Construction. Divide polygon ABODE into A by draw- ing diagonals EB and EO.
At A construct ZB'A'E'==Z.A', and draw line B'E', making Z A B'E' = ZABE, meeting A'E' at E'.
Then, A A' B'E' will be similar to A ABE. (?)
In like manner, construct AB'O'E' similar to ABOE, and A O'D'E' similar to A ODE.
Then, polygon A' B' O'D'E' will be similar to polygon ABODE. (§ 266)
296. Def. A straight line is said to be divided by a given point in extreme and mean ratio when one of the seg- ments (§ 250) is a mean proportional between the whole line and the other segment.
D A C B
Thus, line AB is divided internally in extreme and mean ratio at 0 if
AB:AO=AO:BO',
and externally in extreme and mean ratio at D if AB'.AD = AD'. BD.
PLANE GEOMETRY. — BOOK m.
To
Prop. XXXVII. Problem. divide a given straight line in extreme and mean
ratu
'296).
Given line AB.
Required to divide AB in extreme and mean ratio. Construction. Draw line BE ± AB, and equal to i AB. With ^ as a centre and EB as a radius, describe O BFG. Draw line AE cutting the circumference at F and G. On AB take AC = AF; on BA produced, take AD = AG. Then, AB is divided at O internally, and at D externally,
in extreme and mean ratio.
•
Proof. Since AG is a secant, and AB a tangent,
AG:AB = AB: AF. (§ 283)
.-. AG:AB = AB'.AG. (1)
.-. AG-AB:AB = AB-AC:AG. (?)
.-. AB:AG-AB = ACiBC. (?)
But by cons., AB =2BE = FG. (2)
.-. AG-AB = AG-FG = AF= AC. Substituting, AB : AC = AC : BO. (3)
Therefore, AB is divided at C internally in extreme and mean ratio.
Again, from (1),
AG + AB : AG = AB-{- AC : AB. (?)
But, AG-^AB^AD + AB = BD.
And by (2), AB -\- AC = FG + AF == AG.
SIMILAR POLYGONS. 161
„•. BD:AG = AG:AB.
.'. AB : AG = AG : BD. ' (?)
.-. AB:AD==AD:BD.
Therefore, AB is divided at D externally in extreme and mean ratio.
298. Cor. If AB be denoted by m, and AG by x, propor- tion (3) of § 297 becomes
m:x = x:m — x.
.'. x^ = m (m -x) = m^- mx. (§ 232) Or, a? + Ttix = m^.
Multiplying by 4, and adding m^ to both members,
4 ic^ + 4 mx + m^ = 4 m^ + m^ = 5 m^ Extracting the square root of both members,
2x-\- m = ± mV5. Since x cannot be negative, we take the positive sign before the radical sign ; then,
2x= mV5 — wi.
EXERCISES.
77. To inscribe in a given circle a triangle similar to a given triangle. (§ 261.)
(Circumscribe a O about the given A, and draw radii to the vertices.)
78. To circumscribe about a given circle a triangle similar to a given triangle. (§ 262.)
Book IV.
AREAS OF POLYGONS Prop. I. Theorem.
299. Two rectangles having equal altitudes are to each other as their bases.
Note. The words "rectangle," "parallelogram," "triaugle," etc., in the propositions of Book IV., mean the amount of surface in the rectangle, parallelogram, triangle, etc.
Case I. When the bases are commensurable.
B .a F, , ,G
E
A K Given rectangles ABCD and EFGH, with equal altitudes
AB and EF, and commensurable bases AD and EH.
TnProv« ABCD ^ AD ^
EH Proof.
EFGH Let AK be a common measure of AD and EH, and let it be contained 5 times in AD, and 3 times in EH.
.-. ^ = 1 (1)
EH 3 ^ ^
Drawing Js to AD and EH through the several points of
division, rect. ABCD will be divided into 5 parts, and rect.
EFGH into 3 parts, all of which parts are equal. (§ 114)
ABCD ^ 5
EFGH 3
ABCD ^ AD
EFGH EH
162
From (1) and (2),
(2) (?)
AREAS OF POLYGONS. 163
Case II. When the bases are incommensurable.
" " K ^
Given rectangles ABCD and EFGH, with equal altitudes AB and EF, and incommensurable bases AD and EH.
To Prove ABCD ^ AD ^
EFGH EH
Proof. Divide AD into any number of equal parts, and apply one of these parts to EH as a unit of measure.
Since AD and EH are incommensurable, a certain num- ber of the parts will extend from E to K, leaving a re- mainder KH < one of the equal parts.
Draw line KL A. EH, meeting FG at L.
Then, since AD and EK are commensurable,
4^^ = 4R. (§ 299, Case I.)
EFLK EK ^ ' ^
Now let the number of subdivisions of AD be indefinitely increased.
Then the unit of measure will be indefinitely diminished, and the remainder KH will approach the limit 0.
Then, will approach the limit ^^^^^>
and will approach the limit — — •
EK ^^ EH
By the Theorem of Limits, these limits are equal. (?)
ABCD ^ AD
■ ■ EFGH EH
300. Cor. Since either side of a rectangle may be taken as the base, it follows that
Two rectangles having equal bases are to each other as their altitudes.
161
PLANE GEOMETRY. — BOOK IV.
Prop. II. Theorem.
301. Any two rectangles are to each other as the products of their bases by their altitudes.
M
a\
b b' b'
Given M and N rectangles, with altitudes a and a\ and bases b and b\ respectively.
M _ a xb
]sr~
To Prove
a'xb'
Proof. Let Rhe a. rect. with altitude a and base b'. Then, since rectangles M and E have equal altitudes, they are to each other as their bases. (§ 299)
.-. ^=L (1)
R b' ^ ^
And since rectangles R and ^ have equal bases, they are
to each other as their altitudes. (?)
R^a
" J^ a''
Multiplying (1) and (2), we have
MR M^ axb ^
R N' M a' xb'
(2)
DEFINITIONS.
302. The area of a surface is its ratio to another surface, called the unit of surface, adopted arbitrarily as the unit of measure (§ 180).
The usual unit of surface is a square whose side is some linear unit; for example, a square inch or a square foot.
303. Two surfaces are said to be equivalent (=o=), when their areas are equal.
AREAS OF POLYGONS.
165
304. The dimensions of a rectangle are its base and altitude.
Prop. III. Theorem.
305. The area of a rectangle is equal to the product of its base and altitude.
Note. In all propositions relating to areas, the unit of surface (§ 302) is understood to be a square whose side is the linear unit.
N
Given a and h, the altitude and base, respectively, of rect. 3f ; and N the unit of surface, i.e., a square whose side is the linear unit.
To Prove that, if N is the unit of surface, area M= a xb.
Proof. Since any two rectangles are to each other as the products of their bases by their altitudes (§ 301),
N 1x1 But since N is the unit of surface, the ratio of il!f to JV" is the area of M. (§ 302)
.-. area Jtf= a x b.
306. Sch. I. The statement of Prop. III. is an abbrevia- tion of the following :
If the unit of surface is a square whose side is the linear unit, the number which expresses the area of a rectangle is equal to the product of the numbers which express the lengths of its sides.
An interpretation of this form is always understood in every proposition relating to areas.
166
PLANE GEOMETRY. — BOOK IV.
307. Cor. The area of a square is equal to the square of its side.
308. Sch. II. If the sides of a rec- tangle are multiples of the linear unit, the truth of Prop. III. may be seen by dividing the figure into squares, each equal to the unit of surface.
Thus, if the altitude of rectangle A is 5 units, and its base 6 units, the figure can be divided into 30 squares.
In this case, 30, the number which expresses the area of the rectangle, is the product of 6 and 5, the numbers which express the lengths of the sides.
Prop. IV. Theorem.
309. The area of a parallelogram is equal to the product of its base and altitude.
E B F G
A. b D
Given O ABCD, with its altitude DF= a, and its base AD = h.
To Prove area ABCD = axb.
Proof. Draw line AE II DF, meeting CB produced at E. Then, AEFD is a rectangle. (?)
In rt. A ABE and DOF,
AB = DC, and AE = DF. (?)
.-. AABE = ADCF. (?)
Now if from the entire figure ADOE we take A ABE,
there remains O ABCD\ and if we take AZ)(7i^, there
remains rect. AEFD.
.-. area ABCD = area AEFD =axb. (§ 305)
AREAS OF POLYGONS. ^q^
310. Cor. I. Two parallelograms having equal bases and equal altitudes are equivalent (§ 303).
311. Cor. II. 1. Two parallelograms having equal alti- tudes are to each other as their bases.
2. Two parallelograms having equal bases are to each other as their altitudes.
3. Any two parallelograms are to each other as the products of their bases by their altitudes.
Prop. V. Theorem.
312. The area of a triangle is equal to one-half the product of its base and altitude.
B E h C
■ Given A ABC, with its altitude AE = a, and its base BC=b.
To Prove area ABC = ^axb.
(By § 108, AC divides CJ ABCD into two equal A.)
313. Cor. I. Two triangles having equal bases and equal altitudes are equivalent.
314. Cor. II. 1. Two triangles having equal altitudes are to each other as their bases.
2. Two triangles having equal bases are to each other as their altitudes.
3. Any two triangles are to each other as the products of their bases by their altitudes.
315. Cor. III. A triangle is equivaleyit to one-half of a parallelogram having the same base and altitude.
168 PLANE GEOMETEY. — BOOK IV.
Prop. VI. Theorem.
316. The area of a trapezoid is equal to one-half the sum of its bases multiplied by its altitude.
A E b B
Given trapezoid ABCD, with its altitude DE equal to a, and its bases AB and DC equal to b and 6', respectively. To Prove area ABCD = a x ^ (6 + &').
(The trapezoid is composed of two A whose altitude is a, and bases b and V, respectively.)
317. Cor. Since the line joining the middle points of the non-parallel sides of a trapezoid is equal to one-half the sum of the bases (§ 132), it follows that
The area of a trapezoid is equal to the product of its alti- tude by the line joining the middle points of its non-parallel sides.
318. Sch. The area of any polygon may be obtained by finding the sum of the areas of the triangles into which the polygon may be divided by drawing diagonals from any one of its vertices.
But in practice it is better to draw the /^^""^--^
longest diagonal, and draw perpendicu- / \ \\
lars to it from the remaining vertices of /„_.].._ _ ] \
the polygon. The polygon will then be X. j 'y^
divided into right triangles and trape- ^\[/^
zoids; and by measuring the lengths of
the perpendiculars, and of the portions of the diagonal
which they intercept, the areas of the figures may be found by §§ 312 and 316.
i
AREAS OF POLYGONS. \QQ
Prop. VII. Theorem.
319. Two similar triangles are to each other as the squares of their homologous sides. C
Oiven AB and A'B' homologous sides of similar A ABO and A'B'C, respectively.
rn. -o ABC AB"
|
*'' " A'B'C A'B'' |
||
|
Proof. Draw altitudes CD and CD'. |
||
|
. ABC AB X CD ' ' A'B'C A'B' X CD' |
(§ |
314, 3) |
|
AB ^ CD A'B' '^ CD' |
(1) |
|
|
J.. CD _ AB ^""^^ CD'- A'B'' |
(§ 264) |
|
|
Substituting this value in (1), |
||
|
ABC _ AB ,, AB _ |
AB' |
A'B'C A'B' A'B' A!B''
320. Sch. Two similar triangles are to each other as the squares of any two homologous lines.
EXERCISES.
1. If the area of a rectangle is 7956 sq. in., and its base 3^ yd., find its perimeter in feet.
2. If the base and altitude of a rectangle are 14 ft. 7 in., and 5 ft. 3 in., respectively, what is the side of an equivalent square ?
3. Find the dimensions of a rectangle whose area is 168, and perimeter 52.
(Let X represent the base.)
170 PLANE GEOMETRY.— BOOK IV.
Prop. VIII. Theorem.
321. 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.
Given Z A common to A ABC and AB'C.
To Prove ABC_^ABxAC^
AB'C AB'xAC
Proof. Draw line B'C.
Then A ABC and AB'C, having the common vertex 0, and their bases AB and AB' in the same str. line, have the same altitude.
And A AB'C and AB'C, having the common vertex 5, and their bases AC and AC in the same str. line, have the same altitude.
AB'C ^ AC ' * AB'C AC Multiplying these equations, we have
ABC AB'C ABC ^ AB x AC AB'C AB'C' ^ AB'C AB' x AC'
EXERCISES.
4. The area of a rectangle is 143 sq. ft. 75 sq. in., and its base is 3 times its altitude. Find each of its dimensions.
(Let X represent the altitude.)
5. The hypotenuse of a right triangle is 5 ft. 5 in., and one of its legs is 2 ft. 9 in. Find its area.
AREAS OF POLYGONS.
171
Prop. IX. Theorem.
322. Two similar polygons are to each other as the squares of their homologous sides.
E4-
Given AB and A'B' homologous sides of similar polygons AC and AC, whose areas are /fand K\ respectively.
K AB"
To Prove
K'
A'B'
Proof. Draw diagonals EB, EC, E'B', and E'C. Then, A ABE is similar to A A'B'E'.
ABE A&
In like manner,
AB'E' jr^'
(§ 267) (§ 319)
and
|
BCE |
BC^ |
AB" |
|
B'CE' |
B'C' |
A'B'' |
|
CDE |
CD' |
AB' |
|
C'D'E' |
CD'' |
A'B'' |
|
ABE |
BCE |
CDE |
A'B'E' B'CE' C'D'E' ABE + BCE + CDE ABE
A'B'E' + B'CE' + C'D'E' A'B'E'
(§ 253, 2)
(?)
(§ 240)
K ABE
AB'
K' A'B'E' A^>
323. Cor. Two similar polygons are to each other as the squares of their perimeters. (§ 268)
172
PLANE GEOMETRY. — BOOK IV.
Prop. X. Problem.
324. To express the area of a triangle in terms of its three sides.
Given sides BC, CA, and AB, of A ABC, equal to a, b, and c, respectively.
Required to express area ABC in terms of a, b, and c. Solution. Let C be an acute Z, and draw altitude AD.
.: c' = a'-\-b'-2axCD. (§277) Transposing, 2a x CD = a^ -\' b^ — c?.
... CD = ^^±^^^.
2a
(§ 273)
AU = AC'-CD"
= (AC -\- CD) {AC -CD)
^{2ab + a^ + b^- c") (2 ab-a^-b''-\- c")
^ r(^ + bY - c^l [c^ - (g - 5)T 4a2
_ (g + 6 + c) (g + ^ — c) (c + « — &) (c — g + ^) /-IN
Now let a + 6 + c = 2 s.
^j^2^2g(2s-2c)(2s-26)(2s-2a)
^16g(s-a)(g-6)(g-c) 4a2
AREAS OF POLYGONS. 173
... AI? - ^^^^^ ~ ^^ ^^ -b)(s- c) a .', area ABC =\axAD (?)
= Vs(s — a){s — h) (s — c).
325. Sch. Let it be required to find the area of a tri- angle whose sides are 13, 14, and 15.
Let a = 13, b = 14, and c = 15 ; then
s = i(13 + 14 + 15) = 21.
Whence, s— a = 8, s — 6 = 7, and s — c = 6.
Then, the area of the triangle is
V21 x8x7x6 = V3x7x23x7x2x3
= V2* X 32 X 72 = 22 X 3 X 7 = 84.
EXERCISES.
6. Find the area of a triangle whose sides are 8, 13, and 15.
7. The area of a square is 693 sq. yd. 4 sq. ft. ; find its side.
8. If the altitude of a trapezoid is 1 ft. 4 in., and its bases 1 ft. 1 in. and 2 ft. 5 in., respectively, what is its area ?
9. If, in figure of Prop. VII., AB = 9, A'B' = 7, and the area of A'B'C is 147, find area ABC.
10. If the sides of triangle ABC are AB = 25, BC= 17, and CA = 28, find its area, and the length of the perpendicular from each vertex to the opposite side.
11. Find the length of the diagonal of a rectangle whose area is 2640, and altitude 48.
12. Find the lower base of a trapezoid whose area is 9408, upper base 79, and altitude 96.
13. The area of a rhombus is equal to one-half the product of its diagonals. (§117.)
14. The diagonals of a parallelogram divide it into four equivalent triangles.
15. Lines drawn to the vertices of a parallelogram from any point in one of its diagonals divide the figure into two pairs of equivalent triangles. (Ex. 63, p. 67.)
16. The area of a certain triangle is 2|: times the area of a similar triangle. If the altitude of the first triangle is 4 ft. 3 in., what is the homologous altitude of the second ? (§ 320.)
174
PLANE GEOMETRY.— BOOK IV.
326. Sch. Since the area of a square is equal to the square of its side (§ 307), we may state Prop. XXIV., Book III., as follows :
In any right triangle, the square described upon the hypotenuse is equivalent to the sum of the squares described upon the legs.
The theorem in the above form may be proved as follows :
|
y^ "i- |
\ |
|
; |
P |
B
F ME
Given ABEF, AGGH, and BCKL squares described upon hypotenuse AB, and legs AG and BG, respectively, of rt. AABG. To Prove area ABEF= area AGGH+ area BGKL. Proof. Draw line GD ± AB, and produce it to meet EF at Jf ; also, draw lines BH and GF. Then in A ABH and AGF, by hyp.,
AB = AF and AH= AG. Also, ZBAH = ZGAF,
for each is equal to a rt. Z -\- Z BAG.
.'. A ABH = A AGF. (?)
Now A ABH has the same base and altitude as square AGGH.
.-. Sivesi ABH =i3LTe3i AGGH. (§ 315)
And A AGF has the same base and altitude as rect. ADMF.
AREAS OF POLYGONS.
176
But, area ABH = area ACF.
.'. isiTesiACGH=i3iTea.ADMF, (?)
or area ^OG^ir= area ^/)Jfi^. (1)
Similarly, by drawing lines AL and CE, we may prove
area BCKL = area BDME. (2)
Adding (1) and (2), we have
area ACGH + area BCKL = area ABEF.
327. Sch. The theorem of § 326 is supposed to have been first given by Pythagoras, and is called after him the Pythagorean Theorem.
Several other propositions of Book III. may be put in the form of statements in regard to areas ; as, for example. Props. XXV. and XXVI.
EXERCISES.
17. If EF is any straight line drawn through the centre of parallelogram A BCD, meeting sides AD and BC at E and F, respectively, prove triangles BEF and CED equivalent. (Ex. 61, p. 6Q.) A
(Prove BEDF a O by § 112.)
18. The side of an equilateral triangle is 5 ; find its area. 21, p. 151.)
19. The altitude of an equilateral triangle is 3 ; find its area,
C
(Ex.
C
20. Two triangles are equivalent if they have two sides of one equal respectively to two sides of the other, and the included angles supplementary. D'
21. One diagonal of a rhombus is five-thirds of the other, and the difference of the diagonals is 8 ; find its area. (Ex. 13, p. 173.)
22. If D and E are the middle points of sides BC and AC, respec- tively, of triangle ABC, prove triangles ABD and ABE equivalent. (§ 80.)
176 PLANE GEOMETRY.— BOOK IV.
23. If E is the middle point of CD, one of the j non-parallel sides of trapezoid ABCB, and a par- allel to AB drawn through E meets EC bX F and AD at G, prove parallelogram ABFG equivalent to the trapezoid. ^
24. The sides AB, BC, CD, and DA of quadrilateral ABCD are 10, 17, 13, and 20, respectively, and the diagonal ^C is 21. Find the area of the quadrilateral.
25. Find the area of the square inscribed in a circle whose radius is 3.
(The diagonal is a diameter, by § 157.)
26. The area of an isosceles right triangle is 81 sq. in.; find, its hypotenuse in feet.
(Represent one of the equal sides by x.)
27. The area of an equilateral triangle is 9V3 ; find its side. (Represent the side by x.)
28. The area of an equilateral triangle is 16\/3 ; find its altitude. (Represent the altitude by x.)
29. The base of an isosceles triangle is 66, and each of the equal sides is 53 ; find its area.
/\
30. The area of a triangle is equal to one-half y>\-\ the product of its perimeter by the radius of the K^\ >. inscribed circle. k ^'1^, J\
^ D ^^B
31. The area of an isosceles right triangle is equal to one-fourth the area of the square described upon the base. (§ 307.)
32. If angle A of triangle ABC is 30°, prove
area ABC = ^ AB x AC, (Draw CD ±AB; then CD may be found by Ex. 104, p. 71.)
33. A circle whose diameter is 12 is inscribed in a quadrilateral whose perimeter is 50. Find the area of the quadrilateral.
(Compare Ex. 30, p. 176.)
34. Two similar triangles have homologous sides equal to 8 and 15, respectively. Find the homologous side of a similar triangle equiva- lent to their sum. (§ 319.)
35. If E is any point within parallelogram ABCD, triangles ABE and CDE are together equivalent to one-half the parallelogram.
(Draw through ^ a || to AB.)
AREAS OF POLYGONS.
177
36. The non-parallel sides, AB and CD, of a trapezoid are each 25 units in length, and the sides AD and BC are 33 and 19 units, respectively. Find the area of the trapezoid.
(Draw through ^ a || to CD, and a ± to AD.)
37. If the area of a polygon, one of whose sides is 15 in., is 375 sq. in., what is the area of a similar polygon whose homologous side is 18 in.?
38. If the area of a polygon, one of whose sides is 36 ft., is 648 sq. ft., what is the homologous side of a similar polygon whose area is 392 sq.ft.?
39. If one diagonal of a quadrilateral bisects the other, it divides the quadrilateral into two equivalent triangles.
(To prove A ABC =c= A ACD.)
40. Two equivalent triangles have a com- mon base, and lie on opposite sides of it. Prove that the base, produced if necessary, bisects the line joining their vertices.
(To prove CD = CD.)
41. If the sides of a triangle are 15, 41, and 52, find the radius of the inscribed circle. (Ex. 30, p. 176.)
find its
42. The area of a rhombus is 240, and its side is 17 diagonals. (Ex. 13, p. 173.)
(Represent the diagonals by 2 x and 2 y.)
43. The sum of the perpendiculars from any point within an equilateral triangle to the three sides is equal to the altitude of the triangle.
^ Te "C
44. The longest sides of two similar polygons are 18 and 3, respec- tively. How many polygons, each equal to the second, will form a polygon equivalent to the first ? (§ 322.)
45. If the sides of a triangle are 25, 29, and 36, find the diameter of the circumscribed circle. (§ 287.)
(The altitude of a A equals its area divided by one-half its base.)
178 PLANE GEOMETRY.— BOOK IV.
46. If a is the base, and 6 one of the equal sides of an isosceles triangle, prove its area equal to ^aVTpTT^i.
47. The sides AB and AC of triangle ABC are 15 and 22, respec- tively. From a point D in AB^ a parallel to BC ia drawn meeting AC At E, and dividing the triangle into tvsro equivalent parts. Find AD and AE. (§ 319.)
48. The segments of the hypotenuse of a right triangle made by a perpendicular drawn from the vertex of the right angle, are 5| and 9|, respectively ; find the area of the triangle.
49. Any straight line drawn through the centre of a parallelogram, terminating in a pair of opposite sides, divides the parallelo- gram into two equivalent quadrilaterals. (Ex. 61, p. 66.)
50. If JS is the middle point of CD, one of the non-parallel sides of trapezoid ABCD, prove triangle ABE equivalent to \ABCD.
(Draw through JS" a || to AB.)
51. The sides of triangle ABC are AB = IZ, BC=U, and CA = 15. If AD is the bisector of angle A, meeting BC at D, find the areas of triangles ABD and ACD. (§§ 249, 325.)
52. The longest diagonal AD of pentagon ABCDE is 44, and the perpendiculars to it from B, C, and E are 24, 16, and 15, respectively. If AB = 25, CD = 20, and AE = 17, what is the area of the penta- gon ? (§318.)
53. The sides of a triangle are proportional to the numbers 7, 24, and 25, respectively. The perpendicular to the third side from the vertex of the opposite angle is 13 |i. Find the area of the triangle.
(Represent the sides by 7 x, 24 x, and 25 x, respectively ; the A is a rt. A by Ex. 63, p. 154.)
54. If E and F are the middle points of sides AB and AC, respec- tively, of a triangle, and D is any point in BC, prove quadrilateral AEDF equivalent to one-half triangle ABC.
(Vro\e A DEF =0=1 A ABC, by aid of Ex. 64, p. 67.)
55. If E, F, Cr, and H are the middle points of sides AB, BC, CD, and DA, respectively, of quadrilateral ABCD, prove EFCH a parallelo- gram equivalent to one-half ABCD.
(By Ex. 64, p. 67, area EBF - \ area ABC.) ^
Note. For additional exercises on Book IV. , see p, 229.
AREAS OF POLYGONS.
179
CONSTRUCTIONS.
Prop. XI. Problem.
328. To construct a square equivalent to the sum of two given squares.
|
M |
c |
||
|
N |
A B
Given squares M and N.
Required to construct a square =c=M-\-N.
Construction. Draw line AB equal to a side of M.
At A draw line ACA.AB, and equal to a side of N\ and draw line EG.
Then, square P, described with its side equal to EC, will be =0= Jf+JV.
Proof. In rt. A ABC, BG^ = Jff + AC"^- (?)
.-. area P — area M-\- area N. (§ 307)
329. Cor. By an extension of the above method, a square may be constructed equivalent to the sum of any number of given squares.
Given three squares whose sides are equal to ^
m, n, and p, respectively.
Required to construct a square =o= the sum of the given squares.
Construction. Draw line AB = m.
Draw line AC 1. AB, and equal to n, and line^O. ^^ '" ^
Draw line CD ± EC, and equal to p, and line ED.
Then, the square described with its side equal to ED will be ^ the sum of the given squares.
(The proof is left to the pupil.)
P/
CL
180
PLANE GEOMETRY.— BOOK IV.
Prop. XII. Problem.
330. To construct a square equivalent to the difference of two given squares.
N
B
Given squares M and N, M being > N.
Required to construct a square =o- M— N.
Proof. Draw the indefinite line AD.
At A draw line AB _L AD, and equal to a side of W.
With 5 as a centre, and with a radius equal to a side of M, describe an aj-c cutting AD at C, and draw line BC.
Then, square P, described with its side equal to AC, will be ^M-N.
Proof. In rt. A ABC, AC'' = BG^ - Aff. (?)
.-. area F= area M— area N. (?)
Prop. XIII. Problem.
331. To construct a square equivalent to a given paral- lelogram.
K If
C '
A E
B
F G
Given CJABCD.
Required to construct a square =c= ABCD.
Construction. Draw line DE ± AB, and construct line FG a mean proportional between lines AB and DE (§ 292).
Then, square FGHK, described with its side equal to FG, will be ^ ABCD.
AREAS OF POLYGONS. 181
Proof. By cons., AB: FG = FG: DE.
.-. FG^ = ABxDE. (?)
.'. Sivesi FGHK = sivesi ABCD. (?)
332. Cor. 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.
Ex. 56. To construct a triangle equivalent to a given square, having given its base and an angle adjacent to the base.
(Take for the required altitude a third proportional to one-half the given base and the side of the given square.)
Prop. XIV. Problem.
333. To construct a rectangle equivalent to a given square^ having the sum of its base and altitude equal to a given line.
M
X>-
N
Given square M, and line AB.
Required to construct a rectangle =c= M, having the sum of its base and altitude equal to AB.
Construction. With AB as a diameter, describe semi- circumference ADB.
Draw line AC _L AB, and equal to a side of M.
Draw line CF II AB, intersecting arc ADB at D, and line DE A. AB.
Then, rectangle N, constructed with its base and altitude equal to BE and AE, respectively, will be =c= M.
Proof. AE:DE=DE: BE. (§ 270, 1)
.-. AE xBE = DE' = AC\ (?)
.-. area N= area M. (?)
182 PLANE GEOMETRY.— BOOK IV,
Prop. XV. Problem.
334. To construct a rectangle equivalent to a given square, having the difference of its base and altitude equal to a given line.
\ )p. ..^
N
- E
Given square M, and line AB.
Required to construct a rectangle =0= M, having the differ- ence of its base and altitude equal to AB.
Construction. With AB as a diameter, describe O ADB.
Draw line AC J- AB, and equal to a side of M.
Through centre 0 draw line CO, intersecting the circum- ference at D and E.
Then, rectangle N, constructed with its base and altitude equal to CE and CD, respectively, will be =0= M.
Proof. CE-CD = DE = AB. (?)
That is, the difference of the base and altitude of JV is equal to AB.
Again, AC is tangent to O ADB at A. (?)
.-. CDxCE= GA". (§ 282)
.-. area JV= area M. (?)
EXERCISES.
57. To construct a triangle equivalent to a given triangle, having given its base.
(Take for the required altitude a fourth proportional to. the given base, and the base and altitude of the given A.) How many different A can be constructed ?
58. To construct a rectangle equivalent to a given rectangle, hav- ing given its base.
AREAS OF POLYGONS.
183
59. To construct a square equivalent to twice a given square. (§ 307.)
Prop. XVI. Problem.
335. To construct a square having a given ratio to a given square.
C.
M
m
/ i N
\
|
\A |
N |
|
\^\ |
|
|
n B |
!/ I A"~m"'l)"
Given square M, and lines m and n.
Required to construct a square having to M the ratio n:m.
Construction. On line AB, take AD = m and DB — n.
With AB as a diameter, describe semi-circumference ACB.
Draw line DC1.AB, meeting arc ACB at G, and lines AC and BC
0\i AC take CE equal to a side of M; and draw line EF II AB, meeting BC at F.
Then, square N, constructed with its side equal to GF, will have to M the ratio n-.m.
Proof. Z ACB is a rt. Z. (?)
Then since CD is ± AB,
AG^ AB X AD AD m /^ 271 2)
(?)
|
BC' ABxBD BD |
n |
|
since EF is || AB, GE AG GF BC |
|
|
ge' aW |
_m |
|
cf' bg' |
n |
|
area M m |
area iV n
(?)
184 PLANE GEOMETRY.— BOOK IV.
Prop. XVII. Problem.
336. To construct a triangle equivalent to a given polygon. A
O D Given polygon ABGDE. Required to construct a A =c= ABODE.
Construction. Take any three consecutive vertices, as Ay B, and C, and draw diagonal AC; also, line BF )| AC, meet- ing DC produced at F, and line AF.
Then, AFDE is a polygon =c= ABODE, having a number of sides less by one.
Again, draw diagonal AD ; also, line EG || AD, meeting CD produced at G, and line AG.
Then, AFG is a A =0= ABODE.
Proof. A ABC and AFC have the same base AC. And since their vertices B and F lie in the same line || to AC, they have the same altitude. (§ 80)
.-. area JjB(7 = area ^F(7. (?)
Adding area AODE to both members, we have
area ABODE = area AFDE.
Again, A AED and AGD have the same base AD, and the same altitude.
.-. Sivesi AED = Sivesi AGD. (?)
Adding area AFD to both members, we have
area AFDE = area AFG.
.-. area ABODE = area AFG. (?)
AREAS OF POLYGONS.
185
Note. By aid of §§ 336 and 332, a square may be constructed equivalent to a given polygon.
Prop. XVIII. Problem.
337. To construct a polygon similar to a given polygon, and having a given ratio to it.
m
Given polygon AC, and lines m and n.
Required to construct a polygon similar to AC, and hav- ing to it the ratio n : m.
Construction. Construct A'B^ the side of a square having to the square described upon AB the ratio n : m. (§ 335)
Upon side A!B\ homologous to AB, construct polygon
A^O similar to polygon AC.
Then, A^O will have to ^C the ratio n : m. Proof. Since AC is similar to A'C, AC Aff
But by cons.,
|
A'C |
A'B'' |
|
AB' |
_m^ |
|
A'B'' |
n |
|
AC A'C |
n |
(§ 295)
(§ 322)
(?)
Ex. 60. To construct an isosceles triangle equivalent to a given triangle, having its base co- incident with a side of the given triangle.
186 PLANE GEOMETRY. — BOOK IV.
Prop. XIX. Problem.
338. To construct a polygon similar to one of two given polygons^ and equivalent to the other.
Given polygons M and N.
Required to construct a polygon similar to M, and =0 N.
Construction. Let AB be any side of M.
Construct m, the side of a square =0= M, and n, the side of a square =0= N. (Note, p. 185)
Construct A'B\ a fourth proportional to m, n, and AB.
Upon side A'B\ homologous to AB, construct polygon P similar to M. (§ 295)
Then, P =c= JV.
Proof. Since M is similar to P,
area M AB
(?)
(?)
area P A'B'
But by cons., m : n = AB : A'B', or ——- : = — • ^ ' ' A'B' n
area M _'m^_ area M
area P n^ area If
.: area P = area M
EXERCISES.
61. To construct a triangle equivalent to a given square, having given its base and the median drawn from the vertex to the base.
(Draw a |1 to the base at a distance equal to the altitude of the A. ) What restriction is there on the values of the given lines ?
62. To construct a rhombus equivalent to a given parallelogram, having one of its diagonals coincident with a diagonal of the paral- lelogram. (Ex. 60.)
AREAS OP POLYGONS. Jg*^
63. To draw through a given point within a parallelogram a straight line dividing it into two equivalent parts. (Ex, 49, p, 178.)
64. To construct a parallelogram equivalent to a given trapezoid, having a side and two adjacent angles coincident with one of the non- parallel sides and the adjacent angles, respectively, of the trapezoid (Ex. 23, p. 176.)
65. To construct a triangle equivalent to a given triangle, having given two of its sides. (Ex. 67.)
(Let m and n be the given sides, and take m as the base.) Discuss the solution when the altitude is < n. = n. > w.
66. To construct a right triangle equivalent to a given square, having given its hypotenuse. (Ex. 96, p. 119.)
(Find the altitude as in Ex. 56. )
What restriction is there on the values of the given parts ?
67. To construct a right triangle equivalent to a given triangle, having given its hypotenuse.
What restriction is there on the values of the given parts ?
68. To construct an isosceles triangle equivalent to a given tri- angle, having given one of its equal sides equal to m.
(Draw a II to the given side at a distance equal to the altitude.) Discuss the solution when the altitude is < m. = m. > m.
69. To draw a line parallel to the base of a /\^ triangle dividing it into two equivalent parts. / \.
(§ 319.) b/ Xc
(^AABC and AB'C are similar.) g/ \^
70. To draw through a given point in a side of a parallelogram a straight line dividing it into two equivalent parts.
71. To draw a straight line perpendicular to the bases of a trape- zoid, dividing the trapezoid into two equivalent parts.
(A str. line connecting the middle points of the bases divides the trapezoid into two equivalent parts.)
72. To draw through a given point in one of the bases of a trape- zoid a straight line dividing the trapezoid into two equivalent parts.
(A str. line connecting the middle points of the bases divides the trapezoid into two equivalent parts.)
73. To construct a triangle Similar to two given similar triangles, and equivalent to their sum.
(Construct squares equivalent to the A.)
74. To construct a triangle similar to two given similar triangles, and equivalent to their difference.
Book Y.
REGULAR POLYGONS. -MEASUREMENT OP THE CIRCLE.
339. Def. A regular polygon is a polygon which is both equilateral and equiangular.
Prop. I. Theorem.
340. A circle can he circumscribed about, or inscribed in, any regular polygon.
Given regular polygon ABODE.
To Prove that a O can be circumscribed about, or inscribed in, ABODE.
Proof. Let 0 be the centre of the circumference described through vertices A, B, and 0 (§ 223).
Draw radii OA, OB, 00, and OD.
In A OAB and OOD, OB = 00. (?)
And since, by def., polygon ABODE is equilateral,
AB=OD.
188
REGULAR POLYGONS. j^og
Again, since, by def., polygon ABODE is equiangular,
Z.ABC=Z.BGD. And since A OBC is isosceles,
ZOBC=ZOCB. (?)
.-. Z ABC - Z OBC=Z BCD - Z OOB. Or, Z Oi^^ = Z OCD.
.-. AOAB = AOCD. (?)
.-. 0^=0Z>. (?)
Then, the circumference which passes through A, B, and C also passes through D.
In like manner, it may be proved that the circumference which passes through B, C, and D also passes through E.
Hence, a O can be circumscribed about ABCDE.
Again, since AB, BC, CD, etc., are equal chords of the circumscribed O, they are equally distant from 0. (§ 164)
Hence, a O described with 0 as a centre, and a line OF ± to any side AB as a radius, will be inscribed in ABCDE.
341. Def. The centre of a regular polygon is the common centre of the circumscribed and inscribed circles.
The angle at the centre is the angle between the radii drawn to the extremities of any side ; as AOB.
The radius is the radius of the circumscribed circle, OA. The apothem is the radius of the inscribed circle, OF.
342. Cor. From the equal A OAB, OBC, etc., we have
ZAOB = Z BOO = Z COD, etc. (?)
But the sum of these A is four rt. A. (§ 35)
Whence, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides.
EXERCISES. Find the angle, and the angle at the centre, 1. Of a regular pentagon.
190 PLANE GEOMETRY. — BOOK V.
2. Of a regular dodecagon.
3. Of a regular polygon of 32 sides.
4. Of a regular polygon of 25 sides.
Prop. II. Theorem.
343. If the circumference of a circle he divided into any number of equal arcs,
I. Their chords form a regular inscribed polygon. II. Tangents at the points of division form a regular cir- cumscribed polygon.
L A F
Given circumference ACD divided into five equal arcs, AB, BC, CD, etc., and chords AB, BC, etc.
Also, lines LF, FG, etc., tangent to O ACD at A, B, etc., respectively, forming polygon FGHKL.
To Prove polygons ABCDE and FGHKL regular.
Proof. Chord AB = chord BC = chord CD, etc. (§ 158)
Again, arc BCDE = arc CDEA = arc DEAB, etc., for each is the sum of three of the equal arcs AB, BC, etc. .-. Z EAB = Z ABC = Z BCD, etc. (§ 193)
Therefore, polygon ABCDE is regular. (§ 339)
Again, in A ABF, BCG, CDH, etc., we have AB = BO=CD,QtG.
Also, since arc AB = arc BC= arc CD, etc., we have
Z BAF=ZABF= Z CBG = Z BCG, etc. (§ 197)
Whence, ABF, BCG, etc., are equal isosceles A. (§§ 68, 96)
REGULAR POLYGONS. j^gj
and BF= BG=CG= OH, etc. (§ 66)
.-. FG=GH=HK, etc.
Therefore, polygon FGHKL is regular. (?)
344. Cor. I. 1. If from the middle point of each arc sub- tended by a side of a regular inscribed polygon lines be drawn to its extremities, a regular inscribed polygon of double the number of sides is formed.
2. If at the middle 2?oint of each arc included between two consecutive points of contact of a regular circwnscribed poly- gon tangents be drawn, a regular circumscribed polygon of double the number of sides is formed.
345. Cor. II. An equilateral 2'>olygon' inscribed in a circle is regular; for its sides subtend equal arcs. (?)
Prop. III. Theorem.
346. Tangents to a circle at the middle points of the arcs subtended by the sides of a regular inscribed polygon, form a regular circumscribed polygon.
A'
Given ABODE a regular polygon inscribed in OAG, and A'B'O'D'E' a polygon whose sides A'B', B'C', etc., are tangent to O AC at the middle points F, G, etc., of arcs AB, BO, etc., respectively.
To Prove A'B'O'D'E' a regular polygon.
(Arc AF= arc BF= arc BG = arc OG, etc., and the propo- sition follows by § 343, II.)
192
PLANE GEOMETRY. — BOOK V.
Prop. IV. Theorem.
347. Regular polygons of the same number of sides are similar.
D
D'
(The polygons fulfil the conditions of similarity given in § 252.)
Prop. V. Theorem.
348. The perimeters of two regular polygons of the same number of sides are to each other as their radii, or as their apothems.
F' B'
Given P and P' the perimeters, R and R' the radii, and r and r' the apothems, respectively, of regular polygons AC and A^O of the same number of sides.
To Prove
P P'
R R'
Proof. Let 0 be the centre of polygon AC, and 0' of A'C, and draw lines OA, OB, O'A', and O'B'. Also, draw line OF ± AB, and line O'F' _L A'B'. Then, OA = R, OA' = R', OF = r, and O'P' = r'. Now in isosceles A OAB and OA'B',
ZAO£ = ZA'0'B', (§342)
BEGULAE POLYGONS.
193
And since OA = OB and O'A' = O'B', we have
OA ^ OB
O'A' O'B'' Therefore, A OAB and O'A'B' are similar. (§ 261)
^^ ^ "* (§§253,11,264)
A'B' R' r' But polygons AC and A'O are similar. P ^ AB ' ' P' A'B''
" P' R' r''
(§ 347) (§ 268)
(?)
349. Cor. Let ^denote the area of polygenic, and J^T' of A'C.
^^ ^ (§ 322)
But,
A'B' R
K' i2'2
That is, the areas of two regular polygons of the same number of sides are to each other as the squares of their radii, or as the squares of their apothems.
Prop. VI. Theorem.
350. The area, of a regular polygon is equal to one-half the product of its perimeter and apothem.
D
A F B Given the perimeter equal to P, and the apothem OF equal to r, of regular polygon AC. To Prove area AC = ^Pxr.
(A OAB, OBC, etc., have the common altitude r.)
194
PLANE GEOMETRY. — BOOK V.
Prop. VII. Problem. 351. To inscribe a square m a given circle.
B
Given O AG. Required to inscribe a square in O AC. Construction. Draw diameters AC and BD ± to each other, and chords AB, BC, CD, and DA. Then, ABCD is an inscribed square. (The proof is left to the pupil ; see § 343, I.)
352. Cor. Denoting radius OA by R, we have
AB' = 6A^ + OB'' =^2 R\ (§ 272)
.-. AB = BV2.
That is, the side of an inscribed square is equal to the radius of the circle multiplied by V2.
Prop. VIII. Problem.
353. To inscribe a regular hexagon in a given circle. B^ . C
Given O AG.
REGULAR POLYGONS. I95
Required to inscribe a regular hexagon in O ^0.
Construction. Draw any radius OA.
With ^ as a centre, and AG as a radius, describe an arc cutting the given circumference at B, and draw chord AB.
Then, AB is a side of a regular inscribed hexagon.
Hence, to inscribe a regular hexagon in a given O, apply the radius six times as a chord.
Proof. Draw radius OB ; then, A OAB is equilateral. (?)
Therefore, A OAB is equiangular. (§ 95)
Whence, Z AOB is one-third of two rt. A. (?)
Then, Z AOB is one-sixth of four rt. A, and arc AB is one-sixth of the circumference. (§ 154)
Then, AB is a side of a regular inscribed hexagon.
(§ 343, I.)
354. Cor. I. The side of a regular inscribed hexagon is equal to the radius of the circle.
355. Cor. II. If chords be drawn joining the alternate vertices of a regular inscribed hexagon, there is formed an inscribed equilateral triangle.
356. Cor. ni. The side of an in- ^ — ^^ scribed equilateral triangle is equal / ^.^ \\
to the radius of the circle multiplied . L^__ .^ (j
by V3. r J
Given AB a side of an equilateral A y^ ^x
inscribed in O AD whose radius is B. j)
To Prove AB = R V3.
Proof. Draw diameter AO, and chord BC, then, BC is
a side of a regular inscribed hexagon. (§ 355)
Now ABC is art. A. (§ 195)
.: AB'=AC'-BC' (?)
=={2Ey-R' (§354)
= AR'-Ii^ = 3Ii'.
.: ulB = i2V3.
196
PLANE GEOMETRY. — BOOK V.
Prop. IX. Problem. 357. To inscribe a regular decagon in a given circle.
Given OAG. Required to inscribe a regular decagon in O AC. Construction. Draw any radius OA. and divide it inter- nally in extreme and mean ratio at ilf (§ 297), so that
OA'.OM=OM:AM. (1)
With ^ as a centre, and OM as a radius, describe an arc cutting the given circumference at B, and draw chord AB. Then, AB is a side of a regular inscribed decagon. Hence, to inscribe a regular decagon in a given O, divide the radius internally in extreme and mean ratio, and apply the greater segment ten times as a chord. Proof. Draw lines OB and BM. In A OAB and ABM, ZA = ZA. And since, by cons., 0M= AB, the proportion (1) becomes
OA:AB = AB:AM. Therefore, A OAB and ABM are similar.
.-. ZABM=ZAOB. Again, A OAB is isosceles. Hence, the similar AABMh isosceles, and AB = BM=OM, .'. ZOBM=ZAOB. .-. Z ABM-h Z OBM= Z AOB -f- Z AOB.
(§261) (?) (?)
(Ax. 1) (?)
REGULAR POLYGONS. I97
Or, Z0BA = 2ZA0B. (2)
But since A OAB is isosceles,
2 Z OBA + Z AOB = 180°. (§ 84)
Then, by (2), 5 ZAOB = 180°, or Z AOB = 36°. Therefore, ZAOB is one-tenth of four rt. A, and AB is a side of a regular inscribed decagon. (?)
358. Cor. I. If chords be drawn joining the alternate ver- tices of a regular inscribed decagon, there is formed a regular inscribed pentagon.
359. Cor. II. Denoting the radius of the O by R, we have
AB = 0M= -^(V^-1). (§ 298)
This is an expression for the side of a regular inscribed decagon in terms of the radius of the circle.
Prop. X. Problem.
360. To construct the side of a regular pentedecagon in- scribed in a given circle.
Given arc MN.
Required to construct the side of a regular inscribed polygon of fifteen sides.
Construction. Construct chord AB a side of a regular inscribed hexagon (§ 353), and chord AC a side of a regular inscribed decagon (§ 357), and draw chord BC.
Then, BC is a side of a regular inscribed pentedecagon.
Proof. By cons., arc BC is l — ^, or -^, of the circum- ference.
Hence, chord BO is a side of a regular inscribed pente- decagon. (?)
198 PLANE GEOMETRY.— BOOK V.
361. Sch. I. By bisecting arcs AB, BC, etc., in the figure of Prop. VII., we may xjonstruct a regular inscribed octagon (§ 343, I.); and by continuing the bisection, we may con- struct regular inscribed polygons of 16, 32, 64, etc., sides.
In like manner, by aid of Props. VIII., IX., and X., we may construct regular inscribed polygons of 12, 24, 48, etc., or of 20, 40, 80, etc., or of 30, 60, 120, etc., sides.
362. Sch. II. By drawing tangents to the circumference at the vertices of any one of the above inscribed regular polygons, we may construct a regular circumscribed polygon of the same number of sides. (§ 343, II.)
EXERCISES.
5. The angle at the centre of a regular polygon is the supplement of the angle of the polygon. (§ 127.)
6. The circumference of a circle is greater than the perimeter of any inscribed polygon.
7. An equiangular polygon circumscribed about a circle is regular. (§ 202.)
If r represents the radius, a the apothem, s the side, and k the area,
8. In an equilateral triangle, a = ^r, and k = fr^v'S.
9. In a square, a = ^ r\/2, and k = 2 r^.
10. In a regular hexagon, a = ^rVS, and k = fj-^VS.
11. In an equilateral triangle, r = 2 a, s = 2a v'3, and k = 3 a^VS.
12. In a square, r = a \/2, s = 2a, and k = ia"^.
13. In a regular hexagon, r = | aa/S, and k = 2 a^VS.
14. In an equilateral triangle, express r, a, and k in terms of s.
15. In a square, express r, a, and k in terms of s.
16. In a regular hexagon, express a and k in terms of s.
17. In an equilateral triangle, express r, a, and s in terms of k.
18. In a square, express r, a, and s in terms of k.
19. In a regular hexagon, express r and a in terms of k.
20. The apothem of aij equilateral triangle is one-third the altitude of the triangle,
REGULAR POLYGONS.
199
21. The sides of a regular polygon circumscribed about a circle are bisected at the points of contact. (§ 94.)
22. The radius drawn from the centre of a regular polygon to any vertex bisects the angle at that vertex. (§ 44.)
23. The diagonals of a regular pentagon are equal. (§03.)
B
24. The figure bounded by the five diagonals of a regular pentagon is a regular pentagon. h^ — ^ — X^ — ^c
(Prove, by aid of § 164, that a O can be in- scribed in FGHKL ; then use Ex. 7, p. 198.)
25. The area of a regular inscribed hexagon is a mean propor- tional between the areas of an inscribed, and of a circumscribed equilateral triangle.
(Prove, by aid of Exs. 8, 10, and 11, p. 198, that the product of the areas of the inscribed and circumscribed equilateral A is equal to the square of the area of the regular hexagon.)
26. If the diagonals AC and BE of regular pentagon ABODE intersect at F, prove BE = AE -\- AF. (Ex. 23.)
27. In the figure of Prop. IX., prove that OM is the side of a regular pentagon inscribed in a circle which is circumscribed about triangle OBM.
(ZO^M = 36°.)
28. The area of the square inscribed in a sector whose central angle is a right angle is equal to one- half the square of the radius.
(To prove area ODCE = ^ 6C^.)
29. The square inscribed in a semicircle is equivalent to two-fifths of the square inscribed in the entire circle.
(By Ex. 9, p. 198, the area of the square in- scribed in the entire O is 2 OB^ ; we then have to prove area ABCD = ^oi2 OB^ = 4 OB^.)
30. The diagonals AC, BD, CE, DF, FA, and FB, of regular hexagon ABCDEF, form a regular hexagon whose area is equal to one- third the area of ABCDEF.
(The apothem of GHKLMN is equal to the apothem of A ACE, which may be found by Ex. 8, p. 198.)
200 PLANE GEOMETRY. — BOOK V.
MEASUREMENT OF THE CIRCLE. Prop. XI. Theorem.
363. If a regular polygon he inscribed in, or circumscribed about, a circle, and the number of its sides be indefinitely increased,
I. Its perimeter approaches the circumference as a limit.
II. Its area approaches the area of the circle as a limit.
— ^
Given p and P the perimeters, and Tc and K the areas, of two regular polygons of the same number of sides respec- tively inscribed in, and circumscribed about, a O.
Let C denote the circumference, and S the area, of the ©.
I. To Prove that, if the number of sides of the polygons be indefinitely increased, P and p approach the limit C.
Proof. Let A'B' be a side of the polygon whose perimeter is P, and draw radius OF to its point of contact.
Also, draw lines OA' and OB' cutting the circumference at A and B, respectively, and chord AB.
Then, AB is a side of the polygon whose perimeter is p.
(§ 342)
Now the two polygons are similar. (§ 347)
.-. P:p=OA': OF. (§348)
... P-p:p=OA'-OF:OF. (?)
.-. (P-p) X OF=p X (OA - OF). (?)
MEASUREMENT OF THE CIRCLE. 201
But p is always < the circumference of the O. (Ax. 4)
Also, OA' - OF is < A'F. (§ 62)
'-' P-P<^y^A'F. (1)
Now, if the number of sides of each polygon be indefi- nitely increased, the polygons continuing to have the same number of sides, the length of each side will be indefinitely diminished, and A'F will approach the limit 0.
Then, by (1), since -— is a constant, P — p will approach the limit 0. ^^
But the circumference of the O is < the perimeter of the circumscribed polygon ; * and it is > the perimeter of the inscribed polygon. (Ax. 4)
Then the difference between each perimeter and the cir- cumference, OT P— C and C — p, will approach the limit 0.
Therefore, P and p will each approach the limit C.
II. To Prove that K and k approach the limit S.
Proof. Since the given polygons are similar,
K-.k^OA^iOF". (§349)
.-. K-k:k=aA^ -OF^-.OF". (?)
.'. {K-k)x OF' = kx (OA'' - OF'). (?)
.-. K-k = ^^xiOA'''-OF') = -^,x^F\ (?) OF OF-
Now, if the number of sides of each polygon be indefi- nitely increased, the polygons continuing to have the same number of sides, A'F will approach the limit 0.
Then, ^ x A!F\ being always < ^ x A'F\ will 02^ OF
approach the limit 0.
Whence, K —k will approach the limit 0. But the area of the O is evidently < K, and > k. Then, K— S and S — k will each approach the limit 0. Therefore, ^and k will each approach the limit S.
* For a rigorous proof of this statement, see Appendix, p. 380.
202 PLANE GEOMETRY. — BOOK V.
364. Cor. 1. If a regular polygon be inscribed in a circle, and the number of its sides be indefinitely increased, its apo- them approaches the radius of the circle as a limit
2. If a regular polygon be circumscribed about a circle, and the number of its sides be indefiyiitely increased, its radius approaches the radius of the circle as a limit.
Prop. XII. Theorem.
365. TJie circumferences of two circles are to each other as their radii.
Given C and C the circumferences of two (D whose radii are M and M', respectively.
To Prove | = f
Proof. Inscribe in the © regular polygons of the same number of sides; P and P' being the perimeters of the polygons inscribed in (D whose radii are R and B', re- spectively.
.-. P:P' = R:B'. (§348)
.'. PxB' = P' xE. (?)
Now let the number of sides of each inscribed polygon be indefinitely increased, the two polygons continuing to have the same number of sides.
Then, P x B' will approach the limit C x B', and P' X B will approach the limit C x B. (§ 363, 1.)
By the Theorem of Limits, these limits are equal. (?)
.-. CxB'=C'xB,ov ^ = ^. (§ 234)
MEASUREMENT OF THE CIRCLE. 203
366. Cor. I. Multiplying the terms of the ratio —^ by 2, we have
C 2R
C 2 R' Now let D and D' denote the diameters of the © whose radii are B and E'j respectively.
That is, the circumferences of two circles are to each other as their diameters.
367. Cor. n. The proportion (1) of § 366 may be written
§ = §• (§235)
That is, the ratio of the circumference of a circle to its diameter has the same value for every circle.
This constant value is denoted by the symbol tt.
.-. %=^. (1)
It is shown by methods of higher mathematics that the ratio TT is incommensurable ; hence, its numerical value can only be obtained approximately.
Its value to the nearest fourth decimal place is 3.1416.
368. Cor. m. Equation (1) of § 367 gives
G = 7rD.
That is, the circumference of a circle is equal to its diameter multiplied by tr.
We also have 0 = 2 irR.
That is, the circumference of a circle is equal to its radius multiplied by 2Tr. ^
369. Def. In circles of different radii, similar arcs, simi- lar segments, and similar sector's are those which correspond to equal central angles.
204 PLANE GEOMETRY. — BOOK V.
Prop. XIII. Theorem.
370. The area of a circle is equal to one-half the product of its circumference and radius.
Given R the radius, C the circumference, and S the area, of a O.
To Prove S = iCxR.
Proof. Circumscribe a regular polygon about the O.
Let P denote its perimeter, and K its area.
Then since the apothem of the polygon is R,
K=^PxR. (§350)
Now let the number of sides of the circumscribed polygon be indefinitely increased.
Then, K will approach the limit S,
and \PxR will approach the limit ^0 x R. (§ 363)
By the Theorem of Limits, these limits are equal. (?) .-. S = iCxR.
371. Cor. I. We have 0=2TrR. (§ 368)
.-. S = 'kRxR = ttR\
That is, the area of a circle is equal to the square of its radius multiplied by tt.
Again, S = ^tt x 4.R' = ^tt x (2 R)\
Now let D denote the diameter of the O. .-. S = \^L^.
That is, the area of a circle is equal to the square of its diameter multiplied by Jtt.
MEASUREMENT OF THE CIRCLE. 205
372. Cor. II Let S and S' denote the areas of two (D whose radii are R and B', and diameters D and D', respect- ively.
S ^ttR' ^ R^ ' ' S' ttR'^ R'^'
That is, the areas of two circles are to each other as the squares of their radii, or as the squares of their diameters.
373. Cor. III. Tlie area of a sector is equal to one-half the product of its arc and radius.
Given s and c the area and arc, respectively, of a sector of a O whose area, circumference, and radius are S, G, and R, respectively.
To Prove s = \cx R.
Proof. A sector is the same part of the O that its arc is of the circumference.
But, 1=*^- (§^^^)
.'. s = \cx R.
374. Cor. rV. Since similar sectors are like parts of the (D to which they belong (§ 369), it follows that
Similar sectors are to each other as the squares of their radii.
EXERCISES.
31. Find the circumference and area of a circle whose diameter is 5.
32. Find the radius and area of a circle whose circumference is
25 TT.
33. Find the diameter and circumference of a circle whose area is 289 TT.
34. The diameters of two circles are 64 and 88, respectively. What is the ratio of their areas?
206 PLANE GEOMETRY. — BOOK V.
Prop. XIV. Problem.
375. Given p and P, the perimeters of a regular inscribed and of a regular circumscribed polygon of the same number of sides, to find p' and P', the perimeters of a regular inscribed and of a regular circumscribed polygon having double the num- ber of sides.
A' M FN B'
O
Solution. Let AB be a side of the polygon whose perim- eter is p, and draw radius OF to middle point of arc AB.
Also, draw radii OA and OB cutting the tangent to the O at P at points A' and B', respectively ; then, A'B' is a side of the polygon whose perimeter is P. (§ 342)
Draw chords AF and BF; also, draw AM and BN tan- gents to the O at ^ and B, meeting A'B' at M and JV, re- spectively.
Then AF and MW are sides of the polygons whose perimeters are p' and P', respectively. (§ 344)
Hence, if n denotes the number of sides of the polygons whose perimeters are p and P, and therefore 2 n the number of sides of the polygons whose perimeters are p' and P', we have
AB=^, A'B' = -, AF=^, and MN=^. (1)
n n 2n 2n
Draw line 0M\ then OJf bisects Z A! OF, (§ 175)
.-. A!M'. MF= OA' : OF. (§ 249)
But OA' and OF are the radii of the polygons whose perimeters are P and p, respectively.
.-. P:p=OA':OF. (§348)
MEASUREMENT OF THE CIRCLE. 207
.-. P:p = A'M:MF. (?)
.-. P+p:p = A'M-\-MF:MF. (?)
Or P+p^A^F^^A^B'
p MF ^MN
An Clearing of fractions,
P'(P + p) = 2Pxp.
P + P Again, in isosceles A ABF and AFM,
Z ABF = Z AFM. (§ § 193, 197)
Therefore, A ABF and AFM are similar. (§ 255)
AF^MF
' ' AB AF'
(2)
(?)
.-. AF^ = ABxMF. (?)
.-. p'^=p X p.
.-. p'=v5irp. (3)
Prop. XV, Problem.
376. To compute an approximate value of it (§ 367).
Solution. If the diameter of a O is 1, the side of an inscribed square is iV2 (§ 352); hence, its perimeter is 2^2.
Again, the side of a circumscribed square is equal to the diameter of the 0 ; hence, its perimeter is 4.
We then put in equation (2), Prop. XIV., P = 4, and i) = 2 V2 = 2.82843.
208
PLANE GEOMETRY. — BOOK V.
2Pxp
3.31371.
p+p
We then put in equation (3), Prop. XIV., p = 2.82843, and P' = 3.31371.
.-. y=Vi>xP' = 3.06147. These are the perimeters of the regular circumscribed and inscribed octagons, respectively.
Repeating the operation with these values, we put in (2), P= 3.31371, and p = 3.06147. 2Pxp
P'
3.18260.
P + P We then put in (3), p = 3.06147 and P' = 3.18260.
.-. p'=^pxP' = 3.12145. These are, respectively, the perimeters of the regular cir- cumscribed and inscribed polygons of sixteen sides. In this way, we form the following table :
|
No. OF |
Perimetee of |
Perimeter of |
|
Sides. |
Reg. CiRC. Polygon. |
Eeg. Insc. Polygon. |
|
4 |
4. |
2.82843 |
|
8 |
3.31371 |
3.06147 |
|
16 |
3.18260 |
3.12145 |
|
32 |
3.15172 |
3.13655 |
|
64 |
3.14412 |
3.14033 |
|
128 |
3.14222 |
3.14128 |
|
256 |
3.14175 |
3.14151 |
|
512 |
3.14163 |
3.14167 |
The last result shows that the circumference of a O whose diameter is 1 is > 3114157, and < 3.14163.
Hence, an approximate value of tt is 3.1416, correct to the fourth decimal place.
Note. The value of tt to fourteen decimal places is 3.14159265358979.
MEASUREMENT OF THE CIRCLE. 209
EXERCISES.
35. The area of a circle is equal to four times the area of the circle described upon its radius as a diameter,
36. The area of one circle is 2| times the area of another. If the radius of the first is 15, what is the radius of the second ?
37. The radii of three circles are 3, 4, and 12, respectively. What is the radius of a circle equivalent to their sum ?
38. Find the radius of a circle whose area is one-half the area of a circle whose radius is 9.
39. If the diameter of a circle is 48, what is the length of an arc of 85° ?
40. If the radius of a circle is 3 V3, what is the area of a sector whose central angle is 152° ?
41. If the radius of a circle is 4, what is the area of a segment whose arc is 120° ? (tt = 3.1416.)
(Subtract from the area of the sector whose central A is 120°, the area of the isosceles A whose sides are radii and whose base is the chord of the segment. )
42. Find the area of the circle inscribed in a square whose area is 13.
43. Find the area of the square inscribed in a circle whose area is 196 TT.
44. If the apothem of a regular hexagon is 6, what is the area of its circumscribed circle ?
45. If the length of a quadrant is 1, what is the diameter of the circle? (7r = 3.1416.)
46. The length of the arc subtended by a side of a regular inscribed dodecagon is | tt. What is the area of the circle ?
47. The perimeter of a regular hexagon circumscribed about a circle is 12 V3. What is the circumference of the circle ?
48. The area of a regular hexagon inscribed in a circle is 24 VS. What is the area of the circle ?
49. The side of an equilateral triangle is 6. Find the areas of its inscribed and circumscribed circles.
50. The side of a square is 8. Find the circumferences of its inscribed and circumscribed circles.
51. Find the area of a segment having for its chord a side of a regular inscribed hexagon, if the radius of the circle is 10. (tt =3. 1416.)
210 PLANE GEOMETRY. — BOOK V.
52. A circular grass-plot, 100 ft. in diameter, is surrounded by a walk 4 ft. wide. Find the area of the walk.
53. Two plots of ground, one a square and the other a circle, con- tain each 70686 sq. ft. How much longer is the perimeter of the square than the circumference of the circle ? (tt = 3.1416.)
54. A wheel revolves 55 times in travelling ft. What is its diameter in inches ?
If r represents the radius, a the apothem, s the side, and k the area, prove that
55. In a regular octagon,
s = r V2- \/2, a = i r V2-I- V2, and A; = 2 r^ ^/2. (§ 375)
56. In a regular dodecagon,
s = r V2- V3, a = irV2+ V3, and k = Sr^.
57. In a regular octagon,
s = 2aiV2- l),r = aV4:-2V2, andk = 8a^(V2- 1).
58. In a regular dodecagon,
s = 2 a (2 - V3), r = 2 a V2 - Va, and A; = 12 a^ (2 - VS).
59. In a regular decagon, a = \r VlO 4- 2 Vb. (§ 359.) (Find the apothem by § 273.)
60. What is the number of degrees in an arc whose length is equal to that of the radius of the circle ? (tt = 3.1416.)
(Represent the number of degrees by x.)
61. Find the side of a square equivalent to a circle whose diameter is 3. (Tr = 3.1416.)
62. Find the radius of a circle equivalent to a square whose side is 10. (7r = 3.1416.)
63. Given one side of a regular hexagon, to construct the hexagon.
64. Given one side of a regular pentagon, to construct the pentagon. (Draw a O of any convenient radius, and construct a side of a
regular inscribed pentagon. )
65. In a given square, to inscribe a regular octagon.
(Divide the angular magnitude about the centre of the square into eight equal parts.)
66. In a given equilateral triangle to inscribe a regular hexagon.
67. In a given sector whose central angle is a right angle, to inscribe a square.
Note. For additional exercises on Book V., see p. 231.
APPENDIX TO PLANE GEOMETRY.
MAXIMA AND MINIMA OF PLANE FIGURES.
Prop. I. Theorem.
377. Of all triangles formed with two given sides, that in which these sides are perpendicular is the maximum.
Given, in A ABC and A'BC, AB = A'B, and AB ± BG. To Prove area ABC > area A' BO.
Proof. Draw A'D ± BO-, then, A'B>A'D. .'. AB>A'D. Multiplying both members of (1) by ^BO, iBOxAB>iBOxA'D. .'. area J.5(7 > area J^'^C
(§46)
(1)
(§ 312)
378. Def. Two figures are said to be isoperimetric when they have equal perimeters.
211
212
PLANE GEOMETRY. — APPENDIX.
Prop. II. Theorem.
379. Of isopenmetric triangles having the same base, that which is isosceles is the maximum.
Given ABC and A'BC isoperimetric A, having the same base BC, and A ABC isosceles. To Prove area ABC > area A'BC.
Proof. Produce BA to D, making AD = AB, and draw line CD.
Then, /. BCD is a rt. Z ; for it can be inscribed in a semi- circle, whose centre is A and radius AB. (§ 195) Draw lines AF and A'G 1. to (72); take point E on CD so that A:E = A"C, and draw line BE. Then since A ABC and A'BC are isoperimetric, AB^AC = A'B + A'C = A'B + A'E. .-. A'B + A'E =AB-\-AD= BD. But, A'B + ^'^ > BE. .'. BD>BE. .'. CD>CE.
Now AF and J.'6r are the Js from the vertices to the bases of isosceles AACD and A'CE, respectively. .-. CF=^ CD, and CG = i CE. .-. CF>CG. Multiplying both members of (1) by i BG, iBCx CF>iBCx CG. .'. Sivea, ABC >2iTea, A'BC.
(Ax. 4)
(§51)
(§94) (1)
(?)
MAXIMA AND MINIMA OF PLANE FIGURES. 213
380. Cor. Of isoperimetric triangles, that which is equi- lateral is the maximum.
For if the maximum A is not isosceles when any side is taken as the base, its area can be increased by making it isosceles. (§ 379)
Then, the maximum A is equilateral.
Prop. III. Theorem.
381. Of isoperimetric polygons having the same number of sides, that which is equilateral is the maximum.
E D
Given ABODE the maximum of polygons having the given perimeter and the given number of sides.
To Prove ABODE equilateral.
Proof. If possible, let sides AB and BO be unequal.
Let AB'C be an isosceles A with the base AO, having its perimeter equal to that of A ABO.
.-. area AB'O > area ABO. (§ 379)
Adding area AODE to both members,
area AB'ODE > area ABODE.
But this is impossible ; for, by hyp., ABODE is the maxi- mum of polygons having the given perimeter. Hence, AB and BO cannot be unequal. In like manner we have
BO = OD = DE, etc. Then, ABODE is equilateral.
214
PLANE GEOMETRY. — APPENDIX.
Prop. IV. Theorem.
382. Of isoperimetric equilateral polygons having the same number of sides, that which is equiangular is the maximum.
Given AB, BC, and CD any three consecutive sides of the maximum of isoperimetric equilateral polygons having the same number of sides.
To Prove Z ABO = Z BCD.
Proof. There may be three cases :
1. ABC + BCD = 180°. (Fig. 1.)
2. ABC + BCD > 180°. (Fig. 2.)
3. ABC + BCD <1S0°. (Fig. 3.)
H
Fig. 1. Fig. 2. Fig. 3.
If possible, let Z ABC be >Z BCD, and draw line AD. In Fig. 1.
Let E be the middle point of BC] and draw line EF, meeting AB produciBd at F, making EF = BE. Produce FE to meet CD at G. Then in A BEF and CEG, by hyp., BE = CE.
Also, Z BEF = Z CEO. (?)
MAXIMA AND MINIMA OP PLANE FIGURES. 215
And, ZEBF=ZC,
for each is the supplement of Z B. (§ 33^ 2)
.-. ABEF=ACEG. (§§86,68)
.-. BE =EF=CE= EG, and BF= CG. (§ 66) In Fig. 2. Produce AB and DC to meet at H. ' Since, by hyp., Z ABC > Z BCD, Z CBH< Z J5(7^.
.'.BH>GH. (§99)
Lay off, on BH, FH = CH; and on DH, GH=BH; and draw line FG cutting BC Sit E.
.-. A FGH = A BCH. (§ 63)
.-. ZCBH=ZFGH. (§ 66)
Then, in A BEF and CEG, Z EBF = Z CGE. Also, Z 5^i^ = Z (7J5;G^. (?)
And BF=:CG,
since BF=BH- FH, and CG = GH - CH.
.'. A BEF= A CEG. (§§ 86, 68)
.-. BE = CJ^; and jE;i^= EG. (§ 66)
/w iFYgr. 3. Produce BA and (7Z> to meet at K. Since, by hyp., Z ^^(7 > Z ^(7D, (7/r > ^JT. (?)
Lay off, on ir^ produced, FK= CK; and on CK, GK=BK; and draw line FG cutting BC at E.
.'. ABCK=AFGK. (?)
.-. ZF=ZC. (?)
Then, in A ^^i^ and O^G^, ZF==ZG. Also, Z J5^i^ = Z (7^(^. (?)
And BF=CG,
since JBi?^ = 2^7f - BK, and CG = CK - GK.
.'. ABEF^ACEG. (?)
.-. 5^; = CE and ^F = ^^ (?)
216 PLANE GEOMETRY. — APPENDIX.
Then since, in either figure, BO -\- CG = BF + FG, and A BEF= A CEG, quadrilateral AFGD is isoperimetric with, and =0= to, quadrilateral ABCD.
Calling the remainder of the given polygon P, it follows that the polygon composed of AFGD and P is isoperimetric with, and =0= to, the polygon composed of ABCD and P; that is, the given polygon.
Then the polygon composed of AFGD and P must be the maximum of polygons having the given perimeter and the given number of sides.
Hence, the polygon composed of AFGD and P is equi- lateral. (§ 381)
But this is impossible, since AF is >DG.
Hence, Z ABC cannot be > Z BCD.
In like manner, Z ABC cannot be < Z BCD.
.-. ZABC = ZBCD.
Note. The case of triangles was considered in § 380. Fig. 3 also provides for the case of triangles by supposing B and K to coincide with A. In the case of quadrilaterals, P = 0.
383. Cor. Of isoperimetric polygons having the same num- ber of sides, that which is regular is the maximum.
Prop. V. Theorem.
384. Of two isoperimetric regular polygons, that which has the greater number of sides has the greater area.
Given ABC an equilateral A, and M an isoperimetric square. To Prove area M > area ABC.
SYMMETRICAL FIGURES. 217
Proof. Let D be any point in side AB of A ABC, Draw line DC, and construct isosceles A CDE isoperi- metric with A BCD, CD being its base.
.-. area CDE > area BCD. (§ 379)
.-. area A DEC > area ABC But, since ADEC and M are isoperimetric,
area M > area ADEC (§ 381)
.-. area illf > area ^5C. In like manner, we may prove the area of a regular pen- tagon greater than that of an isoperimetric square ; etc.
385. Cor. The area of a circle is greater than the area of any polygon having an equal perimeter.
SYMMETRICAL FIGURES. DEFINITIONS.
386. Two points are said to be symmetrical with respect to a third, called the centre of symmetry, when the latter bisects the straight line which joins them.
Thus, if 0 is the middle point of straight line AB, points A and B are symmetrical with respect to 0 ^ 0 b as a centre. ' ^ '
387. Two points are said to be symmetrical with respect to a straight line, called the axis of sym- metry, when the latter bisects at right angles the straight line which joins them.
Thus, if line CD bisects line AB at G
right angles, points A and B are sym- metrical with respect to CD as an axis. h
388. Two figures are said to be symmetrical with respect to a centre, or with respect to an axis, when to every point of one there corresponds a symmetrical point in the other.
D
218
PLANE GEOMETRY.— APPENDIX.
389. Thus, if to every point of triangle ABC there corresponds a symmetrical point of triangle A'B'O', with respect to centre 0, triangle A'B'O is symmetrical to triangle ABC with respect to centre 0.
Again, if to every point of triangle ABC there corresponds a symmetrical point of triangle A'B'C, with respect to axis DE, triangle A'B'C is symmetrical to tri- angle ABC with respect to axis DE.
390. A figure is said to be symmet- rical with respect to a centre when every straight line drawn through the centre cuts the figure in two points which are symmetrical with respect to that centre.
^■^w
-E
391. A figure is said to be symmetrical with respect to an axis when it divides it into two figures which are sym- metrical with respect to that axis.
Pkop. VI. Theorem.
392. Two straight lines which are symmetrical ivith respect to a centre are equal and parallel.
.,""0 -,
Given str. lines AB and AB' symmetrical with respect to centre 0.
To Prove AB and AB' equal and ||.
Proof. Draw lines AA, BB', AB\ and AB.
SYMMETRICAL FIGURES.
219
Then, 0 bisects AA' and BB\
Therefore, AB'AB is a O.
Whence, AB and A'B' are equal and ||.
Prop. VII. Theorem.
(§ 386)
(§ 112)
(?)
393. If a figure is symmetrical with respect to two axes at right angles to each other, it is symmetrical with respect to their intersection as a centre.
Y
|
p |
r |
|||
|
A |
J^ |
/....., |
\p R y^ |
D |
|
H |
> |
Q / < |
0 |
E |
|
Q |
F |
Given figure AE symmetrical with respect to axes XX' and YY', intersecting each other at rt. A at 0.
To Prove AE symmetrical with respect to 0 as a centre.
Proof. Let P be any point in the perimeter of AE.
Draw line PQ ± XX', and line PE ± YY'.
Produce PQ and PR to meet the perimeter of AE at P and P", respectively, and draw lines QR, OP, and OP".
Then since AE is symmetrical with respect to XX',
PQ = P^Q. (§ 387)
But PQ= OR; whence, OR is equal and || to P'Q.
Therefore, OP'QR is a O. (?)
Whence, QR is equal and || to OP'. (?)
In like manner, we may prove OP"RQ a O; and there- fore QR equal and || to OP".
Then since both OP' and OP" are equal and || to QR, P'OP" is a str. line which is bisected at 0.
That is, every str. line drawn through 0 is bisected at that point, and hence AE is symmetrical with respect to 0 as a centre. (§ 390)
220
PLANE GEOMETRY. — APPENDIX.
ADDITIONAL EXERCISES. BOOK I.
1. Every point within an angle, and not in the bisector, is un- equally distant from the sides of the angle.
(Prove by Beductio ad Absurdum.)
2. If two lines are cut by a third, and the sum of the interior angles on the same side of the transversal is less than two right angles, the lines will meet if sufficiently produced.
(Prove by Beductio ad Absurdum.)
3. State and prove the converse of Prop. XXXVII., 11. (Prove Z BAD -\-ZB = 180°.)
4. The bisectors of the exterior angles of a tri- angle form a triangle whose angles are respectively the half-sums of the angles of the given triangle taken two and two. (Ex. 69, p. 67.)
(To prove ZA' = l(Z ABC + Z BCA), etc.)
5. If CD is the perpendicular from C to side AB of triangle ABC, and CE the bisector of angle C, prove ZDCJS^ equal to one- half the difference of angles A and B.
6. If E, F, G, and H are the middle points of sides AB, BC, CD, and DA, respectively, of quadrilateral ABCD, prove JE^i^'G^ // a paral- lelogram whose perimeter is equal to the sum of the diagonals of the quadrilateral. (§ 130.)
7. The lines joining the middle points of the opposite sides of a quadrilateral bisect each other. (Ex. 6, p. 220.)
8. The lines joining the middle points of the opposite sides of a quadrilateral bisect the line joining the middle points of the diagonals,
(EKGL is a ZZ7, and its diagonals bisect each other.)
9. The line joining the middle points of the diagonals of a trapezoid is parallel to the bases and equal to one-half their difference.
ADDITIONAL EXERCISES. 221
10. If D is any point in side AC oi triangle ABC, and E, F, G, and ^the middle points of AD, CD, BC, and AB, respectively, prove EFGH a, parallelogram.
11. If ^ and G are the middle points of sides AB and CD, respec- tively, of quadrilateral ABCD, and K and L the middle points of diagonals A C and BD, respectively, prove A EKL = A GKL.
12. If D and E are the middle points of sides BC and AC, respectively, of triangle ABC, and AD be produced to F and BE to G, making DF = AD and EG = BE, prove that line FG passes through C, and is bisected at that point.
13. If D is the middle point of side BC ot triangle ABC, prove AD<UAB + AC).
(Produce AD to E, making DE = AD.)
14. The sum of the medians of a triangle is less than the perimeter, and greater than the semi-perimeter of the triangle.
(Ex. 13, p. 221, and Ex. 106, p. 71.)
15. If the bisectors of the interior angle at C and the exterior angle at B of triangle ABC meet at D, prove Z BDC =\/.A.
16. If AD and BD are the bisectors of the exterior angles at the extremities of the hypotenuse of right triangle ABC, and DE and DF are drawn perpendicular, respectively, to CA and CB produced, prove CEDE a square.
(Z> is equally distant from AC and BC.)
17. AD and BE are drawn from two of the vertices of triangle ABC to the opposite sides, making Z BAD = Z ABE ; if ^2> = BE, prove the triangle isosceles.
18. If perpendiculars AE, BF, CG, and DH, be drawn from the vertices of parallelogram ABCD to any line in its plane, not inter- secting its surface, prove
AE+CG = BF-^DH.
(The sum of the bases of a trapezoid is equal to twice the line joining the middle points of the non-parallel sides.)
19. If CD is the bisector of angle C of triangle ABC, and DF be drawn parallel to AC meeting BC Bit E and the bisector of the angle exterior to g C at F, prove DE = EF.
222 PLANE GEOMETRY.— APPENDIX.
20. If E and F are the middle points of sides AB and AC, re- spectively, of triangle ABC, and AD the perpendicular from ^ to BC, prove ZEDF= ZEAF. (Ex. 83, p. 09.)
21. If the median drawn from any vertex of a triangle is greater than, equal to, or less than one-half the opposite side, the angle at that vertex is acute, right, or obtuse, respectively. (§ 98.)
22. The number of diagonals of a polygon of n sides is ^ ^'^ ~ •").
23. The sum of the medians of a triangle is greater than three- fourths the perimeter of the triangle.
(Fig. of Prop. LII. Since AO = ^ AD and -BO = f BE, we have AB< I {AD + BE), by Ax. 4.)
24. If the lower base AD of trapezoid ABCD u ^c
is double the upper base BC, and the diagonals /tN.E/'X intersect at E, prove CE = \ AC and BE = a BD. / 'v/'^N^X
(Let F be the middle point of DE, and G of L/g ^\\
AE.) ^^ ^°
25. If 0 is the point of intersection of the p^ medians AD and BE of equilateral triangle ABC, and line OF be drawn parallel to side AC, meet- > ing side BC a,t F, prove that DF is equal to ^ .BC. / (§133.) l^
(Let G be the middle point of Ovl.) d f H ^
26. If equiangular triangles be constructed on the sides of a tri- angle, the lines drawn from their outer vertices to the opposite vertices of the triangle are equal. (§ 63.)
27. If two of the medians of a triangle are equal, the triangle is isosceles.
(Fig. of Prop. LII. Let AD = BE.)
BOOK IL
28. AB and AC are the tangents to a circle from point A, and D is any point in the smaller of the arcs subtended by chord BC. If a tangent to the circle at D meets AB at E and ^C at F, prove the perimeter of triangle Ji-E'i?' constant. (§ 174.)
29. The line joining the middle points of the arcs subtended by sides AB and AC of an inscribed triangle ABC cuts AB at F and AC aXG. Trove AF= AG.
(Z AFG = Z AGF.)
ADDITIONAL EXERCISES.
223
30. If ABCD is a circumscribed quadrilateral, prove the angle between the lines joining the opposite points of contact equal to K^+C). (§202.)
31. If sides AB and EC ot inscribed hexagon ABCDEF are parallel to sides DE and EF, respectively, prove side AF parallel to side CD. (§ 172.)
(Draw line CF, and prove Z. AFC = /. FCD.)
32. If AB is the common chord of two inter- secting circles, and AC and AD diameters drawn from A^ prove that line CD passes through B. (§ 195.) c
33. If AB is a common exterior tangent to two circles which touch each other externally at C, prove Z.ACB a right angle.
(Draw the common tangent at C, meeting AB at D.)
34. If ^B and AC are the tangents to a circle from point A, and D is any point on the greater of the arcs subtended by chord BC, prove the sum of angles ABD and ACD constant.
35. If ^, C, B, and D are four points in a straight line, B being between C and Z>, and EF is a common tangent to the circles described upon AB and CD as diametei-s, prove
ZBAE = ADCF. (We have 0^11 O'F.)
36. ABCD is an inscribed quadrilateral, AD being a diameter oi the circle. If 0 is the centre, and sides AD and BC produced meet at E making CE — OA, prove
ZAOB = SZCED. (ZAOB is an ext. Z of A QBE, and Z BCO of A OCE.)
37. ABCD is a quadrilateral inscribed in a circle. If sides AB and DC produced inter- sect at E, and sides AD and BC produced at F, prove the bisectors of angles E and F perpendicular. (§ 199.)
(Prove arc HxM + arc KL = 180°.)
224 PLANE GEOMETRY. —APPENDIX.
38. If ABCD is an inscribed quadrilateral, and sides AD and BC produced meet at P, the tangent at P to the circle circumscribed about triangle ABP is parallel to CD. (§ 196.)
(Prove Z between the tangent and BP equal to Z PCD.)
39. ABCD is a quadrilateral inscribed in a circle. Another circle is described upon AD as a chord, meeting AB and CD at JS and F, respectively. Prove chords BC and ^P parallel.
(Vrove Z ABC =ZAEF.)
40. If ABCDEFGH is an inscribed octagon, the sum of angles A^ C, F, and G is equal to six right angles. (§ 193.)
41. If the number of sides of an inscribed polygon is even, the sum of the alternate angles is equal to as many right angles as the polygon has sides less two.
(Use same method of proof as in Ex. 40.)
42. If a right triangle has for its hypotenuse the side of a square, and lies without the square, the straight line drawn from the centre of the square to the vertex of the right angle bisects the right angle. (§ 200.)
43. The perpendiculars from the vertices of a triangle to the oppo- site sides are the bisectors of the angles of
the triangle formed by joining the feet of the perpendiculars.
(To prove AD, BE, and CF the bisect- ors of the A ol A DEF. By § 200, a O can be circumscribed about quadrilateral BDOF; then ZODF=ZOBF; in this way, Z ODF =dO°-Z BAC.)
Constructions.
44. Given a side, an adjacent angle, and the radius of the circum- scribed circle of a triangle, to construct the triangle.
What restriction is there on the values of the given lines ?
P
45. To describe a circle of given radius tan- ^_^^ /'''^^N.
gent to a given circle, and passing through a given Z^^, y L-'' A point without the circle. \^_^A '^ /
46. To draw between two given intersecting lines a straight line which shall be equal to one given straight line, and parallel to another.
(Draw a || to one of the intersecting lines.)
ADDITIONAL EXERCISES.
225
47. Given an angle of a triangle, the length of its bisector, and the length of the pei-pendicular from its vertex to the opposite side, to construct the triangle .
(The side opposite the given Z. is tangent to a 0 drawn with the vertex as a centre, and with the ± from the vertex to the opposite side as a radius.)
48. Given an angle of a triangle, and the segments of the oppo- site side made by the perpendicular from its vertex, to construct the triangle. (§ 226.)
B
49. To inscribe a square in a given rhombus.
(Bisect the A between diagonals AG and BB. To prove EFGH a square, prove k. OBE, OBF, ODG, and ODH equal j whence, OE=OF= 0G= OH.)
50. To draw a parallel to side BC of triangle ABC meeting AB and AC in D and E, respec- tively, so that DE may equal EC.
51. To draw a parallel to side BC of tri- angle ABC, meeting AB and AC in D and E, respectively, so that DE may equal the sum of BD and CE.
52. Given an angle of a triangle, the length of the perpendicular from the vertex of another angle to the opposite side, and the radius of the circumscribed circle, to construct the triangle.
(The centre of the circumscribed O is equally distant from the given vertices.)
53. Through a given point without a given circle to draw a secant whose internal and external segments shall be equal. (Ex. 65, p. 103.)
54. Given the base of a triangle, an adjacent angle, and the sum of the other two sides, to con- struct the triangle.
(Lay off AD equal to the sum of the other two
•)
E
226 PLANE GEOMETRY. — APPENDIX.
55. Given the base of a triangle, an adjacent acute angle, and the difference of the other two sides, to construct the triangle.
What restriction is there on the values of the given lines? A-
56. Given the feet of the perpendiculars from the vertices of a triangle to the opposite sides, to construct the triangle. (Ex. 43.)
BOOK III.
57. In any triangle, the product of any two
sides is equal to the product of the segments ^^x^ x'P^"'--
of the third side formed by the bisector of the ^>K^ \ '>
exterior angle at the opposite vertex, minus ^/''^'•A^^X /
the square of the bisector. ^-^ Jr — ^c
(To prove AB x AC = DB x DC - AIT. The work is carried out as in § 288; first prove A ABD and ACE similar.)
58. If the sides of a triangle are AB = 4, AC =5, and BC = 0, find the length of the bisector of the exterior angle at vertex A. (§ 251.)
59. ABC is an isosceles triangle. If the perpendicular to AB at A meets base BC, produced if necessary, at E, and D is the middle point of BE, prove AB a mean proportional between BC and BD. (Ex. 83, p. 69.)
(A ABC and ABD are similar.)
60. If D and E, F and G, and H and K are points on sides AB, BC, and CA, respectively, of triangle ABC, so taken that AD=DE^EB, BF=FG=GC, N^ and CH= HK= KA, prove that lines EF, GH, and KD, when produced, form a triangle equal to ABC.
(By § 248, sides of I\LMN are ||, respectively, to sides of A ^BC.)
61. The square of the common tangent to two circles which are tangent to each other externally is equal to 4 times the product of their radii. (§ 273.)
62. The sides AB and BC of triangle ABC are 3 and 7, respec- tively, and the length of the bisector of the exterior angle B is SVf. Find side AC. (Ex. 67, and § 261.)
63. One segment of a chord drawn through a point 7 units from the centre of a circle is 4 units. If the diameter of the circle is 15 units, what is the other segment ? (§ 280.)
ADDITIONAL EXEKCISES.
227
64. If E is the middle point of one of the parallel sides BC of trapezoid ABCD^ and AE and DE produced meet DC and AB pro- duced at F and 6r, respectively, prove FCr parallel to AD.
(A ADCr and BEG are similar, as also are M^ ADF and CEF.)
65. The perpendicular from the intersection of the medians of a triangle to any straight line in the plane of the triangle, not intersecting its surface, is equal to one-third the sum of the perpendiculars from the vertices of the triangle to the same line.
(The sum of the bases of a trapezoid is equal to twice the line joining the middle points of the non-parallel sides.)
66. If two parallels are cut by three or more straight lines passing through a common point, the corresponding segments are proportional.
CTo prove AE. - B^ - ^^ (10 prove ^,^, - ^,^, - ^^^
OBC, and OCD are similar, respectively ^ OA'B', OB'C, and OCA'.)
AOAB,
to
67. State and prove the converse of Ex. 66.
(Fig. of Ex. 66. To prove that AA', BB\ CC, and DD' pass through a common point. Let AA' and BB' meet at O, and draw OC and OC ; then prove 4 OBC and OB'C similar.)
68. The non-parallel sides of a trapezoid and the line joining the middle points of the parallel sides, if produced, meet in a common point. (Ex. 67.)
69. BD is the perpendicular from the vertex of the right angle to the hypotenuse of right triangle ABC. If E is any point in AB, and EF be drawn perpendicular to AC, and FG perpendicular to AB, prove lines CE and DG parallel.
(A ABC and AEF are similar. By § 271^ 2, AD:CD = A&: BC^, and AG : EG = AF" : EF^- ; AD : CD = AG -.EG.)
70. In right triangle ABC, BC' = 3 AC"". If CD be drawn from the vertex of the right angle to the middle point of AB, prove ZACD equal to 60°. (Ex. 83, p. 69.)
(Frove AC = ^ AB.)
we may prove then, we have
228
PLANE GEOMETRY. — APPENDIX.
71. If i) is the middle point of side BC oi right triangle ABCf and BE be drawn perpendicular to hypotenuse AB, prove
AE^ - BE^ = AC^. (AE = AB — BE; square this by the rule of Algebra.)
72. If BE and CF are medians drawn from the extremities of the hypotenuse of right triangle ABC^ prove
(§ 272.)
4.BE^ + 4:UF^ = bBC^.
73. If ABC and ADC are angles inscribed in a semicircle, and AE and CF be drawn per- pendicular to BD produced, prove
^^2 ^ ^2 ^ ^^2 ^ ^2 ^g 273.)
74. If perpendiculars PF, PD, and P^be drawn from any point P to sides AB^ BC, and C^, respectively, of a triangle, prove
AF^ + B& + OE^ = AE^ + :BP^ + C&.
(§ 273.)
75. If PC is the hypotenuse of right triangle ABC, prove
{AB + PC + C^)2 = 2{AB 4- PC) (PC + CA). (Square AB->rBC-\-CA by the rule of Algebra.)
76. If lines be drawn from any point P to the vertices of rectangle ABCD, prove
PA^ ■\- PC = PB' ^ PD\
77. If AB and AC are the equal sides of an isosceles triangle, and BD be drawn perpendicular to AC, prove 2 AC x CD = BC^.
{AD = AC — CD ; square this by the rule of Algebra.)
78. If AD and BE are the perpendiculars from vertices A and P, respectively, of acute-angled triangle ABC to the opposite sides, prove
AC X AE -\- BC X BD = AB\
(By § 277, 2ACxAE= AB^ + AC^ - BC^ ; and in like manner a value may be found for 2 PC x PP.)
79. The sum of the squares of the sides of a parallelogram is equal to the sum of the squares of its diagonals. (§ 279, I.)
80. To construct a triangle similar to a given triangle, having given its perimeter.
(Divide the perimeter into parts proportional to the sides of the A.)
ADDITIONAL EXERCISES.
229
81. To construct a right triangle, having given its perimeter and an acute angle.
(From any point in one side of the given Z draw a ± to the other side. )
82. To describe a circle through two given points, tangent to a given straight line. (§ 282.)
(To prove O draw with 0 as a centre and OP as a radius tangent to AB, draw BF tan- gent to the O, and prove A OBE = A OBF.)
83. If A and B are points on either side of line CD, and line AB cuts CD at F, find a point E in CD such that
AE:BE = AF: BF. (§ 249.) {EF bisects Z AEB of A ABE.)
BOOK IV.
84. In the figure on p. 174,
(a) Prove lines CF and BH perpendicular.
(If CF and BH meet at S, Z CSH is an ext. Z of A BCS.)
(6) Prove lines AG and i?^ parallel.
(c) Prove the sum of the perpendiculars from H and L to AB pro- duced equal to AB.
(If J. from H meets BA produced at Q, A AHQ = A ACD.)
(d) Prove triangles AFH, BEL, and CGK e?ich equivalent to ABC. (If AF be taken as the base of A AFH, its altitude is equal to CD.)
(e) Prove (7, H, and i in the same straight line. (Prove CH and CL in the same'str. line.)
(/) Prove the square described upon the sum of ^C and BG equivalent to the square described upon AB, plus 4 times A ABC.
(Square AC + BC by the rule of Algebra.)
(g) Prove the sum of angles AFH, AHF, BEL, and BLE equal to a right angle.
(Z AFH^- ZAHF= 180= - Z FAH.)
(h) If FN and EP are the perpendiculars from F and E, respec- tively, to HA and LB produced, prove triangles AFN and BEP each equal to ABC.
(0 Prove :^^ + FH^ + ^^^ = 6 .4^.
(^L is the hypotenuse of rt. AELP, and FH of AFHN; sides PL and ifiVmay be found by (^).)
230
PLANE GEOMETRY. — APPENDIX.
(j) Prove CF^ - C£f = AC^ - BC^.
(k) Prove that lines /IL, BH^ and CM meet at a common point. (Ex. 84, («).)
(Produce DC to T, making CT=BM, and prove AL, BH, and CM the Js from the vertices to the opposite sides of A ABT.)
(I) Prove that lines HG, LK, and MC when produced meet at a common, point.
(Draw GTand KT, and prove A CGT and CKTrt. A)
85. If BE and CF are medians drawn from .a
vertices B and C of triangle J.J5C, intersecting
at D, prove triangl lateral AEDF.
BCD equivalent to quadri- area BCF- area BDF.)
(area BCD
86. If Z> is the middle point of side BG of triangle ABC, E the middle point of AD^ F of BE^ and 6r of CF, prove A ABC equivalent toSA^i^G^.
(Draw CE ; then, area ABC = 2 area BCE.)
87. If jE" and J?' are the middle points of sides AB and CD, respec- tively, of parallelogram ABCD, and AF and CE be drawn intersecting BD in If and L, respectively, and BF and DE intersecting J.C in K and G, respectively, prove GHKL a parallelogram equivalent to \ABCD. (§140.)
(If vie and BD intersect at 3/, AMawd. DE are medians of A ABD.)
88. Any quadrilateral ABCD is equivalent to a triangle, two of whose sides are equal to diagonals AC and BD, respectively, and include an angle equal to either of the angles between A C and BD.
(Produce AC to F, making CF=AE; and BD to G, making DG = BE. To prove quadrilateral ABCD=o=AEFG. ADFG<^AABC.)
89. If through any point E in diagonal AC of parallelogram ABCD parallels to AD and AB be drawn, meeting AB and CD in i^ and H, respectively, and BC and AD in G^ and K, respectively, prove triangles EFG and EIIK equivalent.
90. If E is the intersection of diagonals AC and BD of a quadri- lateral, and triangles ABE and Ci>^ are equivalent, prove sides AD and J5C parallel.
(A ^.BZ) and ^Ci> are equivalent.)
ADDITIONAL EXERCISES.
231
91. Find the area of a trapezoid whose parallel sides are 28 and 36, and non-parallel sides 15 and 17, respectively.
(By drawing through one vertex of the upper base a II to one of the non-parallel sides, one Z of the figure may be proved a rt. Z, by Ex. 63, p. 154.)
92. If similar polygons be described upon the legs of a right triangle as homologous sides, the polygon described upon the hypotenuse is equivalent to the sum of the polygons described upon the legs.
(Find, by § 322, the ratio of the area of the polygon described upon each leg to the area of the polygon described upon the hypotenuse.)
93. If JE", F, G^ and H are the middle points of sides AB, BC, CD, and DA, respectively, of a square, prove that lines AG, BH, CE, and DF form a square equivalent to \ABCD.
(First prove A ADG = A ABH; then, by § 85, 1, Z NKL may be proved a rt. Z. By § 131, each side of KLMN may be proved equal to AK. From similar A J.HZTand ADG, AKmd^y be proved equal to
AD
94. HE is any pomt in side BC oi parallelogram ABCD, and DE be drawn meeting AB produced at F, prove triangles ABE and CEF equivalent.
(A ABE + A CDE <^ A CDF.)
95. If i> is a point in side AB of triangle ABC, find a point E m AC such that triangle ADE shall be equivalent to one-half triangle ABC. ^
(ADEF^^ACEF) D
What restriction is there on the position of D ? B
BOOK V.
96. The area of the ring included between two concentric circles is equal to the area of a circle, whose diameter is that chord of the outer circle which is tangent to the inner.
(To prove area of ring = i ttAC .)
97. An equilateral polygon circumscribed about a circle is regular if the number of its sides is odd. (§ 345.)
(The polygon can be inscribed in a O.)
232
PLANE GEOMETRY. — APPENDIX.
98. An equiangular polygon inscribed in a circle is regular if the number of its sides is odd. (§ 345.)
(The polygon can be proved equilateral.)
99. If a circle be circumscribed about a right triangle, and on each of its legs as a diameter a semicircle be described exterior to the triangle, the sum of the areas of the crescents thus formed is equal to the area of the triangle. (§ 272.)
(To prove area AECG -\- area BFCH equal to area ABC.)
100. If the radius of the circle is 1, the side, apothem, and diagonal of a regular inscribed pentagon are, respectively,
I V(10 - 2 \/5), ^ (1 + V5), and | \/(10 + 2 V5). (In Fig. of Prop. IX,, the apothem of a regular inscribed pentagon is the distance from 0 to the foot of a ± from B to OA, and its side is twice this ±. The diagonal is a leg of a rt. A whose hypotenuse is a diameter, and whose other leg is a side of a regular inscribed decagon.)
101. The square of the side of a regular inscribed pentagon, minus the square of the side of a regular inscribed decagon, is equal to the square of the radius. (Ex. 100, and § 359.)
102. The sum of the perpendiculars drawn to the sides of a regu- lar polygon from any point within the figure is equal to the apothem multiplied by the number of sides of the polygon.
(The Js are the altitudes of ^ which m.ake up the polygon.)
103. In a given equilateral triangle to in- scribe three equal circles, tangent to each other, and each tangent to one, and only one, side of the triangle.
(By § 174, the (D touch the Js at the same points.)
104. In a given circle to inscribe three equal circles, tangent to each other and to the given circle.
SOLID GEOMETRY.
Book YI.
LINES AND PLANES IN SPACE. DIEDRAL ANGLES. POLYEDRAL ANGLES.
394. Def. A plane is said to be determined by certain lines or points when one plane, and only one, can be dra\fn through these lines or points.
395.
I.
IL
IIL
IV.
Prop. I. Theorem.
A plane is determined
By a straight line and a point without the line.
By three points not in the same straight line.
By two intersecting straight lines.
By two parallel straight lines.
I. Given point C without str. line AB.
To Prove that a plane is determined by AB and C.
Proof. If any plane MN be drawn through AB, it may be revolved about AB as an axis until it contains point C.
Hence, a plane can be drawn through line AB and point C; and it is evident that but one such plane can be drawn.
233
234
SOLID GEOMETRY.— BOOK VI.
II. Given A, B, and C three points not in the same str. line.
To Prove that a plane is determined by A, B, and C.
Proof. Draw line AB\ then a plane, and only one, can be drawn through line AB and point C.
[A plane is determined by a str. line and a point without the line.]
(§ 395, I)
Then, a plane, and only one, can be drawn through A, B, and C.
III. Given AB and BC intersecting str. lines.
To Prove that a plane is determined by AB and BC.
Proof. A plane, and only one, can be drawn through line AB and point C.
[A plane is determined by a str. line and a point without the line. ]
(§ 395, I)
And since this plane contains points B and C, it must contain line BC.
[A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.] (§ 9)
Then, a plane, and only one, can be drawn through AB and BC.
IV. Given lis AB and CD.
To Prove that a plane is determined by AB and CD. Proof. The lis AB and CD lie in the same plane. [Two str. lines are said to be || when they lie in the same plane, and cannot meet however far they may be produced.] (§ 52)
LINES AND PLANES IN SPACE.
235
And only one plane can be drawn through AB and point C. [A plane is determined by a str. line and a point witliout the line.]
(§ 395, I) Then, a plane, and only one, can be drawn through AB and CD.
Prop. II. Theorem. 396. TJie intersection of two planes is a straight line.
Given line AB the intersection of planes MN and PQ. To Prove AB a str. line.
Proof. Draw a str. line between points A and B. This str. line lies in plane MJSf, and also in plane FQ. [A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.] (§ 9)
Then it must be the intersection of planes MN and FQ. Hence, the line of intersection AB is a str. line.
397. Defs. If a straight line meets a plane, the point of intersection is called the foot of the line.
A straight line is said to be perpendicular to a plane when it is perpendicular to every straight line drawn in the plane through its foot.
A straight line is said to be parallel to a plane when it cannot meet the plane however far they may be produced.
A straight line which is neither perpendicular nor parallel to a plane, is said to be oblique to it.
Two planes are said to be parallel to each other when they cannot meet however far they may be produced.
236
SOLID GEOMETRY.— BOOK VI.
398. Sch. The following is given for convenience of reference :
A perpendicular to a plane is perpendicular to every straight line drawn in the plane through its foot.
Prop. III. Theorem.
399. At a gircen point in a plane, one perpendicular to the plane cayi he drawn, and hut one.
Given point P in plane MN.
To Prove that a ± can be drawn to MN at P, and but one.
Proof. At any point A of indefinite str. line AB, draw lines AC and AD ± to AB.
Let RS be the plane determined hj AC and AD.
Let AE be any other str. line drawn through point A in plane RS\ and draw line CD intersecting AC, AE, and AD at C, E, and D, respectively.
Produce BA to B', making AB' = AB, and draw lines BC, BE, BD, B'C, B'E, and B'D.
In A BCD and B'CD,
CD = CD.
And since AC and AD are ± to BB' at its middle point,
BC = B'CsindBD = B'D.
[If a J- be erected at the middle point of a str. line, any point in the ± is equally distant from the extremities of the line.] (§ 41, I)
LINES AND PLANES IN SPACE. 237
.-. ABCD = AB'CD.
[Two k. are equal when the three sides of one are equal respectively to the three sides of the other.] (§ 69)
Now revolve A BCD about CD as an axis until it coin- cides with A 5'(7Z>.
Then, point B will fall at point B', and line BE will coin- cide with line B'E ; that is, BE = B'E.
Hence, since points A and E are each equally distant from B and B', line AE is ± BB'.
[Two points, each equally distant from the extremities of a str. line, determine a _L at its middle point. ] (§ 43)
But AE is any str. line drawn through A in plane US.
Then, AB is ± to every str. line drawn through A in plane BS.
Whence, ABh ± to plane RS.
[A str. line is said to be ± to a plane when it is ± to every str. line drawn in the plane through its foot.] (§ 397)
Now apply plane BS to plane MN so that point A shall fall at point P; and let AB take the position PQ.
Then, PQ will he ±MN.
Hence, a J_ can be drawn to MN at P,
If possible, let PT be another _L to plane MN at P; and let the plane determined by PQ and PT intersect JOT in line HK.
Then, both PQ and PT are ± HK.
[A J_ to a plane is _L to every str. line drawn in the plane through its foot.] (§398)
But this is impossible ; for, in plane HKT, only one _L can be drawn to HK at P.
[At a given point in a str. line, but one _L to the line can be drawn.]
(§25) Then only one ± can be drawn to MN at P.
400. Cor. I. A straight line perpeyidicular to each of two straight lines at their point of intersection is perpendicular to their plane.
238 SOLID GEOMETRY. — BOOK VI.
401. Cor. I. Since E is any point in plane BS, it fol- lows that
If a plane is perpejidicular to a straight line at its middle point, any point in the plane is equally distant from the ex- tremities of the liyie.
Prop. IV. Theorem.
402. All the perpeiidiculars to a straight line at a given point lie in a plane perpendicular to the line.
Given AC, AD, and AE any three Js to line AB at A.
To Prove that they lie in a plane ± to AB.
Proof. Let MN be the plane determined by ^C and AD. Then, plane MNis ±AB.
[A str. line ± to each of two str. lines at their point of intersection is ± to their plane.] (§ 400)
Let the plane determined by AB and AE intersect MN in line AE' ; then, AB _L AE'.
[A ± to a plane is ± to eveiy str. line drawn in the plane through its foot.] (§398)
But in plane ABE, only one _L can be drawn to AB at A.
[At a given point in a str. line, but one ± to the line can be drawn.]
(§25) Then, AE' coincides with AE, and AE lies in plane MN. . But AC, AD, and AE are any three Js to AB at A. Therefore, all the Js to AB at A lie in a plane J_ AB.
403. Cor. I. Tlirough a given point in a straight line, a plane can he drawn perpendicular to the line, and hut one.
LINES AND PLANES IN SPACE.
239
404. Cor. II Through a given point without a straight line, a plane can he drawn perpendicular to the line, and hut one.
Given point C without line AB.
To Prove that a plane can be drawn through G ± AB, and but one.
Proof. Draw line CB _L AB, and let BD be any other J_ to AB at B.
Then, the plane determined by BO and BD will be a plane drawn through G _L AB.
[A str. line _L to each of two str. lines at their point of intersection is ± to their plane.] (§ 400)
Again, every plane through G 1. AB must intersect the plane determined by AB and J5(7 in a line from G J_ AB.
[A ± to a plane is J. to every str. line drawn in the plane through its foot.] (§ 398)
But only one ± can be drawn from G to AB.
[From a given point without a str. line, hut one ± can be drawn to the line. ] (§ 45)
Then, every plane through G ± AB must contain BG, and be _L to AB at B.
But only one plane can be drawn through B A. AB.
[Through a given point in a str. line, but one plane can be drawn _L to the line.] (§403)
Hence, but one plane can be drawn through G X AB.
405. Cor. III. (Converse of § 401.) Any point equally distant from the extremities of a straight line lies in a plane perpendicidar to the line at its mid- dle point.
Given plane MN ± to line ^B at its middle point G, and point D equally distant from A and B.
To Prove that D lies in MN.
(By § 43, GD±AB; then use § 402.)
240
SOLID GEOMETRY. — BOOK VI.
Note. It follows from §§401 and 405 that
The locus (§ 141) of points in space equally distant from the ex- tremities of a straight line is a plane perpendicular to the line at its middle point.
Prop. V. Theorem.
406. If from a point in a perpendicular to a plane, oblique lines be drawn to the plane,
I. Two oblique lines cutting off equal distances from the foot of the perpendicular are equal.
II. Of two oblique lines cutting off unequal distances from the foot of the perpendicular, the more remote is the greater.
I. Given line AB _L to plane MN at B, and AC and AD
oblique lines meeting MN at equal distances from B.
To Prove AC = AD.
Proof. Draw lines BC and BD.
In A ABC and ABD, AB = AB.
Also, ZABC = ZABD.
[A ± to a plane is ± to every str. line drawn in the plane through its foot.] (§ 398)
And by hyp., BC = BD.
.'. A ABC = A ABD.
[Two A are equal when two sides and the included Z of one are equal respectively to two sides and the included Z of the other.] (§ 63)
.-. AC = AD.
[In equal figures, the homologous parts are equal.] (§ 66)
LINES AND PLANES IN SPACE. 241
. II. Given line AB ± to plane J/iVat B, and AC and AE oblique lines from A to MNj AE meeting MM at a greater distance from B than AO.
To Prove AE > AC.
Proof. Draw lines BC and BE.
On i?^ take BF= BC, and draw line AF.
Since ^F and AC meet JIO^ at equal distances from B, AF = AC.
[If from a point in a ± to a plane, oblique lines be drawn to the plane, two oblique lines cutting off equal distances from the foot of the ± are equal.] (§406,1)
But, AB ± BE.
[A J_ to a plane is ± to every str. line drawn in the plane through its foot.] (§398)
.-. AE > AF.
[If oblique lines be drawn from a point to a str. line, of two oblique lines cutting off unequal distances from the foot of the ± from the point to the line, the more remote is the greater.] (§ 49, II)
.-. AE > AC.
Prop. VI. Theorem.
407. (Converse of Prop. V.) If from a point in a perpen- dicular to a plane, oblique lines he drawn to the plane,
I. Two equal oblique lines cut off equal distances from the foot of the perpendicular.
II. Of two unequal oblique lines, the greater cuts off the greater distance from the foot of the perpendicular.
I. Given line AB ± to plane MN at B, AC and AD equal oblique lines from A to MM, and lines BC and BD. (Fig. of Prop. V.)
To Prove BC = BD.
(Prove A ABC and ABD equal.)
II. Given line AB ± to plane MN at B, and AC and AE oblique lines from A to MN, AE being > AC', also, lines BC and BE.
242
SOLID GEOMETRY. — BOOK VI.
To Prove BE > BC.
(The proof is left to the pupil.)
Prop. VII. Theorem.
408. If through the foot of a perpendicular to a plane a line be drawn at right angles to ayiy line in the plane, the line drawn from its intersection with this line to any point in the perpendicular will he perpendicular to the line in the plane.
Given line AB ± to plane MN at A, line AE _L to any line CD in MN, and line BE from E to any point B in AB.
To Prove BE ± CD.
Proof. On CD take EC = ED.
Draw lines AC, AD, BC, and BD. .: AC=AD.
[If a ± be erected at the middle point of a str. line, any point in the _L is equally distant from the extremities of the line.] (§ 41, I)
.-. BC=BD.
[If from a point in a ± to a plane, oblique lines be drawn to the plane, two oblique lines cutting off equal distances from the foot of the ± are equal.] (§ 406, I)
Then since each of the points B and E is equally distant from G and D,
BE ± CD.
[Two points, each equally distant from the extremities of a str. line, determine a ± at its middle point.] (§ 43)
409. Cor. I. From a given point without a plane, one per- pendicular to the plane can he drawn, and hut one.
LINES AND PLANES IN SPACE.
243
Given point A without plane MIT.
To Prove that a ± can be drawn from A to MN, and but one.
Proof. Let DE be any line in plane MN; draw line AF _L DE, line BF in plane MN ± DE, line AB ± BF, and line BE.
Now EF is J_ to the plane determined by AF and BF.
[A str. line ± to each of two str. lines at their point of intersection is ± to their plane.] . (§ 400)
Then since BF is drawn through the foot of EF, ± to line AB in plane ABE, we have BE _L AB.
[If through the foot of a ± to a plane a line be drawn at rt. A to any line in the plane, the line drawn from its intersection with this line to any point in the _L will be ± to the line in the plane.] (§ 408)
Then AB, being ± to BE and BF, is _L to MN.
[A str. line ± to each of two str. lines at their point of intersection is ± to their plane.] (§ 400)
If possible, let AC be another ± from A to MN; then A ABC will have two rt. A.
[A ± to a plane is ± to every str. line drawn in the plane through its foot.] (§398)
But this is impossible.
Hence, but one _L can be drawn from A to MN.
410. Cor. II. TJie perpendicular is the shortest line that can he clraivn from a point to a plane.
Given AB the J_ from point A to plane MN, and AC any other str. line from A to MJSf. (Fig. of § 409.) To Prove AB < AC.
Proof. Draw line BC ; then, AB ± BC
[A ± to a plane is ± to every str. line drawn in the plane through its foot.] (§ 398)
.-. AB<AC.
[The ± is the shortest line that can be drawn from a point to a str. line.] (§ 46)
244
SOLID GEOMETRY.— BOOK VI.
Note. The distance of a point from a plane signifies the length of the perpendicular from the point to the plane.
Prop. VIII. Theorem.
411. If two straight lines are parallel, a plane drawn through one of them, not coinciding with the plane of the is parallel to the other.
A] \B
\m i
Given line AB II to line CD, and plane MN drawn through CD, not coinciding with the plane of the lis.
To Prove AB II MN.
Proof. The lis AB and CD lie in a plane which intersects JfiV^inline CD.
Hence, if AB meets MN, it must be at some point of CD.
But since AB is II CD, it cannot meet CD (§ 52).
Then AB and MN cannot meet, and are I! (§ 397).
Prop. IX. Theorem.
412. If a straight line is parallel to a plane, the intersec- tion of the plane with any plane drawn through the lirie is parallel to the line.
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Given line AB II to plane MN; and line CD the intersec- tion of MN with any plane AD drawn through AB. To Prove AB li CD.
{AB and CD lie in the same plane, and cannot meet.)
LINES AND PLANES IN SPACE.
245
413. Cor. If a line and a plane are parallel, a parallel to the line through any point of the plane lies in the plane.
Given line AB II to plane MN; and line CI) through any point C of MN II to AB. (Fig. of Prop. IX.)
To Prove that CD lies in MJS^.
Proof. The plane determined by line AB and point G intersects MN in a line II to AB.
[If a str. line is || to a plane, the intersection of the plane with any- plane drawn through the line is || to the line.] (§ 412)
But through C, only one II can be drawn to AB. [But one str. line can be drawn through a given point |1 to a given str. line.] (§ 53)
Whence, CD lies in MN.
Prop. X. Theorem.
414. If two parallel planes are cut by a third plane, the intersections are par'ollel.
Given II planes MN and PQ cut by plane AD in lines AB and CD, respectively. To Prove AB II CD.
{AB and CD lie in the same plane, and cannot meet.)
415. Cor. Parallel lines included between parallel planes are eqtial.
Given AC and BD II lines included between II planes MN and PQ. (Fig. of Prop. X.)
(Prove AC =BD by %^ 414 and 107.)
246
SOLID GEOMETRY. —BOOK VI.
Prop. XI. Theorem.
416. Through any given straight line, a plane can he drawn parallel to any other straight line.
A B
Given lines AB and CD.
To Prove that a plane can be drawn through CD II AB.
(Draw line GE II AB; then use § 411.)
Note. If AB is || (7Z>, an indefinitely great number of planes can be drawn through CD \\ AB (§411) ; otherwise, bilt one such plane can be drawn, for every plane drawn through CD \\ AB must contain CE (§ 413), and but one plane can be drawn through CD and CE.
Prop. XII. Theorem.
417. TJirough a given p)oint a plane can he drawn parallel to any two straight lines in space.
B C
Given poiut A and lines BC and DE. To Prove that a plane can be drawn through A II to BC and DE.
(The proof is left to the pupil ; see § 411.)
Note. If BC and DE are ||, an indefinitely great number of planes can be drawn through A \\ to BC and DE (§ 411) ; otherwise, but one such plane can be drawn.
LINES AND PLANES IN SPACE.
247
Prop. XIII. Theorem. 418. Two perpendiculars to the same plane are parallel. AK G
' Given lines AB and CD ± to plane MN at B and D, respec- tively.
To Prove AB II CD.
Proof. Let A be any point of AB, and draw line AD.
Also, draw line BD, and line DF in plane MN _L BD.
.'. CD±DF. [A ± to a plane is JL to every str. line drawn in the plane through its foot.] (§ 398)
Also, AD ± DF.
[If through the foot of a ± to a plane a line be drawn at rt. A to any line in the plane, the line drawn from its intersection with this line to any point in the ± will be ± to the line in the plane.] (§ 408)
Then, CD, AD, and BD, being ± to DF at D, lie in the same plane.
[All the Js to a str. line at a given point lie in a plane ± to the line.] (§ 402)
Then, since points A and B lie in the plane of the lines AD, BD, and CD, AB lies in this plane.
[A plane is a surface such that the str. line joining any two of its points lies entirely in the surface.] (§9)
That is, AB and CD lie in the same plane. Again, AB and CD are ± BD.
[A ± to a plane is ± to every str. line drawn in the plane through its foot.]
.-. AB II CD.
[Two Js to the same str. line are ||.]
(§54)
248
SOLID GEOMETRY. — BOOK VI.
419. Cor. I. If one of two parallel lines is perpendicular to a jjla7ie, the other is also perpendicular to the pjlane.
Given lines AB and CD II, and a c
AB A. to plane MN.
To Prove CD _L MK
Proof. A _L from C to MN will / b be II AB. j^
[Two Js to the same plane are jj.] (§418)
But through C, only one II can be drawn to AB.
[But one str. line can be drawn through a given point 11 to a given str. line.] (§ 53)
.-. CD1.MN.
420. Cor. II. If each of tiuo straight lines is parallel to a third straight line, they are parallel to each other.
Given lines AB and CD II line EF. To Prove AB II CD. (Draw plane MN 1. EF, and prove AB II CD by §§ 418 and 419.)
Prop. XIV. Theorem. 421. Two planes perpendicular to the same straight line
are parallel.
My
N
Given planes MN and PQ ± to line AB. To Prove MN II PQ.
(Prove as in § 54 ; by § 404, but one plane can be drawn through a given point ± to a given str. line.)
LINES AND PLANES IN SPACE. 249
Prop. XV. Theorem.
422. If each of tivo hitersecting lines is parallel to a plane, their plane is parallel to the given plane.
Given lines AB and AC, in plane MN, II to plane PQ.
To Prove . MN II PQ.
Proof. Draw line AD J_ PQ, and lines DE and DF II to AB and AC, respectively; then DE and DF lie in plane PQ.
[If a line and a plane are II, a || to the line through any point of the plane lies in the plane.] (§413)
Whence, AD is _L to DE and DF.
[A ± to a plane is ± to every str. line drawn in the plane through its foot.] (§398)
Therefore, AD is ± to AB and AC.
[A str. line ± to one of two ||s is ± to the other.] (§ 56)
.-. ADA.MN. [A str. line ± to each of two str. lines at their point of intersection is ± to their plane.] (§ 400)
.-. MNWPQ.
[Two planes ± to the same str, line are ||.] (§ 421)
EXERCISES.
1. What is the locus (§ 141) of the perpendiculars to a given straight line at a given point ?
2. What is the locus of points in space equally distant from the circumference of a given circle ?
3. A line parallel to a plane is everywhere equally distant from it. (Fig. of Prop. IX. Draw lines AC and BD±MN. To prove
AC=BD.)
250 SOLID GEOMETRY. — BOOK VI.
Prop. XVI. Theorem.
423. A straight line perpendicular to one of two parallel planes is perpendicular to the other also.
Given MN and PQ II planes, and line AD ± PQ.
To Prove AD 1. MN.
Proof. Pass two planes through AD, intersecting MN in lines AB and AC, and PQ in lines DE and DF, respectively.
Then, AB II DE, and AC 11 DF.
[If two II planes are cut by a third plane, the intersections are ||.]
(§414)
But AD is ± to DE and DF.
[A ± to a plane is _L to every str. line drawn in the plane through its foot.] (§ 398)
Whence, AD is _L to AB and AC.
[A str. line ± to one of two ||s is ± to the other.] (§ 56)
.'. AD1.MN.
[A str. line ± to each of two str. lines at their point of intersection Is J_ to their plane.] (§ 400)
424. Cor. I. Two parallel planes are everywhere equally distant. (Note, p. 244.)
Given MN and PQ II planes. (Fig. of Prop. XVI.)
To Prove MN and PQ everywhere equally distant.
Proof. All lines which are ± to both planes are II.
[Two Js to the same plane are ||.] (§ 418)
Therefore, these lines are all equal.
[II lines included between II planes are equal.] (§ 415)
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LINES AND PLANES IN SPACE. 251
425. Cor. II. Through a given point a plane can be drawn parallel to a given plane, and but one.
Given point A and plane PQ.
To Prove that a plane can be drawn through A II FQ, and but one.
Proof. Draw line AB ± PQ.
Through A pass plane MNX AB.
Then J»/iVr will be II PQ.
[Two planes ± to the same str. line are ||.] (§421)
If another plane could be drawn through A II PQ, it would be ±AB.
[A str. line J_ to one of two || planes is ± to the other also.] (§ 423)
It would then coincide with MN.
[Through a given point in a str. line, but one plane can he drawn ± to the line.] (§403)
Then but one plane can be drawn through A II PQ.
EXERCISES.
4. "What is the locus of points in space equally distant from the vertices of a given triangle ?
5. "What is the locus of points in space equally distant from a given plane ?
6. "What is the locus of points in space equally distant from two parallel planes ?
7. A line parallel to each of two intersecting planes is parallel to their intersection.
(Pass a plane through AB II PB ; then use § 412.)
8. If two planes are parallel to a third plane, they are parallel to each other. (§§423,421.)
9. Line AB is perpendicular to plane MN at A. A line is drawn from A meeting any line CD of plane MN at E. If line BE is per- pendicular to CD, prove AE perpendicular to CD.
(Fig. of Prop. VIL)
252 SOLID GEOMETRY.— BOOK VI.
Prop. XVII. Theorem.
426. If two angles not in the same plane have their sides parallel and exteiiding in the same direction, they are equal, and their planes are parallel.
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Given A BAC and B'A'C in planes MN and PQ, respec- tively, with AB and AC II respectively to A'B' and A'C, and extending in the same direction.
To Prove Z BAC = Z B'A'C, and MW II PQ.
Proof. Lay off AB = A'B' and AC = A'C, and draw lines AA', BB', CC, BC, and B'C.
Then since AB is equal and II to A'B', ABB' A' is a O.
[If two sides of a quadrilateral are equal and ||, the figure is a O.J
(§ 110) Whence, A A' is equal and II to BB'.
[The opposite sides of a /Z7 are equal.] (§ 106, I)
Similarly, ACCA' is a O, and AA' is equal and II to CC.
Then, BB' is equal and 11 to CC.
[If each of two str. lines is II to a third str. line, they are || to each other.] (§ 420)
Whence, BB'CC is a O, and BC=B'C. .'. AABC=AA'B'C.
[Two A are equal when the three sides of one are equal respectively to the three sides of the other.] (§ 69)
.-. Z BAC =Z B'A'C.
[In equal figures, the homologous parts are equal. ] (§ 66)
Again, lines AB and AC are II to plane PQ.
[If two str. lines are jj, a plane drawn through one of them, not coinciding with the plane of the jjs, is || to the other.] (§ 411)
LINES AND PLANES IN SPACE.
253
.-. MN II pq.
[If each of two intersecting lines is || to a plane, their plane is || to the given plane.] (§422)
Prop. XVIII. Theorem.
427. If tico straight lines are cat hy three parallel planes, the corresponding segments are j^roportional
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Given II planes MN, PQ, and RS intersecting lines AC and AC in points A, B, 0, and A', B', C, respectively.
To Prove AB^A^^
BC B'C Proof. Draw line AC ; and through AC and AC pass a plane intersecting PQ and RS in lines BD and CC, respec- tively.
.-. BD II CC. [If two 11 planes are cut by a third plane, the intersections are ||.]
(§ 414)
" BC~ DC' ^ ^
[A II to one side of a A divides the other two sides proportionally.]
(§244)
In like manner, AD^A^ ^2)
DC B'C ^ ^
From (1) and (2), || = ||'.
[Things which are equal to the same thing, are equal to each other. ]
(Ax. 1)
254
SOLID GEOMETRY.— BOOK VI.
DIEDRAL ANGLES. DEFINITIONS.
428. A diedral angle is the amouiit of divergence of two planes which meet in a straight line.
The line of intersection of the planes is called the edge of the diedral angle, and the planes are called its faces.
Thus, in the diedral angle between planes BD and BF, BE is the edge, and BD and BF the faces.
A diedral angle may be designated by two letters on its edge ; or, if several diedral angles have a common edge, by four letters, one in each face and two on the edge, the letters on the edge being named between the other two.
Thus, we may read the above diedral angle BE, or ABEC.
Two diedral angles are said to be adjacent when they have the same edge, and a common face between them; as, ABEC and CBED.
Two diedral angles are said to be vertical when the faces of one are the extensions of the faces of the other.
429. A plane angle of a diedral angle is the angle be- tween two straight lines drawn one in each face, perpendicular to the edge at the same point.
Thus, if lines AB and J.C be drawn in faces DE and DF, respectively, of diedral angle DG, perpendicular to DG at A, Z BAG is a plane angle of the diedral angle.
430. Let BAG and BA^G (Fig. of § 429) be plane A of diedral Z DG] then, AB II A'B^ and AG II A'G. (§ 54)
.-. Z BAG = Z B'A'G. (§ 426)
That is, all plane angles of a diedral angle are equal.
DIEDRAL ANGLES.
255
431. A plane perpendicular to the edge of a diedral angle intersects the faces in lines perpendicular to the edge (§ 398) ; hence, a plane perpendicular to the edge of a diedral angle intersects the fa.ces in lines which include the plane angle of the diedral angle (§ 429).
432. Two diedral angles are equal when their faces may be made to coincide.
Prop. XIX. Theorem.
433. Two diedral angles are equal if their plane angles are equal.
c
E'
Given ABC and A'B'C plane A of diedral ABB and B'D', respectively, and Z.ABC = ZA'B'C'.
To Prove diedral ZBD = diedral ZB'D'.
Proof. Apply diedral Z B'D' to BD in such a way that A'B' shall coincide with AB, and B'C with BC.
Now BD and B'D' are ± to the planes of A ABC and A'B'C, respectively. (§ 400)
Whence, B'D' will coincide with BD. (§ 399)
Then, A'D' will coincide with AD, and CD' with CD.
(§ 395, III)
Hence, B'D' and BD are equal. (§ 432)
434. Cor. I. (Converse of Prop. XIX.) If two diedral angles are equal, their plane angles are equal. (Fig. of Prop. XIX.)
(Apply ^'Z)' to BD so that face A'D' shall coincide with AD, and CD' with CD, point B' falling at B.)
256
SOLID GEOMETRY. —BOOK VI.
435. Cor. II. If two planes intersect, the vertical diedral angles are equal.
Eor their plane angles are equal. (§ 40)
436. Defs. If a plane meets another plane in such a way as to make the adjacent diedral angles equal, each is called a right diedral angle, and the planes are said to be perpendicu- lar to each other.
Thus, if plane PQ be drawn meeting plane MN in such a way as to make diedral A PRQM and PRQN equal, each of these is a right diedral Z, and MN and PQ are ± to each other.
Prop. XX. Theorem.
437. Through a given line m a plane, a plane can he drawn perpendicular to the given plane, and hut one.
(Prove as in § 25.)
Prop. XXI. Theorem.
438. If two planes are perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other.
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DIEDRAL ANGLES. 257
Given planes PQ and MN ±, intersecting in line QR, and line AB in plane PQ ± QR.
To Prove AB _L MN.
Proof. Draw line C'BC in plane MN 1. QR.
Then, ABG and ^5C' are plane A of diedral zi PRQN and PRQM, respectively. (§ 429)
Now, if two planes are JL to each other, the adj. diedral A are equal (§ 436).
That is, diedral A PRQN= diedral A PRQM.
.-. A ABC = A ABC. (§434)
Whence, A ABC is a rt. A. (§ 24)
Then ^^, being ± to 56' and J5Q at B, is ± Jf^. (§ 400)
439. Cor. I. If two planes are perpendicular to each other, a perpendicular to one of them at any point of their intersec- tion lies in the other.
Given planes PQ and MN ±, intersecting in line QR, and line AB drawn from any point B oi QR ± MN. (Fig. of Prop. XXI.)
To Prove that AB lies in PQ.
Proof. If a line be drawn in PQ from point B ± QR, it will be ± MN. (§ 438)
But from point B but one ± can be drawn to MN. (§ 399) Therefore, AB lies in PQ.
440. Cor. II. If two planes are perpendicidar to each other, a perpendicular to one of them from any point of the other lies in the other.
Given planes PQ and MN ±, intersecting in line QR, and line AB drawn from any point A of PQ J_ MN. (Fig. of Prop. XXI.)
To Prove that AB lies in PQ. (The proof is left to the pupil.)
258
SOLID GEOMETRY. — BOOK VI.
Prop. XXII. Theorem.
441. If a straight line is perpendicular to a pla^ie, every plane draivn through the line is perpendicular to the plane.
Given line AB ± plane MN, and PQ any plane drawn through AB.
To Prove PQ ± MN.
Proof. Let line QR be the intersection of PQ and MN, and draw line C'BC in plane MN ± QR.
We have AB ± BQ. (§ 398)
Then, A ABC and ABO are plane A of diedral A PRQN
and PRQM, respectively. (§ 429)
But A ABC and ABC are rt. A. (§ 398)
.-. Z ABC = A ABC. (§ 26)
.-. diedral A PRQN=diedr2i\ A PRQM. (§ 433)
.-. PQ±MN (§436)
Prop. XXIII. Theorem.
442. A plane perpendicular to each of two intersecting planes is perpendicular to their intersection.
DIEDRAL ANGLES.
259
Given planes PQ and RS _L to plane MJST, and intersect- ing in line AB.
To Prove AB ± MN.
(By § 439, a ± to JfiV^at B lies in both PQ and RS.)
Prop. XXIV. Theorem.
443. Every point in the bisecting plane of a diedral angle is equally distant from its faces.
Given P any point in bisecting plane BE of diedral Z ABDC, and lines PM and PJSf ± to AD and OZ), .respec- tively.
To Prove PM=PN.
Proof. Let the plane determined by PM and PN inter- sect planes AD, BE, and CD in lines FM, FP, and FN, respectively.
Plane PMFNis _L to planes AD and CD. (§ 441)
Then, plane PMFN is ± BD. (§ 442)
Whence, Pi^Jf and PFN are plane Zs of diedral A ABDE
and CBDE, respectively. (§ 431)
.-. ZPFM=^ZPFN. (§434)
In A PP3/ and PEN, PF = PF.
And, APFM=Z.PFN.
Also, zl PJltfP and PNE are rt. A (§ 398)
.-. APFM=APFK (§ 70)
.-. PM=PN. (?)
260 SOLID GEOMETRY.— BOOK VI.
444. Cor. I. (Converse of Prop. XXIV.) A7iy point ivhich is ivithin a diedi'ol angle, and equally distant from its faces, lies in the bisecting plane of the diedral angle.
Given point P within diedral Z.ABDC, equally distant from AD and CD, and plane BE determined by BD and P. (Fig. of Prop. XXIV.)
To Prove that BE bisects diedral Z ABDC.
(Prove A PFM and PEN equal ; then Z PFM = Z PEN, and the theorem follows by § 433.)
445. Cor. II. It follows from §§ 443 and 444 that
The locus of points in space equally distant from the faces of a diedral angle is the plane bisecting the diedral angle.
Pkop. XXV. Theorem.
446. Through a given straight li?ie ivithout a plane, a plane can be drawn perpendicular to the given plane, and but one.
Given line AB without plane MN.
To Prove that a x^lane can be drawn through AB ± MN, and but one.
Proof. Draw line ACA-MN, and let AD be the plane determined by AB arid AC; then, AD1.MN. (§ 441)
If more than one plane could be drawn through AB _L MN, their common intersection, AB, would be ± MN. (§ 442)
Hence, but one plane can be drawn through ABA.MN, unless AB is ± MN.
Note. If Hne AB is ± MN., an indefinitely great number of planes can be drawn through AB 1.MN (§ 441).
DIEDRAL ANGLES.
261
447. Def s. The projection of a point on a plane is the foot of the perpendicular drawn from the point to the plane.
The projection of a line on a plane is the line which con- tains the projections of all its points.
448. Cor. The iwojection of a straight line on a plane is a straight line.
Given line CD the projection (§ 447) of str. line AB on plane MN. (Fig. of Prop. XXV.)
To Prove CD a str. line.
Proof. Draw a plane through ABA-MK
The Js to MN from all points of AB will lie in this plane. (§ 440)
Therefore, CD is a str. line. (§ 396)
Prop. XXVI. Theorem.
449. The angle between a straight hne and its projection on a plane is the least angle which it makes with any line drawn in the plane through its foot.
N
Given line BC the projection of line AB on plane MN, and BD any other line drawn through B in MN. To Prove Z ABC < Z ABD.
Proof. Lay off BD = BC, and draw lines AC and AD. In A ABC and ABD, AB = AB. And by hyp., BC = BD.
Also, AC<AD. (§410)
.-. ZABC<ZABD. (§92)
Note. Z ABC is called the angle between line AB and plane MN.
262
SOLID GEOMETRY. — BOOK VI.
Prop. XXVII. Theorem.
450. Two straight lines, not in the same plane, have one coynmon j^^^'P^ndicular, and but one; and this line is the shortest line that can be drawn between them.
Given lines AB and CD, not in the same plane.
To Prove that one common ± to AB and CD can be drawn, and but one ; and that this line is the shortest line that can be drawn between AB and CD.
Proof. Through CD draw plane MN II AB. (§ 416)
Through AB draw plane AH 1. MN, and produce their
intersection to meet CD at G.
Draw line AG in plane AH 1. GH
(§ 446) then, AG ± MN.
(§ 438)
(§ 398)
(§ 412)
(§56)
.-. AGXCD. Also, GH II AB.
.'. AG±AB. Then, AG is a common J_ to AB and CD. If possible, let EKhe another common ± to AB and CD,
and draw line EF II AB, and line KL in plane AH J_ GH. Then, EF lies in plane MN. (§ 413)
Also, EK is _L to ED and EF. (§ 56)
Whence, EK is _L MN. (?)
But KL is also.X JWiV. (§ 438)
We should then have two Js from K to MN, which is
impossible. (§ 409)
Hence, but one common ± can be drawn to AB and CD. Again, EK > KL. (§ 410)
DIEDRAL ANGLES.
263
.'.EK>AG. (§80)
Hence, AG \^ the shortest line between AB and CD.
Prop. XXVIII. Theorem.
451. Two diedral angles are to each other as their plane angles.
Case I. When the plane angles are commensurable.
i>W~
-T-^-.-
3^
Dl^-^:
Given ABC and A'B'C, plane A of diedral A ABDC and A'B'IJ'C, respectively, and commensurable.
To Prove
ABDC Z ABC
A'B'D'C ZA'B'C
Proof. Let Z ABE be a common measure of A ABC and A'B'C; and suppose it to be contained 4 times in ZABC and 3 times in Z^'5'0'.
ZABC ^4 ...
'* Z A'B'C 3 ^ ^
Passing planes through edges BD and B'D', and the several lines of division of A ABC and A'B'C, respectively, diedral ZABDC will be divided into 4 parts, and diedral Z A'B'D'C into 3 parts, all of which parts are equal. (§ 433)
ABDC 4
A'B'D'C 3
T. ... WON ABDC ZABC Prom (1) and (2),^,^^, = ^-^^,
(2)
264 SOLID GEOMETRY.— BOOK VI.
Case II. Whe7i the plane angles are incommensurdble.
c
Given ABC and ABC plane A of diedral A ABDC and A'B'D'C, respectively, and incommensurable.
To Prove ABDC ^ Z ABC ^
A'B'D'C AA'B'C
Proof. Let Z ABC be divided into any number of equal parts, and let one of these parts be applied to Z A'B'C as a unit of measure.
Since A ABC and A'B'C are incommensurable, a certain number of the parts will extend from A'B' to B'E, leaving a remainder ZEB'C < one of the equal parts.
Pass a plane through B'D' and B'E\ then since the plane A of diedral AA'B'D'E and ABDC are commensurable,
ABDC ^Z ABC ^ (§451, Case I) A'B'D'E ZA'B'E ^ ' ^
Now let the number of subdivisions of Z ABC be indefi- nitely increased.
Then the unit of measure will be indefinitely diminished, and the remainder Z EB'C will approach the limit 0.
^, ABDC -n 1 ^v. T -4. ABDC
Then will approach the limit
and
A'B'D'E ZABC
will approach the limit
A'B'D'C ZABC
Z A' B'E '■ •■ Z A'B'C
By the Theorem of Limits, these limits are equal. (§ 188)
ABDC ^ ZABC ' ' A'B'D'C Z A'B'C'
DIEDRAL ANGLES.
265
Note. It follows from § 451 that the plane angle may be taken as the measure of the diedral angle ; thus, if the plane angle contains n degrees, the diedral angle may be regarded as being of n degrees.
EXERCISES.
10. A straight line and a plane perpendicular to the same straight line are parallel.
(Fig. of Prop. IX. Let plane determined by ^B and AC inter- sect MN in CD.)
11. If two planes are parallel, a line parallel to one of them through any point of the other lies in the other.
(Fig. of Prop. X. Given planes MN and PQ ||, and AB through any point A of MN II PQ. Prove that AB lies in MN by § 413.)
12. If a straight line is parallel to a plane, any plane perpendicular to the line is perpendicu- lar to the plane.
(Draw line CD ± QE, and prove it ± MN)
13. If two parallels meet a plane, they make equal angles with it. (Given AB \\ CD ; to prove ZABA' =ZCDC'.)
14. If a straight line intersects two parallel planes, it makes equal angles with them.
15. The angle between perpendiculars to the faces of a diedral angle from any point within the angle is the supplement of its plane angle.
(Prove ZBDC the plane Z of diedral ZPQBS. )
16. If each of two intersecting planes be cut by two parallel planes, not parallel to their inter- section, their intersections with the parallel ^ planes include equal angles.
(To prove ZABC=Z DEF. )
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266 SOLID GEOMETRY.— BOOK VI.
POLYBDRAL ANGLES. DEFINITIONS.
452. A polyedral angle is a figure composed of three or more triangles, called /aces, having for their
bases the sides of a polygon, and for their 9
common vertex a point without its plane ; //l\
asO-ABCD. / \\
The common vertex, 0, is called the ver- / IId \ tex of the polyedral angle, and the polygon, "^ vT / ^^^^ ABCD, the base ; the vertical angles of the ^^^^^
triangles, AOB, BOC, etc., are called the face angles, and their sides, OA, OB, etc., the edges.
Note. The polyedral angle is not regarded as limited by the base ; thns, the face AOB is understood to mean the indefinite plane between the edges OA and OB produced indefinitely.
A triedral angle is a polyedral angle of three faces. Two polyedral angles are called vertical when the edges of one are the prolongations of the edges of the other.
453. A polyedral angle is called convex when its base is a convex polygon (§ 121).
454. Two polyedral angles are equal when they can be applied to each other so that their faces shall coincide.
455. Two polyedral angles are said to be symmetrical when the face and diedral
angles of one are equal respec- a y^
tively to the homologous face // \ / \\
and diedral angles of the other, / / \ / \ \
if the equal parts occur in the A^—/—--^C c^^^^Ar--^^' reverse order. jb ^b'
Thus, if face A AOB, BOO, and COA are equal respectively to face AA'O'B', B'O'C, and COA', and diedral A OA, OB, and OC to diedral A OA', O'B', and O'C, triedral A 0-ABC and O'-A'B'C are symmetrical.
POLYEDRAL ANGLES.
267
It is evident that, in general, two symmetrical polyedral angles cannot be placed so that their faces shall coincide.
Prop. XXIX. Theorem. 456. Two vertical polyedral angles are symmetrical.
B'
Given 0-ABC s^nd O-A'B'C (Fig. 1) vertical triedral A. To Prove 0-ABC Siud O-A'B'C symmetrical.
Proof. Face A AOB, BOC, etc., are equal, respectively, to face A A' OB', B'OC, etc. (§ 40)
Again, diedral A OA and OA' are vertical ; for AOB and A' OB' are portions of the same plane, as also are AOC and A'OC ; in like manner, diedral A OB and OB' are vertical ; etc.
Then, diedral A OA, OB, etc., are equal, respectively, to diedral A OA', OB', etc. (§ 435)
But the equal parts of the triedral A occur in the reverse order; as may be seen by conceiving O-A'B'C moved II to itself to the right, and then revolved, as shown in Fig. 2, about an axis passing through 0, until face OA'C comes into the same plane as before ; edge OB' being on this side of, instead of beyond, plane OA'C.
Hence, 0-ABC and O-A'B'C are symmetrical (§ 455).
In like manner, the theorem may be proved for any two polyedral A.
268 SOLID GEOMETRY. — BOOK VI.
Ex. 17. If two parallel planes are cut by a third plane, the al- ternate-interior diedral angles are equal.
(Prove the plane A of the alt. -int. diedral A equal.)
Prop. XXX. Theorem.
457. The sum of any two face angles of a triedral angle is greater than the third.
Note. The theorem requires proof only in the case where the third face angle is greater than either of the others.
Given in triedral Z 0-ABC,
face ZAOC> face Z.AOB or face Z BOO. To Prove Z AOB -\- Z BOC > Z AOC.
Proof. In face AOC draw line OD equal to OB, making Z AOD = Z AOB] and through B and D pass a plane cut- ting the faces of the triedral Z in lines AB, BG, and CA, respectively.
In AAOB and AOD, OA = OA.
And by cons., OB = OD,
and ZAOB = ZAOD.
.-. AAOB = AAOD. (?)
.-. AB = AD. (?)
Now, AB-\-BC>AD-^ DO. (Ax. 4)
Or, since AB = AD, BC > DC.
Then, in A BOG and GOD, OG = OG.
POLYEDRAL ANGLES. 269
Also, OB = OD, and BC > DC.
.-. ZB0C>ZC0IJ. (§ 91)
Adding Z.AOB to the first member of this inequality, and its equ^l Z AOD to the second member, we have ^AOB+ZBOC>AAOD + ZCOD. .: ZAOB-hZBOOZAOa
Prop. XXXI. Theorem.
458. The sum of the face angles of any convex polyedral angle is less than four right angles.
Given 0-ABCDE a convex polyedral Z.
To Prove Z AOB -f- Z BOC + etc. < 4 rt. A.
Proof. Let ABCIJE he the base of the polyedral Z.
Let 0' be any point within polygon ABODE, and draw lines O'A, O'B, O'O, O'D, said O'E.
Then, in triedral ZA-EOB, .-
Z OAE + Z OAB > Z O'AE + Z O'AB. (§ 457)
Also, Z OB A -h Z OBC > Z O'B A + Z O'BC; etc.
Adding these inequalities, we have the sum of the base A of the A whose common vertex is 0 > the sum of the base A of the A whose common vertex is 0'.
But the sum of all the A of the A whose common vertex is 0 is equal to the sum of all the A of the A whose com- mon vertex is 0'. (§ 84)
Hence, the sum of the zi at 0 is < the sum of the ^ at 0'.
Then, the sum of the Zs at 0 is < 4 rt. A. (§ 35)
270
SOLID GEOMETRY.— BOOK VI.
Prop. XXXII. Theorem.
459. If two triedrdl angles have the face angles of oyie equal respectively to the face angles of the other, their homolo- gous diedral angles are equal.
0' O'
Fig. 1
Given, in triedral A 0-ABC and 0'-^'J5'(7',
AAOB = AA'0'B', ZBOC= Z B'O'C, and ZCOA = ZC'0'A'. To Prove diedral ZOA = diedral Z O'A'. Proof. Lay off OA, OB, OC, O'A', O'B', and O'C all equal, and draw lines AB, BC, CA, A'B', B'C, and C'A'.
.-. A OAB = A O'A'B'. (§ 63)
.-. AB = A'B'. (§ 66)
Similarly, BC = B'C and CA = C'A'.
.: A ABC = A A'B'C. (§ 69)
.-. ZEAF=ZE'A'F'. (?)
On OA and O'A' take AD = A'D'. Draw line DE in face OAB± OA.
Since A OAB is isosceles, Z OAB is acute, and hence DE will meet AB ; let it meet AB at E.
Also, draw line DF in face OACJ- OA, meeting AC slI F; and lines D'E' and D'F' in faces O'A'B' and 0'A'C±0'A', meeting A'B' and A'C at E' and jP', respectively. Draw lines EF and E'F. Then, in rt. A ADE and A' D'E', AD = ^'Z>'.
POLYEDRAL ANGLES.
271
And since A OAB = A O'A'B',
Z DAE = Z D'A'E'. (?)
.: AADE = AA'D'E'. - (§89)
.-. AE = A'E', and DE = D'E'. (?)
Similarly, AF= A'F', and DF = D'F'.
Then, in A AEF and A'E'F',
AE = A'E', AF=A'F', and Z EAF= Z E'A'F'.
.'. A AEF = A A'E'F'. (?)
.-. EF= E'F'. (?)
Then, in A DEF and D'E'F',
DE = D'E', DF= D'F', and EF= E'F'.
.: A DEF = A D'E'F. (?)
.-. ZEDF=ZE'D'F'. ' (?)
But, EDF and E'D'F' are the plane A of diedral A OA
and 0'^', respectively. (§ 429)
.-. diedral ZOA = diedral Z 0'^'. (§ 433)
Note. The above proof holds for Fig. 3 as well as for Fig. 2 ; in Figs. 1 and 2, the equal parts occur in the same order, and in Figs. 1 and 3 in the reverse order.
460. Cor. If two triedral angles have the face angles of one equal respectively to the face angles of the other,
1. They are equal if the equal parts occur in the same order. For if triedral Z O'-A'B'C (Fig. 2) be applied to 0-ABC
so that diedral A O'A' and OA coincide, point 0' falling at 0, then since Z A'O'C = Z AOC, and Z A'O'B' =Z AOB, O'B' will coincide with OB, and O'C with OC.
2. They are symmetrical if the equal parts occur in the reverse order.
EXERCISES.
18. If ^C is the projection of line AB upon plane 3IN, and BD and BE be drawn in the plane making ZCBD^ZCBE, prove ZABD = ZABE.
(Lay off BD = BE, and draw lines AD, AE. CD, and CE. Prove /^ABD and ABE equal.)
272
SOLID GEOMETRY. —BOOK VI.
19. If a plane be drawn through a diagonal of a parallelogram, the perpendiculars to it from the extremities of the other diagonal are equal.
(Given plane EF through diagonal ^C of CJABCD; toipro\eBG=DH. Prove rt. i^BGO^' and Z)ifO equal. ) q-
20. Two triedral angles are equal when a face angle and the ad- jacent diedral angles of one are equal respectively to a face angle and the adjacent diedral angles of the other, and similarly placed.
21. D is any point in perpendicular AF from A to side BC of triangle ABC. If line DE be drawn perpendicular to the plane of ABC, and line GH through E parallel to BC, prove line AE perpendicular to GH.
(Prove BC± to plane AED by § 438.)
22. A is any point in face EG of diedral ZDEFG. If ^Cbe drawn perpendicular to edge EF, and AB perpendicular to face DF, prove the plane deter- mined by ylC and BC perpendicular to EF. (Ex. 9.)
23. From any point E within diedral Z CABD, EF a-Tid EG are drawn perpendicular to faces ABC and ABD, respectively, and GH perpendicular to face ABC at H. Prove FH perpendicular to AB.
(Prove that FH lies in the plane of EF and EG.)
24. The three planes bisecting the diedral angles of a triedral angle meet in a common straight line.
(Let planes GAD and QBE intersect in line A OG. Prove G in plane OCF hy ^ 444.)
25. Any point in the plane passing through the bisector of an angle, perpendicular to its plane, is equally distant from the sides of the angle.
26. Any face angle of a polyedral angle is less than the sum of the remaining face angles.
(Divide the polyedral Z into triedral A by passing planes through any lateral edge.)
Book VII.
POLYEDRONS. DEFINITIONS.
461. Apolyedron is a solid bounded by polygons.
The bounding polygons are called the faces of the polye- dron; their sides are called the edges, and their vertices the vertices.
A diagonal of a polyedron is a straight line joining any two vertices not in the same face.
462. The least number of planes which can form a polye- dral angle is three.
Whence, the least number of polygons which can bound a polyedron is four.
A polyedron of four faces is called a tetraedron; of six faces, a hexaedron; of eight faces, an octaedron; of twelve faces, a dodecaedron; of twenty faces, an icosaedron.
463. A polyedron is called convex when the section made by any plane is a convex polygon (§ 121).
All polyedrons considered hereafter will be understood to be convex.
464. The volume of a solid is its ratio to another solid, called the 7init of volume, adopted arbitrarily as the unit of measure (§ 180).
The usual unit of volume is a cube (§ 474) whose edge is some linear unit; for example, a cubic inch or a cubic foot.
465. Two solids are said to be equivalent when their vol- umes are equal.
273
274
SOLID GEOMETRY. — BOOK VII.
PRISMS AND PARALLBLOPIPBDS. DEFINITIONS.
466. A prism is a polyedron, two of whose faces are equal polygons lying in parallel planes, having their homologous sides parallel, the other faces being parallelograms (§ 110).
The equal and parallel faces are called the bases of the prism, and the other faces the lateral faces; the edges which are not sides of the bases are called the lateral edges, and the sum of the areas of the lateral faces the lateral area.
The altitude is the perpendicular distance between the planes of the bases.
467. The following is given for convenience of reference: The bases of a prism are equal.
468. It follows from the definition of § 466 that the lat- eral edges of a prism are equal and parallel. (§ 106, I)
469. A prism is called triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc.
470. A right prism is a prism whose lat- eral edges are perpendicular to its bases.
The lateral faces are rectangles (§ 398). An oblique prism is a prism whose lateral edges are not perpendicular to its bases.
471. A regular prism is a right prism whose base is a regular polygon.
472. A truncated prism is a portion of a prism included between the base, and a plane, not parallel to the base, cutting all the lateral edges.
The base of the prism and the section made by the plane are called the bases of the trun- cated prism.
PRISMS AND PARALLELOPIPEDS.
275
473. A right section of a prism is a section made by a plane cutting all the lateral edges, and perpendicular to them.
474. A parallelopiped is a prism whose bases are parallelograms; that is, all the faces are parallelograms.
A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to its bases.
A rectangular parallelopiped is a right parallelopiped whose bases are rectangles ; that is, all the faces are rectangles.
A cube is a rectangular parallelopiped whose six faces are all squares.
Prop. I. Theorem.
475. The sections of a prism made by two parallel planes which cut all the lateral edges, are equal polygons.
Given II planes OF and C'F' cutting all the lateral edges of prism AB.
To Prove section CDEFG = section C'D'E'F'G'.
Proof. We have CD II CD', DE II D'E\ etc. (§ 414)
... CD= CD', DE = D'E', etc. (§ 107)
Also Z CDE = Z CD'E', Z DEF= Z D'E'F', etc. (§ 426)
Then, polygons CDEFG and C'D'E'FG', being mutually equilateral and mutually equiangular, are equal. (§ 124)
276
SOLID GEOMETRY.— BOOK VII
476. Cor. The section of a prism made by a plane paral- lel to the base is equal to- the base.
Prop. II. Theorem.
477. Two prisms are equal when the faces including a tne- dral angle of one are equal respectively to the faces including a triedral angle of the other, and similarly placed.
Given, in prisms AH and A'H', faces ABCDE, AG, and AL equal respectively to faces A'B'C'D'E', A'G', and A'L'-, the equal parts being similarly placed.
To Prove prism AH = prism A'H'.
Proof. We have AEAB, EAF, and FAB equal respec- tively to AE'A'B', E'A'F', and F'A'B'. (§ 66) .-. triedral Z A-BEF = triedral Z. A'-B'E'F'. (§ 460, 1)
Then, prism A'H' may be applied to prism AH in such a way that vertices A', B', C, U, E', G', F', and L' shall fall at A, B, C, D, E, G, F, and L, respectively.
Now since the lateral edges of the prisms are II, edge OH will fall on CH, B'K' on DK, etc. (§ 53)
And since points 6r', F, and V fall at G, F, and L, respec- tively, planes LH and L'H' coincide. ' (§ 395, II)
Then points H' and K' fall at H and K, respectively.
Hence, the prisms coincide throughout, and are equal.
478. Cor. Two right prisms are equal when they have equal bases and equal altitudes ; for by inverting one of the prisms if necessary, the equal faces will be similarly placed.
PRISMS AND PARALLELOPIPEDS. 277
479. Sch. The demonstration of § 477 applies without change to the case of two truncated prm/is.
Prop. III. Thi:orem.
480. An oblique prism is equivalent to a right prison hav- ing for its base a right section of the oblique prism, and for its altiticde a lateral edge of the oblique prism.
B C
Given FK' a right prism, having for its base FK a right section of oblique prism AD', and its altitude FF' equal to AA', a lateral edge of AD'.
To Prove AD'^FK'.
Proof. In truncated prisms AK and A'K', faces FGHKL and F'G'H'K'L' are equal. (§ 475)
Therefore, A'K' may be applied to AK so that vertices F', G', etc., shall fall at F, G, etc., respectively.
Then, edges A'F', B'G', etc., will coincide in direction with AF, BG, etc., respectively. (§ 399)
But since, by hyp., FF' = AA', we have AF = A'F\
In like manner, BG = B'G', CH= C'H', etc.
Hence, vertices A', B', etc., will fall at A, B, etc., respec- tively.
Then, A'K' and ^j^ coincide throughout, and are equal.
Now taking from the entire solid AK' truncated prism A'K', there remains prism AD'.
And taking its equal AK, there remains prism FK'. .'. AD'^FK.
278
SOLID GEOMETRY.— BOOK VII.
Prop. IV. Theorem.
481. The opposite lateral faces of a parallelopiped are equal and parallel.
D' c'
Given AC and A'C the bases of parallelopiped AC. To Prove faces AB' and DC equal and 11. Proof. AB is equal and II to DC, and AA' to DD'. (§ 106, 1) .-. Z A'AB = Z D'DC, and AB' 11 DC. (§ 426) .-. face AB' = face DC. (§ 113)
Similarly, we may prove AD' and BC equal and II.
482. Cor. Either face of a parallelopiped may be taken as the base.
Prop. V. Theorem.
483. The plane passed through tico diagonally opposite edges of a parallelopiped divides it into two equivalent tria^i- gular prisms.
Given plane AC passing through edges A A' and CC of parallelopiped A'C.
To Prove prism ABC-A' =o prism ACD-A'.
PRISMS AND PARALLELOPIPEDS. 279
Proof. Let EFGH be a right section of the parallelopiped, intersecting plane AA'C'Cin line EG.
Now, face AB' II face DC. (§ 481)
.-. EFW GH . (§ 414)
In like manner, EH II FG, and EFGH is a O.
.-. A EFG = A ^(^jy. (§ 108)
Now, ABC-A is =0= a right prism whose base is EFG
and altitude AA\ and ACD-A' is =0= a right prism whose
base is EGH and altitude AA\ (§ 480)
But these right prisms are equal, for they have equal
bases and the same altitude. (§ 478)
.-. ABC-A^ ^ ACD-A\
Prop. VI. Theorem.
484. Tlie lateral area of a prism is equal to the perimeter of a right section multiplied by a lateral edge.
Given DEFGH a right section of prism AO.
To Prove lat. area AC = (DE + EF -\- etc.) x AA'.
Proof. We have, AA' ± DE. (§ 398)
.-. area AA'B'B = DE x AA'. (§ 309)
Similarly, area BB'CO = EF x BB'
= EFxAA'; etc. (§468)
Adding these equations, we have
lat. area AC = DE x AA' + EF x AA' + etc. = {DE -\-EF+ etc.) x AA'.
280
SOLID GEOMETRY. — BOOK VII.
485. Cor. The lateral area of a right prism is equal to the perimeter of the base m^dtiplied by the altitude.
Prop. VII. Theorem.
486. Two rectangidar parallelopipeds having equal bases are to each other as their altitudes.
Note. The phrase "rectangular parallelepiped " in the above statement signifies the volume of the rectangular parallelopiped.
Case I. When the altitudes are commensurable.
^V^
c^
|
/l--^ |
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/ yT |
/ / |
|
/ / |
-'i |
|
/ / |
|
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/ |
/ |
Given P and Q rect. parallelopipeds, with equal bases, and commensurable altitudes, AAl and BB\
Proof. Let AC be a common measure of AA^ and BB\
and suppose it to be contained 4 times in AA!, and 3 times
in BB\
. ^ = 1 * ' BB' 3*
(1)
Through the several points of division of AA and BB' pass planes ± to lines AA' and BB\ respectively.
Then, rect. parallelopiped P will be divided into 4 parts, and rect. parallelopiped Q into 3 parts, all of which parts
(§ 478)
(2)
(?)
|
will be equal. |
P 4 |
|
From (1) and (2), |
P AA' Q BB' |
PEISMS AND PARALLELOPIPEDS.
281
Case II. Wlien the altitudes are incommensurable.
|
a' |
P |
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A |
/ |
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L |
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A |
/ |
/ |
Given P and Q rect. parallelepipeds, with equal bases, and incommensurable altitudes, AA and BB\
To Prove ^ = M.
Q BW
Proof. Divide AA} into any number of equal parts, and apply one of these parts to BB^ as a unit of measure.
Since AA^ and BB^ are incommensurable, a certain num- ber of the parts will extend from B to (7, leaving a remainder CB' < one of the parts.
Draw plane CDJLBB', and let rect. parallelepiped BD be denoted by Q'.
Then since, by const., AA' and BC are commensurable,
1 = ^- (§486, Case I)
Kow let the number of subdivisions of AA' be indefinitely increased.
Then the length of each part will be indefinitely dimin- ished, and remainder CB' will approach the limit 0.
P P
— will approach the limit — >
Then,
and
AA
AA'
^^==— will approach the limit — —
BC
BB'
AA' BB''
(§ 188)
487. Def. The dimensions of a rectangular parallelo- piped are the three edges which meet at any vertex.
282
SOLID GEOMETRY. — BOOK VII.
488. Sch. The theorem of § 486 may be expressed :
If two rectangular parallelopipeds have two dimensions of
one equal respectively to tivo dimensions of the other, they are
to each other as their third dimensions.
Prop. VIII. Theorem.
489. Two rectangular parallelopipeds having equal altitudes are to each other as their bases.
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p |
Q |
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/ |
/ |
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A |
/ |
/ |
/ |
A |
/ |
|||
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c |
/ |
c |
^j — |
p_ |
||||
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/ |
/ |
\x |
Given P and Q rect. parallelopipeds, with the same alti- tude c, and the dimensions of the bases a, 6, and a', h\ respectively.
P axb
To Prove
(§ 305)
Q a' xb'
Proof. Let i? be a rect. parallelopiped with the altitude c, and the dimensions of the base a' and b.
Then since P and M have each the dimensions b and c, they are to each other as their third dimensions a and a'.
(§ 488)
That is, ?: = -.• (1)
E a' And since R and Q have each the dimensions a' and c,
R^b
Q b'
Multiplying (1) and (2), we have
(2)
R
or
axb
R Q' Q a' xb'
PRISMS AND PARALLELOPIPEDS.
283
490. Sch. The theorem of § 489 may be expressed : Two rectangular parallelopipeds having a dimeyision of one
equal to a dimension of the other, are to each other as the
products of their other two dimensions.
Prop. IX. Theorem.
491. Any two rectangular parallelopipeds are to each other as the products of their three dimensions.
p R
/~7\ Ay\ 47
i f- — W I
/ / / ' / / /
A / /^ / A
k 1/ k 1/ k: 1/
Given P aud Q rect. parallelopipeds with the dimensions a, b, c, and a', 6', c', respectively.
To Prove ^^ ? \^/ V
Q a' X b' X c'
(Let I{ be a rect. parallelopiped with the dimensions a', b', and c, and find values of ^ and :? by §§ 490 and 488.)
EXERCISES.
1. Two rectangular parallelopipeds, with equal altitudes, have the dimensions of their bases 6 and 14, and 7 and 9, respectively. Find the ratio of their volumes.
2. Find the ratio of the volumes of two rectangular parallelopipeds, whose dimensions are 8, 12, and 21, and 14, 15, and 24, respectively.
D_' _C'
3. The diagonals of a parallelopiped bisect each other.
(To prove that AO and ^'C bisect each other. Prove AA'C'Gz. O by § 110.)
284
SOLID GEOMETRY.— BOOK VII.
Prop. X. Theorem.
492. If the unit of volume is the cube whose edge is the linear unit, the volume of a rectangular parallelopiped is equal to the product of its three dimensions.
/ X
|
A |
/ |
|
\ / |
1 A |
Given a, 6, and c the dimensions of rect. parallelopiped P, and Q the unit of volume ; that is, a cube whose edge is the linear unit.
To Prove vol. P=a xh xc.
P _ g-x b X c
Proof. We have
1x1x1 = a X 6 X c. But since Q is the unit of volume,
"= vol. P.
(§ 491)
Q
vol. P= a X b X c.
(§ 464)
493. Sch. I. In all succeeding theorems relating to vol- umes, it is understood that the unit of volume is the cube whose edge is the linear unit, and the unit of surface the square whose side is the linear unit. (Compare § 306.)
494. Cor. I. of its edge.
The volume of a cube is equal to the cube
495. Cor. 11! TJie volume of a rectangular parallelopiped is equal to the product of its base and altitude. (The proof is left to the pupil.)
PRISMS AND PARALLELOPIPEDS.
285
496. Sch. II. If the dimensions of the rectangular parallelopiped are tmdtiples of the linear unit, the truth of Prop. X. may be seen P
by dividing the solid into cubes, each equal to the unit of volume.
Thus, if the dimensions of rectangular parallelopiped P are 5 units, 4 units, and 3 units, respectively, the solid can evi- dently be divided into 60 cubes.
In this case, 60, the number which expresses the volume of the rectangular parallelopiped, is the product of 5, 4, and 3, the numbers which express the lengths of its edges.
|
r — r — t — i' — |
|
|
._.!._! ! |
|
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1 1 1 |
yi.
EXERCISES.
4. Find the altitude of a rectangular parallelopiped, the dimen- sions of whose base are 21 and 30, equivalent to a rectangular paral- lelopiped whose dimensions are 27, 28, and 35.
5. Find the edge of a cube equivalent to a rectangular parallelo- piped whose dimensions are 9 in., 1 ft. 9 in., and 4 ft. 1 in.
6. Find the volume, and the area of the entire surface of a cube whose edge is 3^ in.
7. Find the area of the entire surface of a rectangular parallelo- piped, the dimensions of whose base are 11 and 13, and volume 858.
8. Find the volume of a rectangular parallelopiped, the dimen- sions of whose base are 14 and 9, and the area of whose entire surface is 620.
9. Find the dimensions of the base of a rectangular parallelo- piped, the area of whose entire surface is 320, volume 336, and altitude 4.
(Represent the dimensions of the base by x and y. )
10. How many bricks, each 8 in. long, 2f in. wide, and 2 in.
thick, will be required to build a wall 18 ft. long, 3 ft. high, and
11 in. thick ?
11. The diagonals of a rectangular parallelo- piped are equal.
(Prove A A* C" C a rectangle. )
286
SOLID GEOMETRY. — BOOK VII.
Prop. XI. Theorem.
497. Tlie volume of any parallelopiped is equal to the product of its base and altitude.
Given AE the altitude of parallelopiped AC.
To Prove vol. AC = area ABCD x AE.
Proof. Produce edges AB, A'B\ D'C, and DO.
On AB produced, take FG = AB ; and draw planes FK' and GH' J_ FG, forming right parallelopiped FH'.
.-. FH'^AC. (§480)
Produce edges HG, H'G', K'F, and KF. On HG produced, take NM= HG; and draw planes iVP' and ML' ± NM, forming right parallelopiped LN'.
.'. LN'^FH'. (§ 480)
.-. LN'^AC.
Now since, by cons., FG is _L plane GH', planes LH and
MH' are ±. (§ 441)
Then MM', being J. MN, is ± plane LH. (§ 438)
Whence, Z LMM' is a rt. Z. (§ 398)
Then, LM' is a rectangle. (§ 76) Therefore iiV' is a rectangular parallelopiped.
.-. vol. LN' = area LMNP x JOf' . (§ 495)
.-. vol. AC = area LMNP x JO/'. (1)
PRISMS AND PARALLELOPIPEDS. 287
But rect. LMNP = rect. FGHK\ for they have equal
bases MN and GH, and the same altitude. (§ 114)
Also, rect. FOHK^nABCD ; for they have equal bases
FG and AB^ and the same altitude. (§ 310)
.-. LMNP^ABCD.
Also, JOf ' = AE. (§ 424)
Substituting these values in (1), we have
vol. AC = area ABCD x AE. '
Prop. XII. Theorem.
498. The volume of a triangular prism is equal to the product of its base and altitude.
Given AE the altitude of triangular prism ABC-O. To Prove vol. ABC-O = area ABC x AE. Proof. Construct parallelopiped ABCD-D', having its edges II to AB, BC, and BB', respectively.
.-. vol. ABC-C = 1 vol. ABCD-D' (§ 483)
= i- area ABCD x AE (§ 497) = area ABC X AE. (§ 108)
EXERCISES.
12. Find the lateral area and volume of a regular triangular prism, each side of whose base is 5, and whose altitude is 8.
13. The square of a diagonal of a rectangular parallelopiped is equal to the sum of the squares of its dimensions.
(Fig. of Ex. 11. To prove JJC^ = AA^ + AB^ + AD\)
288 SOLID GEOMETRY. —BOOK VII.
Pkop. XIII. Theorem.
499. The volume of any prism is equal to the product of its base and altitude.
Given any prism.
To Prove its volume equal to the product of its base and altitude.
Proof. The prism may be divided into triangular prisms by passing planes through one of the lateral edges and the corresponding diagonals of the base.
The volume of each triangular prism is equal to the prod- uct of its base and altitude. (§ 498)
Then, the sum of the volumes of the triangular prisms is equal to the sum of their bases multiplied by their common altitude. ,
Therefore, the volume of the given prism is equal to the product of its base and altitude.
500. Cor. I. Two prisms having equivalent bases and equal altitudes are equivalent.
501. Cor. IT. 1. Two prisms having equal altitudes are to each other as their bases.
2. Two prisms having equivalent bases are to each other as their altitudes.
3. Any two prisms are to each other as the products of their bases by their altitudes.
Ex. 14. Find the lateral area and volume of a regular hexagonal prism, each side of whose base is 3, and whose altitude is 9.
PYRAMIDS.
289
PYRAMIDS. DEFINITIONS.
502. A pyramid is a polyedron bounded by a polygon, called the base, and a series of triangles
having a common vertex.
The common vertex of the triangular faces is called the vertex of the pyramid.
The triangular faces are called the lateral faces, and the edges terminating at the vertex the lateral edges.
The sum of the areas of the lateral faces is called the lateral area.
The altitude is the perpendicular distance from the vertex to the plane of the base.
503. A pyramid is called triangular, quadrangular, etc., according as its base is a triangle, quadrilateral, etc.
504. A regular pyramid is a pyramid whose base is a regular polygon, and whose vertex lies in the perpendicular erected at the centre of the base.
505. A truncated pyramid is a portion of a pyramid included between the base and a plane cutting all the lateral edges.
The base of the pyramid and the section made by the plane are called the bases of the truncated pyramid.
506. A frustum of a pyramid is a trun- cated pyramid whose bases are parallel.
The altitude is the perpendicular distance between the planes of the bases.
EXERCISES.
15. Find the length of the diagonal of a rectangular parallelepiped whose dimensions are 8, 9, and 12,
16. The diagonal of a cube is equal to its edge multiplied by V3.
290
SOLID GEOMETRY.— BOOK VII.
Prop. XIV. Theorem. 507. In a regular pyramid, The lateral edges are equal. The lateral faces are equal isosceles triangles. o
I. 11.
B C
(The theorem follows by §§ 406, I, and 69.)
508. Def. The slant height of a regular pyramid is the altitude of any lateral face.
Or, it is the line drawn from the vertex of the pyramid to the middle point of any side of the base. (§ 94, I)
Prop. XV. Theorem.
509. Tlie lateral faces of a frustum of a regular pyramid are equal trapezoids. ^
B C
Given AC a frustum of regular pyramid 0-ABCDE.
To Prove faces AB' and EC equal trapezoids.
Proof. We have A OAB = A OBC. (§ 507, II)
We may then apply A OAB to A OBC in such a way that sides OB, OA, and AB shall coincide with sides OB, 00, and BC, respectively.
PYRAMIDS. 291
Now, A'B' II AB and B'C II BC. (?)
Hence, line A'B' will coincide with line B'C. (§ 53)
Then, AB' and BC coincide throughout, and are equal.
510. Cor. The lateral edges of a frustum of a regular pyramid are equal.
511. Def. The slant height of a frustum of a regular pyramid is the altitude of any lateral face.
Prop. XVI. Theorem.
512. The lateral area of a regular pyramid is equal to the perimeter of its base multiplied by one-half its slant height.
O
B C
Given slant height OH of regular pyramid 0-ABCDE. To Prove
lat. area 0-ABCDE = (AB -\- BC -{- etc.) x i OH. (By § 508, OH is the altitude of each lateral face.)
513. Cor. The lateral area of a frustum of a regidar pyramid is equal to one-half the sum of the perimeters of its bases, multiplied by A'^-'-rr—-~-yJ)' its slant height. / -Ib^I — r\
Given slant height HH' of the frus- ^/-/^/'^" y"--4jD tum of a regular pyramid AD'. h\L \X
B C
To Prove
lat. area AD' = \ {AB + A'B' + BC+B'C -f etc.) x HH', {HH is the altitude of each lateral face.)
292 SOLID GEOMETRY. — BOOK VIT.
EXERCISES.
17. The volume of a cube is 4^^ cu. ft. Find the area of its entire surface in square inches.
18. The volume of a right prism is 2310, and its base is a right triangle whose legs are 20 and 21, respectively. Find its lateral area,
19. Find the lateral area and volume of a right triangular prism, having the sides of its base 4, 7, and 9, respectively, and the altitude 8.
20. The volume of a regular triangular prism is 96 Vs, and one side of its base is 8. Find its lateral area.
21. The diagonal of a cube is 8 VS. Find its volume, and the area of its entire surface.
(Represent the edge by x.)
22. A trench is 124 ft. long, 2| ft. deep, 6 ft. wide at the top, and 5 ft. wide at the bottom. How many cubic feet of water will it con- tain ? (§§ 316, 499.)
23. The lateral area and volume of a regular hexagonal prism are 60 and 15 V3, respectively. Find its altitude, and one side of its base.
(Represent the altitude by x, and the side of the base by y.)
Prop. XVII. Theorem.
514. If a pyramid he cut by a plane parallel to its 6ase, 1. The lateral edges and the altitude are divided propor- tionally.
II. The section is similar to the base.
Given plane ^'C II to base of pyramid 0-ABCD, cutting faces OAB, OBC, OCD, and ODA in lines A^B\ B'C, CD', and D'A', respectively, and altitude OP at P'.
PYRAMIDS.
I. To Prove ^^ = M' = ^' etc. = ^. OA OB OC OP
Proof. Through 0 pass plane MN II ABCD.
OA^OB^^qC _0P' " OA OB 00 op'
293
(§ 427)
II. To Prove section A'B'C'D' similar to ABCD.
Proof. We have A'B' II AB, B'C II BC, etc. (?)
.-. Z A'B'C = Z ABC, Z BOU = ZBCD, etc. (§ 426)
Again, A OA^B\ OB'C, etc., are similar to A OAB, OBC,
etc., respectively. (§ 257)
OA' A'B' OB' B'C
OA AB' OB BC
etc. (1)
AB BC CD ^ '^
Then, polygons A'B'C'D' and ABCD are mutually equi- angular, and have their homologous sides proportional. Whence, A'B'C'D' and ABCD are similar. (§ 252)
515. Cor. I. Since A'B'C'D' and ABCD are similar, area A'B'C'D' A^''
But from (1), § 514,
|
area ABCD |
AB' |
|
|
rii4. |
A'B' |
OA' |
|
AB |
OA OP' OP |
|
|
area |
A'B'C'D' |
OP' |
(§ 322)
(§514,1)
area ABCD Qp^
Hence, the area of a section of a pyramid, parallel to the base, is to the area of the base as the square of its distance from the vertex is to the square of the altitude of the pyramid.
294
SOLID GEOMETRY.— BOOK VII,
516. Cor. II. If two pyramids have equal altitudes and equivalent bases, sections parallel to the bases equally distant from the vertices are equivalent.
Given bases of pyramids 0-^50 and 0'-^'^'(7'=o, and the altitude of each pyramid = fi"; also DEF and D'E'F' sections = to the bases at distance h from 0 and 0', respectively.
To Prove area DEF = area D'E'F'.
Proof. We have
area DEF^ h^ ^^^a^vesi D'E'F' ^ W /.k^kx area^SC H^' aTesiA'B'C ^^' ^ ^
W
But by hyp.,
area DEF _ area D'E'F' area ABC ~ area A'B'O ' area ABC = area A'B'C. area DEF= area D'E'F'.
(?)
Prop. XVIII. Theorem.
517. Two triangular pyramids having equal altitudes and equivalent bases are equivalent.
Given o-abc and o'-a'b'c' triangular pyramids with equal altitudes and =c= bases.
PYRAMIDS.
295
To Prove vol. o^abc = vol. o'-a%'c\
Proof. Place the pyramids with their bases in the same plane, and let PQ be their common altitude.
Divide PQ into any number of equal parts.
Through the points of division pass planes II to the plane of the bases, cutting o-abc in sections def and ghk, and O'-a'b'c' in sections d'e'f and g'h'k', respectively.
.-. def^ d'e'f, and ghk ^ g'h'k'. (§ 516)
With abc, def, and ghk as lowei' bases, construct prisms X, Y, and Z, with their lateral edges equal and II to ad; and with d'e'f and g'h'k' as upper bases, construct prisms T' and Z', with their lateral edges equal and II to a'd'.
.*. prism Y^ prism Y', and prism Zo prism Z'. (§ 500)
Hence, the sum of the prisms circumscribed about o-abc exceeds the sum of the prisms inscribed in o'-a'b'c' by prism X
But, o-abc is evidently < the sum of prisms X, Y, and Z; and it is > the sum of prisms o= to Y' and Z', respec- tively, which can be constructed with def and ghk as upper bases, having their lateral edges equal and II to ad.
Again, o'-a'b'c' is > the sum of prisms Y' and Z' ; and it is < the sum of prisms =c= to X, Y, and Z, respectively, which can be constructed with a'b'c', d'e'f, and g'h'k' as lower bases, having their lateral edges equal and II to a'd'.
That is, each pyramid is < the sum of prisms X, Y, and Z, and > the sum of prisms Y' and Z' ; whence, the differ- ence of the volumes of the pyramids must be < the dif- ference of the volumes of the two systems of prisms, or < volume X.
Now by sufficiently increasing the number of subdivisions of PQ, the volume of prism X may be made < any assigned volume, however small.
Hence, the volumes of the pyramids cannot differ by any volume, however small.
.•. vol. o-abc = vol. o'-a'b'c'.
296 SOLID GEOMETRY.— BOOK VII.
518. Cor. Since vol. o'-a'b'c' is > the total volume of the inscribed prisms, and < the total volume of the cir- cumscribed, the difference between vol. o'-a'b'c' and the total volume of the inscribed prisms is < the difference between the total volumes of the two systems of prisms, or < vol. X] and hence approaches the limit 0 when the number of subdivisions is indefinitely increased.
Prop. XIX. Theorem.
519. A triangular pyramid is equivalent to one-third of a triangular prism having the same base and altitude.
Given triangular pyramid 0-ABC, and triangular prism ABC-ODE having the same base and altitude.
To Prove vol. 0-ABC = i vol. ABC-ODE.
Proof. Prism ABC-ODE is composed of triangular pyra- mid 0-ABC, and quadrangular pyramid 0-ACDE.
Divide the latter into two triangular pyramids, 0-ACE and 0-CDE, by passing a plane through 0, C, and E.
Now, 0-ACE and 0-CDE have the same altitude.
And since CE is a diagonal of O ACDE, they have equal bases, ACE and CDE. (§ 108)
.-. vol. 0-ACE = vol. 0-CDE. (§ 517)
Again, pyramid 0-CDE may be regarded as having its vertex at C, and A ODE for its base.
Then, pyramids 0-ABC and C-ODE have the same altitude. (§ 424)
PYRAMIDS. 297
They have also equal bases, ABC and ODE. (§ 467)
.-. vol. 0-ABC= vol C-ODE. (?)
Then, vol. 0-ABG = vol. 0-ACE = vol. 0-CDE. (?) .-. vol. 0-ABC = ^ vol. ABC-ODE.
520. Cor. The volume of a triangular pyramid is equal to one-third the product of its base and altitude. (§ 498)
Prop. XX. Theorem.
521. The volume of any pyramid is equal to one-third the product of its base and altitude.
(Prove as in § 499.)
522. Cor. 1. Two pyramids having equivalent bases and equal altitudes are equivalent.
2. Two pyramids having equal altitudes are to each other as their bases.
3. Two pyramids having equivalent bases are to each other as their altitudes.
4. Any two pyramids are to each other as the products of their bases by their altitudes.
EXERCISES.
24. The altitude of a pyramid is 12 in., and its base is a square 9 in. on a side. What is the area of a section parallel to the base, whose distance from the vertex is 8 in. ? (§ 515.)
25. The altitude of a pyramid is 20 in., and its base is a rectangle whose dimensions are 10 in. and 15 in., respectively. What is the dis- tance from the vertex of a section parallel to the base, whose area is 54 sq. in. ?
298
SOLID GEOMETRY.— BOOK VII.
Prop. XXI. Theorem.
523. Two tetraedro7is hav'my a triedral angle of one equal to a triedral angle of the other, are to ea^h other as the products of the edges including the equal triedral angles.
C
Given V and F' the volumes of tetraedrons 0-ABO and O-A'B'C, respectively, having the common triedral Z 0.
To Prove Z= OA x OB x PC ^
V OA' X OB' X OC
Proof. Draw lines CP and C'P' ± to face OA'B'. Let their plane intersect face OA'B' in line OPP'. ■ Now, OAB and OA'B' are the bases, and CP and CP the altitudes, of triangular pyramids C-OAB and C'-OA'B', respectively.
V ^ area OAB x CP " V
area OA'B' x CP' area OAB CP
X
area OA'B' CP But area OAB _ OA x OB
' 3ivesi,0A'B'~ OA' x OB''
Also, A OCP and OCP' are rt. A. Then, A OCP and OCP are similar. CP ^OC " CP' oc' Substituting these values in (1), we have
(§ ^^^2, 4)
(1) (§ 321)
398) 256)
(?)
_F
V
OA X OB .. OC OA X OBx OC
OA' X OB' OC OA' X OB' x OC
PYRAMIDS. 299
Prop. XXII. Theorem.
524. The volume of a frustum of a pyramid is equal to the sum of its bases and a mean proportional between its bases, multiplied by one-third its altitude.
O
B
Given B the area of the lower base, b the area of the upper base, and ^ the altitude, of AC, a frustum of any pyramid 0-AC. To Prove vol. AC =(B + b+ VJB x b) x^H. (§ 233) Proof. Draw altitude OP, cutting A'C at Q. Now, vol. AC = vol. 0-AG - vol. 0-A'C
= Bx K^+ OQ)-bx iOQ (§ 521) = BxlH+BxiOQ-bxiOQ = Bx^H+{B-b)x\Oq. (1)
But, B:b = OF':OQ\ (§515)
Taking the square root of each term,
^B:^b=OP'.OQ. (§241)
... VB-Vb:Vb=OP-OQ:OQ (§238) = H:OQ. .'. (VB - Vb) xOQ=VbxH. (§ 232)
Multiplying both members by ( V^ + V^),
(B-b) xOQ= (V^x b -\-b) xH. Substituting this value in (1), we have
vol. AC = BxiII+ ( VS^Tft -\-b)xiH = (B + b+VWxb)xiH,
300 SOLID GEOMETRY.— BOOK VII.
Prop. XXIII. Theorem.
525. The volume of a truncated tria7igular prism is equal to the product of a right section by one-third the sum of the lateral edges. _P
B
Given GHC and DKL rt. sections of truncated triangular prism ABC-DEF. To Prove
vol. ABC-DEF = area GHC x \ {AD + BE + CF). Proof. Draw line DM± KL.
The given truncated prism consists of the rt. triangular
prism GHC-DKL, and pyramids D-EKLF and C-ABHG.
vol. GHC-DKL = area GHC x GD (§ 498)
= area GHC x \ {GD + HK+ CL), (1)
since the lateral edges of a prism are equal (§ 468).
Now DM is the altitude of pyramid D-EKLF. (§ 438)
.-. vol. D-EKLF = area EKLF x ^ DM. (§ 521)
But KL is the altitude of trapezoid EKLF. (§ 398)
.-. vol. D-EKLF = i {KE + LF) x KL x i DM. (§ 316)
Rearranging the factors, we have
vol. D-EKLF ={\KLx DM) x i {KE + IjF)
= area DKL x i {KE + Li^) (§ 312) = area (7irO x ^ (^-E -f- i>i^). (2)
In like manner, we may prove
vol. C-ABHG = area GHC x i {AG + BH). (3)
Adding (1), (2), and (3), the sum of the volumes of the solids GHC-DKL, D-EKLF, and C-ABHG is area GHC x ^ {AG + GD + BH+HK+KE+ CLTLF)- .: vol. ABC-DEF = area GHC x i {AD -^ BE + CF).
PYRAMIDS.
301
526. Cor. Tlie volume of a truncated right triangular prism is equal to the product of its base by one-third the sum of the lateral edges.
EXERCISES.
26. Each side of the base of a regular triangular pyramid is 6, and its altitude is 4. Find its lateral edge, lateral area, and volume.
Let OAB be a lateral face of the regular tri- angular pyramid, and C the centre of the base ; draw line GDJLAB; also, lines OC, AC, and OD.
Now, ^C = — (§ 356) = A
2V3.
lat. edge OA = ^ AC^ + 0C'\% 272) = Vl2 + 16 = V28 = 2 a/7.
.-. slant ht. OD = Vo^^ - Alf{% 273) = V28 - 9 = Vl9. .-. lat. area of pyramid = 9 VlQ (§ 512). Again, CD
VI^
ad' = V12-9 =^/Z. ■
.'. area of base = ^ x 18 x V3 (§ 350) = 9\/3. .-. vol. of pyramid = ^ x 9 VS x 4 (§ 520) = 12 V3.
27. Find the lateral edge, lateral area, and volume of a frustum of a regular quadrangular pyramid, the sides of whose bases are 17 and 7, respectively, and whose altitude is 12.
Let ABB' A' be a lateral face of the frustum, and 0 and 0' the centres of the bases; draw lines OC±AB, 0'C'±A'B', C'D± OC, and A'E±AB; also, lines 00' and CC.
Now, CD=OC- O'C .-. Slant ht. CC
8^-31
= V CD^+ C'D^= V25 + 144 = Vl69 = 13.
.-. lat, area frustum
= K68 + 28) X 13 (§ 513) = 624.
Again, AE = AC - A'C = 8^ - 3^ = 5, and A'E= CC- = 13.
.-. lat. edge AA' = Vae^ + A^' = V25 + 169 = Vl94. Again, area lower base = 17^, area upper base = 7^, and a mean proportional between them = VlT^ x T^ = 17 x 7 = 119. .-. vol. frustum =(289 + 49 + 119) x 4 (§ 524) = 1828.
302 SOLID GEOMETRY.— BOOK VII.
Find the lateral edge, lateral area, and volume
28. Of a regular triangular pyramid, each side of whose base is 12, and whose altitude is 15.
29. Of a regular quadrangular pyramid, each side of whose base is 3, and whose altitude is 5.
30. Of a regular hexagonal pyramid, each side of whose base is 4, and whose altitude is 9.
31. Of a frustum of a regular triangular pyramid, the sides of whose bases are 18 and 6, respectively, and whose altitude is 24,
32. Of a frustum of a regular quadrangular pyramid, the sides of whose bases are 9 and 5, respectively, and whose altitude is 10.
33. Of a frustum of a regular hexagonal pyramid, the sides of whose bases are 8 and 4, respectively, and whose altitude is 12.
34. Find the volume of a truncated right triangular prism, the sides of whose base are 5, 12, and 13, and whose lateral edges are 3, 7, and
5, respectively.
35. Find the volume of a truncated right quadrangular prism, each side of whose base is 8, and whose lateral edges, taken in order, are 2,
6, 8, and 4, respectively.
(Pass a plane through two diagonally opposite lateral edges, divid- ing the solid into two truncated right triangular prisms.)
36. Find the volume of a truncated right triangular prism, whose lateral edges are 11, 14, and 17, having for its base an isosceles triangle whose sides are 10, 13, and 13, respectively.
37. The slant height and lateral edge of a regular quadrangular pyramid are 25 and V674, respectively. Find its lateral area and volume.
38. The altitude and slant height of a regular hexagonal pyramid are 15 and 17, respectively. Find its lateral edge and volume.
(Represent the side of the base by x.)
39. The lateral edge of a frustum of a regular hexagonal pyramid is 10, and the sides of its bases are 10 and 4, respectively. Find its lateral area and volume.
40. Find the lateral area and volume of a frustum of a regular triangular pyramid, the sides of whose bases are 12 and 6, respectively, and whose lateral edge is 5.
PYRAMIDS.
303
41. Find the lateral area and volume of a regular quadrangular pyramid, the area of whose base is 100, and whose lateral edge is 13.
42. The lateral surface of a pyramid is greater than its base. (From foot of altitude draw lines to the vertices of the base ; each A
formed has a smaller altitude than the corresponding lateral face.)
43. If E, F, G, and H are the middle points of edges AB, AD, CD, and BC, respectively, of tetraedron A BCD, prove EFGH a parallelogram. (§ 130.)
44. Two tetraedrons are equal if a diedral angle and the adjacent faces of one are equal, respectively, to a diedral angle and the adjacent faces of the other, if the equal parts are similarly placed.
(Figs, of § 459. Given faces GAB, OAC, and diedral Z OA equal, respectively, to faces O'A B', O'A'C, and diedral Z O'A'.)
Dl. F'
45. The section of a prism made by a plane parallel to a lateral edge is a parallelogram.
(Given section EE'F'F \\ AA . Prove EE' \\ to plane CD' ; then use § 412.)
46. The point of intersection of the diagonals of a parallelopiped is called the centre of the par- allelopiped. (Ex. 3.)
Prove that any line drawn through the centre of a parallelopiped, terminating in a pair of oppo- site faces, is bisected at that point.
47. The volume of a regular prism is equal to its lateral area, multiplied by one-half the apothem of its base. (§ 350.)
48. The volume of a regular pyramid is equal to its lateral area, multiplied by one-third the distance from the centre of its base to any lateral face.
(Pass planes through the lateral edges and the centre of the base.)
49. Find the area of the entire surface and the volume of a trian- gular pyramid, each of whose edges is 2.
50. The areas of the bases of a frustum of a pyramid are 12 and 75, respectively, and its altitude is 9. What is the altitude of the pyramid ?
(Let altitude of pyramid = x ; then x - 9 is the ± from its vertex to the upper base of the frustum ; then use § 515.)
304
SOLID GEOMETRY.— BOOK VII.
51. The bases of a frustum of a pyramid are rectangles, whose sides are 27 and 15, and 9 and 5, respectively, and the line joining their centres is perpendicular to each base. If the altitude of the frustum is 12, find its lateral area and volume.
(From the centre of each base draw Js to two of its sides ; in this way the altitudes of the lateral faces may be found.)
52. A frustum of any pyramid is equivalent to the sum of three pyramids, having for their common altitude the altitude of the frus- tum, and for their bases the lower base, the upper base, and a mean proportional between the bases, of the frustum. (§ 524.)
53. The upper base of a truncated paral- lelopiped is a parallelogram.
(Let planes AC and BD' intersect in 00' ; ^^ prove that 00' bisects A'C and B'D'.)
54. The sum of two opposite lateral edges of a truncated paral- lelopiped is equal to the sum of the other two lateral edges.
(Fig. of Ex. 53. Find the length of 00' in terms of the lateral edges by § 132.)
55. The volume of a truncated parallelo- piped is equal to the area of a right section, multiplied by one-fourth the sum of the lateral
(By proof of § 483, a rt. section of a paral- lelopiped is a O ; divide the solid into two truncated triangular prisms, and apply Ex. 54.)
56. The volume of a truncated parallelopiped is equal to the area of a right section, multiplied by the distance between the centres of the bases.
(By Ex. 54, the distance between the centres of the bases may be proved equal to one-fourth the sum of the lateral edges.)
57. If ABCD is a rectangle, and EF any line not in its plane parallel to AB^ the vol- ume of the solid bounded by figures ABCD, ABFE, CDEF, ADE, and BGF, is
\hx ADx (2AB-hEF), where h is the perpendicular from any point of EF to ABCD. (§ 525.)
PYRAMIDS.
305
58. If ABCD and EFGH are rectangles lying in parallel planes, AB and BC being parallel to EF and FG, respectively, the solid bounded by the figures ABCD, EFGH, ABFE, BCGF, CDHG, and DAEH, is called a rec- tangular prismoid.
ABCD and EFGH are called the bases of the rectangular prismoid, and the perpendicular distance between them the altitude.
Prove the volume of a rectangular prismoid equal to the sum of its bases, plus four times a section equally distant from the bases, multi- plied by one-sixth the altitude.
(Pass a plane through CD and EF, and find volumes of solids ABCD-EF and EFGH-CD by Ex. 57.)
59. Find the volume of rectangular prismoid the sides of whose bases are 10 and 7, and 6 and 5, respectively, and whose altitude is 9.
60. Two tetraedrons are equal if three faces of one are equal, re- spectively, to three faces of the other, if the equal parts are similarly placed. (§ 460, 1.)
61. The perpendicular drawn to the lower base of a truncated right triangular prism from the intersection of the medians of the upper base, is equal to one-third the sum of the lateral edges.
(Let P be the middle point of DL, and draw PQLABC; express LM in terms of PQ and 6^iV^by § 132.)
62. The three planes passing through the lateral edges of a tri- angular pyramid, bisecting the sides of the base, meet in a common straight line.
(Fig. of Ex. 24, p. 272. The intersections of the planes with the base of the pyramid are the medians of the base.)
63. A monument is in the form of a frustum of a regular quad- rangular pyramid 8 ft. in height, the sides of whose bases are 3 ft. and 2 ft., respectively, surmounted by a regular quadrangular pyramid 2 ft. in height, each side of whose base is 2 ft. What is its weight, at 180 lb. to the cubic foot ?
64. Find the area of the base of a regular quadrangular pyramid, whose lateral faces are equilateral triangles, and whose altitude is 5.
(Represent lateral edge and side of base by x.)
306
SOLID GEOMETRY.— BOOK VII.
65. A plane passed through the centre of a parallelopiped divides it into two equivalent solids. (Ex. 55. )
66. The sides of the base, AB, BC, and CA, of truncated right triangular prism ABC-DEF are 15, 4, and 12, respectively, and the lateral edges AD, BE, and CF are 15, 7, and 10, re- spectively. Find the area of upper base DEF.
(Draw EH±CF, and HG and FKJLAD. Find area DEF by § 324.)
67. The volume of a triangular prism is equal to a lateral face, multiplied by one-half its perpendicular distance from any point in the opposite lateral edge.
(Draw a rt. section of the prism, and apply § 525.)
68. The sum of the squares of the four diagonals of a parallelopiped is equal to the sum of the squares of its twelve edges.
(To prove ACi^ + AKI^ + BD^ + Wd^ equal to 4 Zi^ + 4 Js"^ + 4 AD^ Apply Ex. 79, p. 228, to OJAA'C'G.)
A ' B
69. The altitude and lateral edge of a frustum of a regular tri- angular pyramid are 8 and 10, respectively, and each side of its upper base is 2v^. Find its volume and lateral area.
70. If ABCD is a tetraedron, the section made by a plane parallel to each of the edges AB and 0Z> is a parallelogram, (§412.) ^
(Toprove J^iJ'G^lfaO.)
71. In tetraedron ABCD, a plane is drawn through edge CD per- pendicular to AB, intersecting faces ABC and ABD in CE and ED, respectively. If the bisector of Z CED meets CD at F, prove
CF:DF = area ABC : area ABD. (§ 249.)
72. The sum of the perpendiculars drawn to the faces from any point within a regular tetraedron (§ 536) is equal to its altitude.
(Divide the tetraedron into triangular pyramids, having the given point for their common vertex.)
SIMILAR POLYEDRONS.
307
73. The planes bisecting the diedral angles of a tetraedron intersect in a common point.
74. If the four diagonals of a quadrangular prism pass through a common point, the prism is a parallelopiped.
(In Fig. of Ex. 68, let AC, A'C, BD\ and B'D pass through a common point. To prove J. C a parallelopiped. Prove ^C a O.)
SIMILAR POLYEDRONS.
527. Def. Two polyedrons are said to be similar when they have the same number of faces similar each to each and similarly placed, and have their homologous polyedral angles equal.
Prop. XXIV. Theorem.
528. Tlie ratio of any two homologous edges of two similar polyedrons is equal to the ratio of any other two homologous edges.
Given, in similar polyedrons .AF and A'F', edge AB homologous to edge A'B', and edge EF to edge E'F' ; and faces AO and I)F similar to faces A'C and D'F', respec- tively.
AB ^ EF A'B' E'F'' AB CD
To Prove
By § 253, 2,
A'B' CD'
308 SOLID GEOMETRY.— BOOK VII.
529. Cor. I. Any two homologous faces of two similar polyedrons are to each other as the squares of any two homolo- gous edges.
To prove ^^^!^, = MEr, See § 322.^ area A'B'C'D' E'F' J
530. Cor. II. The entire surfaces of two similar polyedrons are to ea^h other as the squares of any two homologous edges.
/^To rove area ABCD + area CDEF etc. _ EF^ \ V ^^ ^ area A'B'C'D' + area C'D'E'F' etc. ~ WF''' J
Prop. XXV. Theorem.
531. Two tetraedrons are similar when the faces including a triedral angle of one are similar, respectively, to the faces including a triedral angle of the other, and similarly placed.
A\
C
Given, in tetraedrons ABCD and A'B'C'D', face ABC similar to A'B'C, ACD to A'C'D', and ADB to A'D'B'. To Prove ABCD and A'B'C'D' similar. Proof. From the given similar faces, we have
B^^AG^CD_^AD^^BD^ m
B'C A'C CD' A'D' B'D'' ^'^
Hence, faces BCD and B'C'D' are similar. (§ 259)
Again, A BAG, CAD, and DAB are equal, respectively,
to AB'A'C, C'A'D', and D'A'B'. (?)
Then, triedral AA-BCD and A'-B'C'D' are equal.
(§ 460, 1) Similarly, any two homologous triedral A are equal. Therefore, ABCD and A'B'C'D' are similar (§ 527).
SIMILAR POLYEDRONS. 3Q9
Prop. XXVI. Theorem.
532. Two tetraedrons are similar when a diedral angle of one is equal to a diedral angle of the other, and the faces including the equal diedral angles similar each to each, and similarly placed.
Given, in tetraedrons ABGD and A'B'C'D\ diedral Z AB equal to diedral Z A'B' ; and faces ABC and ABD similar to faces A'B'C and A'B'D', respectively.
To Prove ABCD and A'B'C'D' similar.
Proof. Apply tetraedron A'B'C'D' to ABCD so that die- dral ZA'B' shall coincide with its equal diedral ZAB, point A' falling at A.
Then since Z B'A'C = Z BAG and Z B'A'D' = Z BAD, edge A'C will coincide with edge AC, and A'D' with AD. .'. ZC'A'D' = ZCAD.
Again, from the given similar faces,
ATP^AJB^^^A^D^ m
AC AB ad' ^'^
Hence, A C'A'D' is similar to A CAD. (§ 261)
Then, the faces including triedral Z A'-B'C'D' are similar respectively to the faces including triedral Z A-BCD, and similarly placed.
Therefore, ABCD and A'B'C'D' are similar. (§ 531)
Ex. 75. If a tetpaedron be cut by a plane parallel to one of its faces, the tetraedron cut off is similar to the givei^i tetraedron.
310 SOLID GEOMETRY.— BOOK VII.
Prop. XXVII. Theorem.
533. Two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each, and simi- larly placed.
Given ^Fand A^F^ similar polyedrons, vertices JL and J.' being homologous.
To Prove that they may be decomposed into the same number of tetraedrons, similar each to each, and similarly placed.
Proof. Divide all the faces of AF, except the ones hav- ing ^ as a vertex, into A ; and draw lines from A to their vertices.
In like manner, divide all the faces of A^F', except the ones having A' as a vertex, into A similar to those in AF, and similarly placed. (§ 267)
Draw lines from A' to their vertices.
Then, the given polyedrons are decomposed into the same number of tetraedrons, similarly placed.
Let ABCF 2aidi A'B'C'F' be homologous tetraedrons.
A ABC and BCF are similar, respectively, to AA'B'C and B'C'F'. (§ 267)
And since the given polyedrons are similar, the homolo- gous diedral A BC and B'C are equal.
Therefore, ABCF and A'B'C'F' are similar. (§ 532)
In like manner, we may prove any two homologous tetraedrons similar.
Hence, the given polyedrons are decomposed into the same number of tetraedrons, similar each to each, and similarly placed.
J
SIMILAR POLYEDRONS. ^H
Prop. XXVIII. Theorem.
534. Two similar tetraedrons are to each other as the cubes of their homologous edges.
C
Given Fand "F the volumes of similar tetraedrons ABCD and A'B'C'V, vertices A and A' being homologous.
To Prove Z^A^.
Proof. Since the triedral A 2it A and A' are equal, V ABxACxAD
V A'B'xA'C'xA'D' AB ^AC_^ AD
523)
A'B' A'C A'D'
-p . AC AB -, AD AB .. ^oq\
^"*' 1^ = 1^'^"^ J^ = ^- ^^'''^
V ^ AB AB AB ^ A^ " V A'B' A'B' A'B' jTb^'
535. Cor. Any two similar polyedrons are to each other as the cubes of their homologous edges.
For any two similar polyedrons may be decomposed into the same number of tetraedrons, similar each to each (§ 533).
Any two homologous tetraedrons are to each other as the cubes of their homologous edges. (§ 534)
Then, any two homologous tetraedrons are to each other as the cubes of any two homologous edges of the polyedrons.
(§ 528)
312 SOLID GEOMETRY. — BOOK VII.
REGULAR POLYBDRONS.
536. Def. A regular poly edron is a polyedron whose faces are equal regular polygons, and whose polyedral angles are all equal.
Prop. XXIX. Theorem.
537. Not more than Jive regular convex polyedrons are possible.
A convex polyedral Z must have at least three faces, and the sum of its face A must be < 360° (§ 458).
1. With equilateral triangles.
Since the Z of an equilateral A is 60°, we may form a con- vex polyedral Z by combining either 3, 4, or 5 equilateral A.
Not more than 5 equilateral A can be combined to form a convex polyedral Z. (§ 458)
Hence, not more than three regular convex polyedrons can be bounded by equilateral A.
2. With squares.
Since the Z of a square is 90°, we may form a convex polyedral Z by combining 3 squares.
Not more than 3 squares can be combined to form a con- vex polyedral Z. (?)
Hence, not more than one regular convex polyedron can be bounded by squares.
3. With regular pentagons.
Since the Z of a regular pentagon is 108°, we may form a convex polyedral Z by combining 3 regular pentagons.
Not more than 3 regular pentagons can be combined to form a convex polyedral Z. (?)
Hence, not more than one regular convex polyedron can be bounded by regular pentagons.
Since the Z of a regular hexagon is 120°, no convex polye- dral Z can be formed by combining regular hexagons. (?)
REGULAR POLYEDRONS. 3;[3
Hence, no regular convex polyedron can be bounded by regular hexagons.
In like manner, no regular convex polyedron can be bounded by regular polygons of more than six sides.
Therefore, not more than five regular convex polyedrons are possible.
Prop. XXX. Theorem. 538. With a given edge, to construct a regular polyedron.
We will now prove, by actual construction, that five regu- lar convex polyedrons are possible :
1. The regular tetraedron, bounded by 4 equilateral A.
2. The regular hexaedron, or cube, bounded by 6 squares.
3. The regular octaedron, bounded by 8 equilateral A.
4. The regular dodecaedron, bounded by 12 regular pen- tagons.
5. The regular icosaedron, bounded by 20 equilateral A.
1. To construct a regular tetraedron. Given line AB.
Required to construct with AB as an a
edge a regular tetraedron. /
Construction. Construct the equilateral / A ABC. /
At its centre E, draw line ED J_ ABC', ^^--^^ and take point D so that AD = AB. ^^
Draw lines AD, BD, and CD.
Then, solid ABCD is a regular tetraedron.
Proof. Since A, B, and C are equally distant from E,
AD=:BD= CD. (§ 406, I)
Hence, the six edges of the tetraedron are all equal. Then,»the faces are equal equilateral A. (§ 69)
And since the A of the faces are all equal, the triedral A whose vertices are A, B, C, and D are all equal. (§ 460, 1) Therefore, solid ABCD is a regular tetraedron. (§ 536)
314
SOLID GEOMETRY. —BOOK VII.
D^-
(§ 460, 1)
2. To construct a regular hexaedron, or ctibe. Given line AB. Required to construct with AB as an
edge a cube. e
Construction. Construct square ABCD ; and draw lines AI], BF, CG, and DH, each equal to AB, and _L ABCD.
Draw lines EF, FG, GH, and HE-, then, solid AG is a cube.
Proof. By cons., its faces are equal squares.
Hence, its triedral A are all equal.
3. To construct a regular octaedron. Given line AB. Required to construct with AB as an
edge a regular octaedron.
Construction. Construct the square ABCD; through its centre 0 draw line EF± ABCD, making OE=OF= OA.
Draw lines EA, EB, EC, ED, FA, FB, FC, and FD ; then solid AEFC is a regular octaedron.
Proof. Draw lines OA, OB, and OD.
Then in rt. A AOB, AOE, and AOF, by cons., OA=OB=OE= OF. .-. A AOB = A AOE = A AOF. .: AB = AE = AF.
Then, the eight edges terminating at E and F are all equal. (§ 406, I)
Thus, the twelve edges of the octaedron are all equal, and the faces are equal equilateral A. (?)
Again, by cons., the diagonals of quadrilateral BEDF are equal, and bisect each other at rt. A.
Hence, BEDF is a square equal to ABCD, and OA is ± to its plane. (§ 400)
Then, pyramids A-BEDF and E-ABCD are equal ; and hence polyedral A A-BEDF and E-ABCD are equal.
(?)
CO
REGULAR POLYEDRONS.
In like manner, any two polyedral A are equal. Therefore, solid AEFC is a regular octaedron. 4. To construct a regular dodecaedron. d'
315
D
Fig. 1.
Given line AB.
Required to construct with AB as an edge a regular do- decaedron.
Construction. Construct regular pentagon ABODE (Fig. 1) ; and to it join five equal regular pentagons, so inclined as to form equal triedral A at A, B, C, D, and E. (§ 460, 1)
Then there is formed a convex surface AK composed of six regular pentagons, as shown in lower part of Fig. 1.
Construct a second surface A'K' equal to AK, as shown in upper part of Fig. 1.
Surfaces AK and A'K' may be combined as shown in Fig. 2, so as to form at i^ a triedral Z equal to that at A, having for its faces the regular pentagons about vertices F and F' in Fig. 1. (§ 460, 1)
Then, solid AK is a regular dodecaedron.
Proof. Since G' falls at G, and diedral Z FG and face A FGH and FGD' (Fig. 2) are equal respectively to the diedral Z and face A of triedral A F, the faces about vertex G will form a triedral Z equal to that at F.
In this way, it may be proved that at each of the vertices H, K, etc., there is formed a triedral Z equal to that at F.
Therefore, solid AKis a regular dodecaedron.
316
SOLID GEOMETRY.— BOOK VII.
5. To construct a regular icosaedron.
E-^ ^F
Fig. 1.
Given line AB.
Fig. 3.
Required to construct with AB as an edge a regular icosaedron.
Construction. Construct regular pentagon ABODE (Fig. 1) ; at its centre 0 draw line OF _L ABODE, making AF= AB, and draw lines AF, BF, OF, DF, and EF.
Then, F-ABODE is a polyedral Z composed of five equal equilateral A. (§§ 406, I, 69)
Then construct two other polyedral A, A-BFEGH and E-AFDKG, each equal to F-ABODE ; and place them as shown in upper part of Fig. 2, so that faces ABF and AEF of A-BFEGH, and faces AEF and DEF of E-AFDKG, shall coincide with the corresponding faces of F-ABODE.
Then there is formed a convex surface GO, composed of ten equilateral A.
Construct a second surface (r'C equal to GO, as shown in lower part of Fig. 2.
Surfaces GO and 6r'C" may be combined as shown in Fig. 3, so that edges GH and HB shall coincide with edges G'H' and H'B\ respectively.
Then, solid G^O is a regular icosaedron.
Proof. Since diedral A AH, E'H', and F'H' are equal to the diedral A of polyedral Z F, the faces about vertices Hand H' form a polyedral Z at iJ equal to that at F.
REGULAR POLYEDRONS.
817
Then, since diedral A FB, AB, HB, and F'B (Fig. 3) are equal to the diedral A of polyedral AF, the faces about vertex B form a polyedral A equal to that at F; and it may be shown that at each of the vertices C, D, etc., there is formed a polyedral A equal to that at F.
Therefore, solid OC is a regular icosaedron.
539. Sch. To construct the regular polyedrons, draw the following figures on cardboard ; cut them out entire, and on the interior lines cut the cardboard half through ; the edges may then be brought together to form the respective solids.
Tetraedron.
Hexaedron.
OCTAEDRON.
DODECAEDRON.
Icosaedron.
EXERCISES.
76. The volume of a pyramid whose altitude is 7 in, is 686 cu. in. Find the volume of a similar pyramid whose altitude is 12 in.
318 SOLID GEOMETRY.— BOOK VII.
77. If the volume of a prism whose altitude is 9 ft. is 171 cu. ft., find the altitude of a similar prism whose volume is 50f cu. ft.
(Represent the altitude by x.)
78. Two bins of similar form contain, respectively, 375 and 648 bushels of wheat. If the first bin is 3 ft. 9 in. long, what is the length of the second ?
79. A pyramid whose altitude is 10 in., weighs 24 lb. At what distance from its vertex must it be cut by a plane parallel to its base so that the frustum cut off may weigh 12 lb. ?
80. An edge of a polyedron is 56, and the homologous edge of a similar polyedron is 21. The area of the entire surface of the second polyedron is 135, and its volume is 162. Find the area of the entire surface, and the volume, of the first polyedron.
81. The area of the entire surface of a tetraedron is 147, and its volume is 686. If the area of the entire surface of a similar tetrae- dron is 48, what is its volume ?
(Let X and y denote the homologous edges of the tetraedrons. )
82. The area of the entire surface of a tetraedron is 75, and its volume is 500. If the volume of a similar tetraedron is 32, what is the area of its entire surface ?
83. The homologous edges of three similar tetraedrons are 3, 4, and 5, respectively. Eind the homologous edge of a similar tetrae- dron equivalent to their sum.
(Represent the edge by x.)
84. State and prove the converse of Prop. XXVII.
85. The volume of a regular tetraedron is equal to the cube of its edge multiplied by jV^2.
86. The volume of a regular tetraedron is 18 V2. Find the area of its entire surface. (Ex. 85.)
(Represent the edge by x.)
87. The volume of a regular octaedron is equal to the cube of its edge multiplied by |V2.
Book VIII.
THE CYLINDER, CONE, AND SPHERE. DEFINITIONS.
540. A cylindrical surface is a surface generated by a moving straight line, which constantly intersects a given plane curve, and in all its positions is parallel to a given straight line, not in the plane of the curve.
Thus, if line AB moves so as to constantly intersect plane curve AD, and is constantly parallel to line MN, not in the plane of the curve, it generates a cylindrical surface.
The moving line is called the generatrix, and the curve the directrix.
Any position of the generatrix, as EF, is called an element of the surface.
A cylinder is a solid bounded by a cylin- drical surface, and two parallel planes.
The parallel planes are called the bases of the cylinder, and the cylindrical surface the lateral surface.
The altitude of a cylinder is the perpen- dicular distance between the planes of its bases.
A light cylinder is a cylinder the elements of whose lateral surface are perpendicular to its bases.
A circular cylinder is a cylinder whose base is a circle. 319
320 SOLID GEOMETRY. — BOOK VIII.
A plane is said to be tangent to a cylinder when it con- tains one, and only one, element of the lateral surface.
541. It follows from the definition of a cylinder (§ 540) that
The elements of the lateral surface of a cylinder are equal and parallel. (§ 415)
Prop. I. Theorem.
542. A section of a cylinder made by a plane passing through an element of the lateral surface is a parallelogram.
Given ABCD a section of cylinder AF, made by a plane passing through AB, an element of the lateral surface.
To Prove section ABCD a O.
Note. It should be observed that, with the above hypothesis, CD simply represents the intersection of plane AC with the cylindrical surface, and may be a curved line ; it must be proved that it is a str. line II AB.
Proof. AD and BC are str. lines, and II. (§§ 396, 414)
Now draw str. line CE in plane AC II AB; then, CE is an element of the cylindrical surface. (§§ 541, 53)
Then since CE lies in plane AC, and also in the cylin- drical surface, it must be the intersection of the plane with the cylindrical surface.
Then, CD is a str. line II AB, and ABCD is a O.
543. Cor. A section of a right cylinder made by a plane perpendicular to its base is a rectangle.
THE CYLINDER.
Prop. II. Theorem. 544. The bases of a cylinder are equal.
^ Q'
321
Given cylinder AB'.
To Prove base A'B' = base AB.
Proof. Let E', F, and G' be any three points in the perim- eter of base A^B\ and draw EE', FF', and GG' elements of the lateral surface.
Draw lines EF, FG, GE, E'F', F'G', and G'E'.
Now, EE' and FF' are equal and II. (§ 541)
Then, EE'F'F is a O. (?)
.-. E'F' = EF. (?)
Similarly, E'G' = EG and F'G' = FG.
.'. AE'F'G' = AEFG. (?)
Then, base A'B' may be superposed upon base AB so that points E', F', and G' shall fall at E, F, and G, respectively.
But E' is any point in the perimeter of A'B'.
Then, every point in the perimeter of A'B' will fall some- where in the perimeter of AB, and base A'B' = base AB.
545. Cor. I. The sections of a circular cylinder made by planes parallel to its bases are equal circles.
For each may be regarded as the upper base of a cylinder whose lower base is a O.
546. Def. The axis of a circular cylinder is a straight line drawn between the centres of its bases.
322
SOLID GEOMETRY. —BOOK VIII.
547. Cor. II. The axis of a circular cylinder is parallel to the elements of its lateral surface.
Given AA' the axis, and BB' an ele- ment of the lateral surface, of circular cylinder BC
To Prove AA' II BB'.
Proof. Let BB'C'G be a section made
by a plane passing through BB' and A ;
then5i5'(7'(7isaO. (§542)
.-. B'C' = Ba (?)
Then since BC is a diameter of O BC, and © BC and B'C are equal, B'C is a diameter of QB'C, and passes through A'.
Hence, AB and ^'5' are equal and II. (?)
Then, ABB' A' is a O. (?)
.-. AA' II 55'.
54B. Cor. III. The axis of a circular cylinder passes through the centres of all sections parallel to the bases.
Prop. III. Theorem.
549. A right circular cylinder may he generated by the revolution of a rectangle about one of its sides as an axis.
Given rect. ABCD.
To Prove the solid generated by the revolution of ABCD about AD as an axis a rt. circular cylinder. Proof. All positions of BC are II AD. Again, AB and CD generate CD ± AD. (§ 402)
Then, these (D are II, and ± BC. (§§ 421, 419)
Whence, ABCD generates a rt. circular cylinder.
THE CYLINDER. 323
550. Defs, From the property proved in § 549, a right circular cylinder is called a cylinder of revolution.
Similar cylinders of revolution are cylinders generated by the revolution of similar rectangles about homologous sides as axes.
Prop. IV. Theorem.
551. A plane drawn through an element of the lateral sur- face of a circular cylinder and a tangent to the base at its extremity, is tangent to the cylinder.
Given AA^ an element of the lateral surface of circular cylinder AB\ line CD tangent to base AB at A, and plane CD' drawn through AA! and CD.
To Prove CD^ tangent to the cylinder.
Proof. Let E be any point in plane CD\ not in AA\ and draw through E a plane II to the bases, intersecting CD' in line EF, and the cylinder in O FH. (§ 545)
Draw axis 00' ; then 00' is II AA'. (§ 547)
Let the plane of 00' and AA' intersect the planes of AB and FH in radii OA and GF, respectively. (§ 548)
Then, OF II OA and FE II AD. (§ 414)
.-. ZGFE = ZOAD. (§426)
But Z OAD is a rt. Z. (§ 1'^^)
Then, FE is ± GF, and tangent to O FH. (§ 169)
Whence, point E lies without the cylinder.
Then, all portions of CD', not in AA', lie without the cylinder, and CD' is tangent to the cylinder.
324 SOLID GEOMETUV. — BOOK VIII.
552. Cor. A plane tangent to a circular cylinder intersects the planes of the bases in lines which are tangent to the bases.
Ex. 1. The sections of a cylinder made by two parallel planes which cut all the elements of its lateral surface are equal.
THE CONE. DEFINITIONS.
553. A conical surface is a surface generated by a moving straight line, which constantly intersects
a given plane curve, and passes through M- --.
a given point not in the plane of the ^ vi ^^^
curve. \\ y^
Thus, if line OA moves so as to con- U^
stantly intersect plane curve ABC, and y"^ \\
constantly passes through point 0, not a/^ V" \c
in the plane of the curve, it generates a n \y
conical surface. ^
The moving line is called the generatrix, and the curve the directrix.
The given point is called the vertex, and any position of the generatrix, as OB, is called an element of the surface.
If the generatrix be supposed indefinite in length, it will generate two conical surfaces of indefinite extent, O-A'B'C and 0-ABC.
These are called the upper and lower nappes, respectively.
A co7ie is a solid bounded by a conical surface, and a plane cutting all its elements.
The plane is called the base of the cone, and /K
the conical surface the lateral surface. / \ \
The altitude of a cone is the perpendicular / [ \
distance from the vertex to the plane of the / 1 A base. ^-^ -^
A circular cone is a cone whose base is a circle.
THE CONE. 325
The axis of a circular cone is a straight line drawn from the vertex to the centre of the base.
A right circular cone is a circular cone whose axis is per- pendicular to its base.
A frustum of a cone is a portion of a cone included between the base and a plane parallel to the base. /^ ^
The base of the cone is called the loicer / (
base, and the section made by the plane the /- -,^ j
upper base, of the frustum. ( V
The altitude is the perpendicular distance \. ^
between the planes of the bases.
A plane is said to be tangent to a cone, or frustum of a cone, when it contains one, and only one, element of the lateral surface.
Prop. V. Theorem.
554. A right circular cone may be generated by the revo- lution of a right triangle about one of its legs as an axis.
Given C the rt. Z of rt. A ABC.
To Prove the solid generated by the revolution of ABC about AC as an axis a right circular cone. (The proof is left to the pupil.)
555. Defs. "From the above property, a right circular cone is called a cone of revolution.
Similar cones of revolution are cones generated by the revolution of similar right triangles about homologous legs as axes.
326 SOLID GEOMETRY.— BOOK VIII.
Prop. VI. Theorem.
556. A section of a cone made by a plane passing through the vertex is a triangle.
Given OCD a section of cone GAB made by a plane pass- ing through vertex 0.
To Prove section OCD a A.
Proof. We have CD a str. line. (§ 396)
Now draw str. lines in plane OCD from 0 to C and D ; these str. lines are elements of the conical surface. (§ 550)
Then, since these str. lines lie in plane OCD, and also in the conical surface, they must be the intersections of the plane with the conical surface.
Then, OC and OD are str. lines, and OCD is a A.
Prop. VII. Theorem.
557. A section of a circular cone made by a plane parallel to the base is a circle.
Given A'B'O a section of circular cone S-ABC, made by a plane II to the base. To Prove A'B'C a O.
THE CONE.
327
Proof. Draw axis OS, intersecting plane xl'B'C at 0'. Let A' and B' be any two points in perimeter A'B'C. Let the planes determined by these points and OS inter- sect the base in radii OA and OB, the section in lines O'A' and O'B', and the lateral surface in lines SA'A and SB'B, respectively.
Then, SA'A and SB'B are str. lines. (§ 556)
Now, O'A' II OA and O'B' 11 OB. (§ 414)
Then, A SO' A' and SO'B' are similar to A.S'ok and
SOB, respectively.
O'A'
OA
^^' and ^'^'
OB O'B'
ob'
SO'
so'
(§ 257) (?)
(?) (§ 143)
OA
But, 0^ = OB.
Then, 0'^' = 0'5' ; and as A' and B' are aw?/ two points in perimeter A'B'C, section A'B'C is sl O.
558. Cor. TJie axis of a circular cone passes through the centre of every section parallel to the base.
Prop. YIII. Theorem.
559. A plane drawn through an element of the lateral sur- face of a circular cone and a tangent to the base at its extremity, is tangent to the cone. .o
Given OA an element of the lateral surface of circular cone OAB, line CD tangent to base AB at A, and plane OCB drawn through OA and OD.
328 SOLID GEOMETRY. — BOOK VIIL
To Prove OCD tangent to the cone. (Prove that E lies without the cone.)
560. Cor. A plane tangent to a circular cone intersects the plane of the base in a line tangent to the base.
THE SPHERE. DEFINITIONS.
561. A sphere is a solid bounded by a surface, all points of which are equally distant from a point within called the centre.
A radius of a sphere is a straight line drawn from the centre to the surface.
A diameter is a straight line drawn through the centre, having its extremities in the surface.
562. It follows from the definition of § 561 that all radii of a sphere are equal.
Also, all its diameters are equal, since each is the sum of two radii.
563. Two spheres are equal when their radii are equal. For they can evidently be applied one to the other so
that their surfaces shall coincide throughout. Conversely, the radii of equal spheres are equal.
564. A line (or a plane) is said to be tangent to a sphere when it has one, and only one, point in common with the surface ; the common point is called the point of contact.
A polyedron is said to be inscribed in a sphere when all its vertices lie in the surface of the sphere ; in this case the sphere is said to be circumscribed about the polyedron.
A polyedron is said to be circumscribed about a sphere when all its faces are tangent to the sphere; in this case the sphere is said to be inscribed in the polyedron.
565. A sphere may be generated by the revolution of a semi- circle about its diameter as an axis.
THE SPHERE.
329
For all points of such a surface are equally distant from the centre of the O. (?)
Prop. IX. Theorem. 566. A section of a sphere made by a plane is a circle.
6
Given ABC a section of sphere AFC made by a plane.
To Prove ABC a O.
Proof. Let 0 be the centre of the sphere, and draw line 00' ± to plane ABC.
Let A and B be any two points in perimeter ABC, and draw lines OA, OB, OA, and O'B.
Now, • OA=OB. (?)
.-. OA=0'B. (§407,1)
But A and B are any two points in perimeter ABC.
Therefore, ABC is a O.
567. Defs. A great circle of a sphere is a section made by a plane passing through the centre; as ABC
A small circle is a section made by a plane which does not pass through the centre.
The diameter perpendicular to a circle of a sphere is called the axis of the circle, and its extremities are called the poles.
568. Cor. I. TJie axis of a circle of a sphere passes through the centre of the circle.
330 SOLID GEOMETRY.— BOOK VIII.
569. Cor. II. All great circles of a sphere are equal. For their radii are radii of the sphere.
570. Cor. III. Every great circle bisects the sphere and its surface.
For if the portions of the sphere formed by the plane of the great O be separated, and placed so that their plane sur- faces coincide, the spherical surfaces falling on the same side of this plane, the two spherical surfaces will coincide throughout ; for all points of either surface are equally dis- tant from the centre.
571. Cor. rV. Any two great circles bisect each other.
For the intersection of their planes is a diameter of the sphere, and therefore a diameter of each O. (§ 152)
572. Cor. V. Between any two points on the surface of a sphere, not the extremities of a diameter, an arc of a great circle, less than a semi-circumference, can be drawn, and but one.
For the two points, with the centre of the sphere, deter- mine a plane which intersects the surface of the sphere in the required arc.
Note. If the points are the extremities of a diameter, an indefi- nitely great number of arcs of great © can be drawn between them ; for an indefinitely great number of planes can be drawn through the diameter.
573. Def. The distance between two points on the surface of a sphere, not at the extremities of a diameter, is the arc of a great circle, less than a semi-circum- ference, drawn between them.
Thus, the distance between points C and D is arc CED, and not arc CAFBD.
574. Cor. VI. An arc of a circle may be drawn through any three points on the surface of a sphere.
For the three points determine a plane which intersects the surface of the sphere in the required arc.
THE SPHERE. 33^
Prop. X. Theorem.
575. All points in the circumference of a circle of a sphere are equally distant from each of its poles.
Given P and P' the poles of O ABC of sphere APC.
To Prove all points in circumference ABC equally distant (§ 573) from P, and also from P'.
Proof. Let A and B be any two points in circumference ABC, and draw arcs of great © PA and PB. Draw axis PP\ intersecting plane ABC at 0. Draw lines OA and OB, and chords PA and PB. Now 0 is the centre of O ABC (§ 568)
.-. 0A= OB. (?)
.-. chord PA = chord PB. (§ 406, I)
.-. arcP^ = arcP^. (§ 157)
But A and B are any two points in circumference ABC.
Therefore, all points in circumference ABC are equally distant from P.
In like manner, all points in circumference ABC are equally distant from P'.
576. Def. The polar distance of a circle of a sphere is the distance (§ 573) from the nearer of its poles, or from either pole if they are equally near, to the circumference.
Thus, in figure of Prop. X, the polar distance of O ABC is arc PA.
332
SOLID GEOMETRY.— BOOK VIII.
577. Cor. All points in the circumfer- ence of a great circle of a sphere are at a quadranfs distance from either pole.
Given P a pole of great Q ABC of sphere APC, B any point in circumfer- ence ABC, and PB an arc of a great O.
To Prove arc PB a quadrant (§ 146).
Proof. Let 0 be the centre of the sphere, and draw radii OB and OP. Then, Z POB is a rt. Z. (§ 398)
Whence, arc PB is a quadrant. (§ 191)
The above proof holds for either pole of the great O.
Note. An arc of a circle may be drawn on the surface of a sphere by placing one foot of the compasses at the nearer pole of the circle, the distance between the feet being equal to the chord of the polar distance.
Prop. XI. Theorem.
578. If a point on the surface of a sphere lies at a quad- rants distance from each of two points in the arc of a great cii'cle, it is a pole of that arc.
Note. The term quadrant^ in Spherical Geometry, usually signi- fies a quadrant of a great circle.
P
Given point P on surface of sphere APC, AB an arc of great O ABC, and PA and PB quadrants.
To Prove P a pole of arc AB.
(PO is ± to OA and OB; then use § 400.)
THE SPHERE.
333
Prop. XII. Theorem.
579. The intersection of two spheres is a circle, whose centre is in the straight line joining the centres of the spheres, and whose plane is perpendicular to that line.
Given two intersecting spheres.
To Prove their intersection a O, whose centre is in the line joining the centres of the spheres, and whose plane is ± to this line.
Proof. Let 0 and 0' be the centres of two ©, whose common chord is AB; draw line 00', intersecting ^5 at C.
Then, 00' bisects AB at rt. A. (§ 178)
If we revolve the entire figure about 00' as an axis, the (D will generate spheres whose centres are 0 and 0'. (§ 565)
And AC will generate a O ± 00', whose centre is C, which is the intersection of the two spheres. (§ 402)
Prop. XIII. Theorem.
580. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.
C
(The proof is left to the pupil ; compare § 169.)
334 SOLID GEOMETRY.— BOOK VIII.
581. Cor. (Converse of Prop. XIII.) A j)lane tangent to a sphere is perpendicular' to the radius drawn to the 2)oint of contact. (Fig. of Prop. XIII.)
(The proof is left to the pupil ; compare § 170.)
Prop. XIV. Theorem.
582. Through four poi7its, not in the same plane, a spherical surface can he made to pass, and hut one.
A
C
Given A, B, C, and D points not in the same plane.
To Prove that a spherical surface can be passed through A, B, C, and D, and but one.
Proof. Pass planes through A, B, C, and D, forming tetraedron ABCD, and let K be the middle point of CD.
Draw lines KE and KF in faces ACD and BCD, respec- tively, ± CD ; and let E and F be the centres of the cir- cumscribed © of A ACD and BCD, respectively. (§ 222)
Then plane EKF is ± CD. (§ 400)
Draw line EG 1. ACD, and line FH1.BCD; then EG and FH lie in plane EKF. (§ 439).
Then EG and FH must meet at some point 0, unless they are II ; this cannot be unless ACD and BCD are in the same plane, which is contrary to the hyp. (§ 418)
Now 0, being in EG, is equally distant from A, C, and D; and being in FH, is equally distant from B, C, and D.
(§ 406, I)
Then 0 is equally distant from A, B, C, and D ; and a spherical surface described with 0 as a centre, and OA as a radius, will pass through A, B, C, and Z).
THE SPHERE. ggc
Now the centre of any spherical surface passing through A, B, C, and D must be in each of the Js EG and FH.
Then as EG and FH intersect in but one point, only one spherical surface can be passed through A, B, C, and D.
583. Defs. The angle between two intersecting curves is the angle between tangents to the curves at their point of intersection.
A spherical angle is the angle between two intersecting arcs of great circles.
Prop. XV. Theorem.
584. A spherical ayigle is measured by an arc of a great circle having its vertex as a pole, included betioeen its sides produced if necessary.
A
C
Given ABC and AB'C arcs of great (D on the surface of sphere AC, lines AD and AD' tangent to ABC and A'BC, respectively, and BB' an arc of a great 0 having ^ as a pole, included between arcs ABC and AB'C.
To Prove that A DAD' is measured by arc BB'.
Proof. Let 0 be the centre of the sphere, and draw diameter AOC and lines OB and OB'.
Now, arcs AB and AB' are quadrants. (§ 577)
Whence, z^^05 and ^05' are rt. Zs. (?)
Therefore, OB II AD and OB' II AD'. (§§ 170, 54)
.-. ZDAD' = ZBOB'. (§426)
But Z BOB' is measured by arc BB'. (?)
Then, Z DAD' is measured by arc BB'.
336 SOLID GEOMETRY. —BOOK VIII.
585. Cor. I. (Fig. of Prop. XV.) Plane BOB' is ± OA.
(§ 400) Then planes ABC and BOB' are ±. (§ 441)
Now a tangent to arc AB at jB is J_ BOB'. (§ 439)
Then it is J_ to a tangent to arc BB' at B. (§ 398)
Then, spherical Z ABB is a rt. Z. (§ 583)
That is, an arc of a great circle drawn from the pole of a great circle is perpendicular to its circumference.
586. Cor. II. The angle between two arcs of great circles is the plane angle of the diedral angle between their planes.
(§ 429)
SPHERICAL POLYGONS AND SPHERICAL PYRAMIDS. Definitions.
587. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles; as ABCD.
The bounding arcs are called the sides of the spherical polygon, and are usually measured in degrees.
The angles of the spherical polygon are the spherical angles (§ 583) between the adjacent sides, and their verti- ces are called the vertices of the spherical polygon.
A diagonal of a spherical polygon is an arc of a great circle joining any two vertices which are not consecutive.
A spherical triangle is a spherical polygon of three sides.
A spherical triangle is called isosceles when it has two sides equal ; equilateral when all its sides are equal ; and right-angled when it has a right angle.
588. The planes of the sides of a spherical polygon form a polyedral angle, whose vertex is the centre of the sphere, and whose face angles are measured by the sides of the spherical polygon (§ 192).
I
THE SPHERE.
337
Thus, in the figure of § 587, the planes of the sides of the spherical polygon form a polyedral angle, 0-ABCD, whose face AAOB, BOC, etc., are measured by arcs AB, BC, etc., respectively.
A spherical polygon is called convex when the polyedral angle formed by the planes of its sides is convex (§ 453).
589. A spherical pyramid is a solid bounded by a spherical polygon and the planes of its sides; as 0-ABCD, figure of § 587.
The centre of the sphere is called the vertex of the spheri- cal pyramid, and the spherical polygon the base.
Two spherical pyramids are equal when their bases are
For they can evidently be applied one to the other so as to coincide throughout.
590. If circumferences of great circles be drawn with the vertices of a spherical triangle as poles, they divide the surface of the sphere into eight spherical triangles.
Thus, if circumference B'C'B" be drawn with vertex A of spherical A ABC as a pole, circumference A'C'A" with 5 as a pole, and circum- ference A'B"A"B' with C as a pole, the surface of the sphere is divided into eight spherical A; A'B'C, A'B"C', A"B'C', and A"B"C' on the hemisphere represented in the figure, the others on the oppo- site hemisphere.
Of these eight spherical A, one is called the polar triangle of ABC, and is determined as follows :
Of the intersections, A' and A", of circumferences drawn with B and C as poles, that which is nearer (§ 573) to A, i.e., ^', is a vertex of the polar triangle; and similarly for the other intersections.
Thus, A'B'C is the polar A of ABC.
338 SOLID GEOMETRY.— BOOK VIII.
591. Two spherical polygons, on the same or equal spheres, are said to be symmetrical when the sides and an- gles of one are equal, respectively, to the sides and angles of the other, if the equal parts occur in the reverse order.
Thus, if spherical A ABC and A'B'C, on the same or equal spheres, have sides AB, BG, and CA equal, A^C' ^i
respectively, to sides A'B', B'C, and B^ ^C
CA', and A A, B, and C to A A', B', and C, and the equal parts occur in the reverse order, the A are symmetrical.
It is evident that, in general, two symmetrical spherical polygons cannot be placed so as to coincide throughout.
Prop. XVI. Theorem.
592. If one spherical triangle is the polar triangle of an- other, then the second spherical triangle is the polar triangle of the first.
A'
Given A'B'C the polar A of spherical A ABC; A, B, and C being the poles of arcs B'C, CA', and A'B', respectively.
To Prove ABC the polar A of spherical A A'B'C.
Proof. B is the pole of arc A'C.
Whence, A' lies at a quadrant's distance from B. (§ 577)
Again, C is the pole of arc A'B'.
Whence, A' lies at a quadrant's distance from C.
Therefore, A' is the pole of arc BC. (§ 578)
Similarly, B' is the pole of arc CA, and C of arc AB.
Then, ABC is the polar A of A'B'C.
THE SPHERE. 33^
For of the two intersections of the circumferences having 5' and C, respectively, as poles, A is the nearer to A!) and similarly for the other vertices. (§ 59O)
Note. Two spherical triangles, each of which is the polar triangle of the other, are called polar triangles.
Prop. XVII. Theorem.
593. In two polar triangles, each angle of one is measured by the supplement of that side of the other of which it is the pole.
X
JD
Given A, B, C, A', B', and C the A, expressed in degrees, of polar A ABO and A'B'C ; A being the pole of B'O, B of C'A', C of A'B\ A' of BC, B' of CA, and C of AB.
Let sides BC, CA, AB, B'C, C'A', and A'B', expressed in degrees, be denoted by a, b, c, a', b', and c', respectively. To Prove
A = 180° - a', B = 180° -b', C = 180° - c', A' = 180° - a, B' = 180° -b, C" = 180° - c. Proof. Produce arcs J^ and AC to meet sltcB'C at D and E, respectively.
Since B' is the pole of arc AE, and C of arc AD, arcs B'E and CD are quadrants. (§ 577)
.•. arc 5'^ + arc CD =180°. Or, arc DE + arc B'C = 180°.
But since A is the pole of arc B'C, arc DE is the measure ofZA. (§584)
.'. A + a' = 180°, or ^ = 180° - a'. In like manner, the theorem may be proved for any Z of either A.
340 SOLID GEOMETRY. — BOOK VIII.
Prop. XVIII. Theorem.
594. Any side of a spherical triangle is less than the sum of the other two sides.
Given AB any side of spherical A ABC.
To Prove AB<AC + Ba
(By § 4.57,ZA0B<ZA0C-\-/:B0C', and these Zs are measured by sides AB, AC, and BC, respectively.)
Prop. XIX. Theorem.
595. The sum of the sides of a convex spherical polygon is less than 360°.
B
Given convex spherical polygon ABCD. To Prove AB + BC+CD + DA< 360°.
(By § 458, sum of AAOB, BOC, COD, and DOA is < 360°.)
Prop. XX. Theorem.
596. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles.
341
B'
a'
Given A, B, and C the A, expressed in degrees, of spheri- cal A ABC.
To Prove A-{-B+C> 180°, and < 540°. Proof. Let A'B'O be the polar A of spherical A ABC, A being the pole of B'O, B of CA', and C of A^B'.
Also, let sides B'C, C'A', and A'B', expressed in degrees, be denoted by a', b', and c', respectively. Then, A = 180° - a',
B = 180° - 6',
C = 180° - c'. (§ 593)
Adding these equations, we have
A + J5+ C= 540° - (a' + 6' + c'). (1)
.-. ^ + 5+C<540°. Again, a' + 6' + c' < 360°. (§ 595)
Whence, by (1), A + B + C> 180°.
597. Cor. A spherical triangle may have one, two, or tlwee right angles, or one, two, or three obtuse angles.
DEFINITIONS.
598. A spherical triangle having two right angles is called 2b bi-rectangular triangle, and one having three right angles a tri-rectangular triangle.
599. Two spherical polygons on the same sphere, or equal spheres, are said to be mutually equilateral, or mutu- ally equiangular, when the sides or angles of one are equal, respectively, to the homologous sides or angles of the other, whether taken in the same or in the reverse order.
342 SOLID GEOMETRY.— BOOK VIII.
Prop. XXI. Theo:^em.
600. If two spherical triangles on the same sphere, or equal spheres, are mutually equiangular, their polar triangles are mutually equilateral.
A' D'
Given ABC and DEF mutually equiangular spherical A on the same sphere, or equal spheres, A A and D being homologous; also, A'B'C the polar A of ABC, and D'E'F' of DEF, A being the pole of B'C, and D of E'F'.
To Prove A'B'C and D'E'F' mutually equilateral.
Proof. A A and D are measured by the supplements of sides B'C and E'F', respectively. (§593)
But by hyp., ZA = AD.
.-. B'C = E'F. (§ 31,2)
In like manner, any two homologous sides of A'B'C and D'E'F' may be proved equal.
Then, A'B'C and D'E'F' are mutually equilateral.
601. Cor. (Converse of Prop. XXI.) If two spherical tri- angles on the same sphere, or equal spheres, are mutually equilateral, their polar triangles are mutually equiangular.
(The proof is left to the pupil ; compare § 600.)
Prop. XXII. Theorem.
602. If two spherical triangles on the same sphere, or equal spheres, have two sides and the included angle of one equal, respectively, to tioo sides and the included angle of the other,
I. They are equal if the equal parts occur in the same order.
THE SPHERE.
343
II. They are symmetrical if the equal parts occur in the reverse order.
i>'
I. Given ABC and DEF spherical A on the same sphere, or equal spheres, having
AB = DE, AC = DF, and ZA = ZD',
the equal parts occurring in the same order.
To Prove A ABC = A DEF.
Proof. Superpose A ABC upon A DEF in such a way that Z A shall coincide with its equal Z D ; side AB fall- ing on side DE, and side AC on side DF.
Then, since AB = DE and AC = DF, point B will fall on point E, and point C on point F.
Whence, arc BC will coincide with arc EF. ■ (§ 572)
Hence, ABC and DEF coincide throughout, and are equal.
II. Given ABC and D'E'F' spherical A on the same sphere, or equal spheres, having
AB = D'E', AC = D'F', and Z ^ = Z Z)' ; the equal parts occurring in the reverse order. To Prove ABC and D'E'F' symmetrical. Proof. Let DEF be a spherical A on the same sphere, or an equal sphere, symmetrical to D'E'F, having DE = D'E', DF = D'F', and ZD = ZD'; the equal parts occurring in the reverse order. Then, in spherical A ABC and DEF, we have AB = DE, AC = DF, and ZA = Z.D', and the equal parts occur in the same order. (Ax. 1)
. •. A ABC = A DEF. (§ 602, I)
Therefore, A ABC is symmetrical to A D'E'F'.
344 SOLID GEOMETRY.— BOOK VIII.
Prop. XXIII. Theorem.
603. If two spherical triangles on the same sphere, or equal spheres, have a side and two adjacent angles of one equal, respectively, to a side and two adjacent angles of the other,
I. They are equal if the equal parts occur in the same order. II. They are symmetrical if the equal parts occur in the reverse order.
(The proof is left to the pupil ; compare § 602.)
Prop. XXIV. Theorem.
604. If two spherical triangles on the same sphere, or equal spheres, are mutually equilateral, they are mutually equiangular.
Given ABC and DEF mutually equilateral spherical A on equal spheres ; sides BC and EF being homologous.
To Prove ABC and DEF mutually equiangular.
Proof. Let 0 and 0' be the centres of the respective spheres, and draw lines OA, OB, OC, O'D, O'E, and O'F.
Now the triedral A 0-ABC and O'-DEF have their ho- mologous face A equal. (§ 192) .-. diedral Z 0A= diedral Z O'D. (§ 459)
But the Z between arcs AB and AC is the plane Z of
diedral Z OA, and the Z between arcs DE and DF is the
plane Z of diedral Z O'D. (§ 586)
. •• ZBAC = Z EDF. (§ 434)
In like manner, any two homologous A of ABC and DEF may be proved equal.
Whence, ABC and DEF are mutually equiangular.
THE SPHERE.
345
Note. The theorem may be proved in a similar manner when the given spherical ^ are on the same sphere.
605. Cor. If two spherical triangles on the same sphere, or equal spheres, are mutually equilateral,
1. They are equal if the equal parts occur in the same order.
2. They are symmetrical if the equal parts occur in the reverse order.
Prop. XXV. Theorem.
606. If two spherical triangles on the same sphere, or equal spheres, are mutually equiangular, they are mutually equi- lateral.
A' D'
Given ABC and DEF mutually equiangular spherical A on the same sphere, or equal spheres.
To Prove ABC and DEF mutually equilateral.
Proof. Let A'B'C be the polar A of ABC, and D'E'F of DEF.
Then, since ABC and DEF are mutually equiangular, A'B'C and D'E'F' are mutually equilateral. (§ 600)
Then A'B'C and D'E'F' are mutually equiangular.
(§ 604)
But ABC is the polar A of A'B'C, and DEF of D'E'F'.
(§ 592)
Then ABC and DEF are mutually equilateral. (§ 600)
607. Cor. I. If two spherical triangles on the same sphere, or equal spheres, are mutually equiangular,
1. They are equal if the equal parts occur in the same order.
2. They are symmetrical if the equal parts occur in the reverse order.
346
SOLID GEOMETRY.— BOOK VIII.
608. Cor. II. If three diameters of a sphere he so drawn that each is perpen- dicular to the other two, the plaries deter- mined by them divide the surface of the sphere into eight equal tri-rectangular tri- angles.
(Prove by § 607, 1. By § 585, each Z of each spherical A is a rt. Z.)
609. Cor. m. The surface of a sphere is eight times the surface of one of its tri-rectangular triangles.
Prop. XXVI. Theorem.
610. In an isosceles spherical triangle the angles opposite the equal sides are equal.
Given, in spherical A ABC, AB = AC. To Prove ZB = ZC.
Proof. Draw AD an arc of a great O, bisecting side BC atZ).
In spherical A ABD and ACD, AD = AD. Also, AB = AC and BD = CD.
Then, ABD and ACD are mutually equiangular. (§ 604) .-. ZB = ZC.
611. Cor. I. An isosceles spherical triangle is equal to the spherical triangle which is symmetrical to it.
Por the equal parts occur in the same order.
612. Cor. n. (Converse of Prop. XXVI.) If two angles of a spherical triangle are equal, the sides opposite are equal.
THE SPHERE.
347
Given, in spherical A ABC, Z.B = Z C.
To Prove AB = AC.
Proof. Let A'B'C be the polar A of ABC', B being the pole of A'O, and C of A^B\ IB
Then, A'C is the sup. of Z B, and A^B' ^
^^ ^ ^' (§ 593)
.-. A'C' = A'B'. (§31,2)
••. ^B' = Z C. (§ 610)
But ABC is the polar A of A'B'C; B' being the pole of AC, and C of AB. (§ 592)
Then AB is the sup. of Z C, and ^0 of Z B'. (?)
.-. ^^ = ^(7. . (?)
Prop. XXVII. Theorem.
613. If two angles of a spherical triangle are unequal, the sides opposite are unequal, and the greater side lies opposite the greater angle.
Given, in spherical A ABC, Z ABC > Z C. To Prove AC > AB.
(Prove by a method analogous to that of § 99. Draw BD an arc of a great O meeting AC at D, and making Z CBD equal to Z C.)
614. Cor. (Converse of Prop. XXVII.) If two sides of a spherical triangle are unequal, the angles opposite are unequal, and the greater angle lies opposite the greater side.
(Prove by Beductio ad Absurdum.)
348 SOLID GEOMETRY. —BOOK VIII.
Prop. XXVIII. Theorem.
615. The shortest line on the surface of a sphere between two given points is the arc of a great circle, not greater than a semi-cii'cumference, which joins the points.
Given points A and B on the surface of a sphere, and AB an arc of a great O, not > a semi-circumference.
To Prove AB the shortest line on the surface of the sphere between A and B.
Proof. Let C be any point in arc AB.
Let' DCF and ECG be arcs of small © with A and B, respectively, as poles, and AC and BC as polar distances.
Now arcs DCF and ECG have only point C in common.
For let F be any other point in arc DCF, and draw arcs of great CD AF and BF.
.'. AF=Aa (§ 575)
But, AF+BF>AC-\- BC. (§ 594)
Subtracting arc AF from the first member of the inequal- ity, and its equal arc AC from the second member,
BF > BC, or BF > BG. (§ 575)
Whence, F lies without small O ECG, and arcs DCF and ECG have only point C in common.
We will next prove that the shortest line on the surface of the sphere from ^ to jB must pass through C
Let ADEB be any line on the surface of the sphere between A and B, not passing through C, and cutting arcs DCF and ECG at D and E, respectively.
Then, whatever the nature of line AD, it is evident that an equal line can be drawn from A to C.
THE SPHERE. 349
In like manner, whatever the nature of line BE, an equal line can be drawn from B to C.
Hence, a line can be drawn from A to B passing through C, equal to the sum of lines AD and BE, and consequently < line ADEB by the portion DE.
Therefore, no line which does not pass through C can be the shortest line between A and B.
But by hyp., C is any point in arc AB.
Hence, the shortest line from A to B must pass through every point of AB.
Then, the arc of a great O AB is the shortest line on the surface of the sphere between A and B.
EXERCISES.
2. If the sides of a spherical triangle are 77°, 123°, and 95°, how many degrees are there in each angle of its polar triangle ?
3. If the angles of a spherical triangle are 86°, 131°, and 68°, how many degrees are there in each side of its polar triangle ?
MEASUREMENT OF SPHERICAL POLYGONS. Definitions.
616. A lune is a portion of the surface of a sphere bounded by two semi-circumfer- ences of great circles ; as ACBD.
The angle of the lune is the angle between its bounding arcs.
617. A spherical wedge is a solid bounded by a lune and the planes of its bounding arcs.
The lune is called the base of the spherical wedge.
618. It is evident that two lunes on the same sphere, or equal spheres, are equal when their angles are equal
619. It is evident that two spherical ivedges in the same sphere, or equal spheres, are equal when the angles of the lanes which form their bases ai'e equal.
350 SOLID GEOMETRY. — BOOK VIII.
Prop. XXIX. Theorem.
620. The sphencal tnangles corresponding to a pair of vertical triedral angles are symmetrical.
Given AOA', BOB', and COC diameters of sphere AC; also, the planes determined by them, intersecting the sur- face in circumferences ABA'B', BCB'C, and CACA'.
To Prove spherical A ABC and A'B'C symmetrical.
Proof. A AOB, BOC, and CO A are equal, respectively, to AA'OB', B'OC, and C'OA'. (§ 40)
Then, AB = A'B', BC = B'C, and CA = CA'. (§ 192)
But the equal parts of ABC and A'B'C occur in the reverse order.
Whence, ABC and A'B'C are symmetrical. (§ 605, 2)
Prop. XXX. Theorem.
621. Two spherical triangles corresponding to a pair of vertical triedral angles are equivalent.
A
Given AOA, BOB', and COC diameters of sphere AB; also, the planes determined by them, intersecting the sur- face in arcs AB, BC, CA, A'B', B'C, and CA'.
THE SPHERE. 351
To Prove area ABC = area A'B'C.
Proof. Let P be the pole of the small O passing through points A, B, and C, and draw arcs of great ® PA, PB, and PC.
.-. PA = PB = PC. (§575)
Draw the diameter of the sphere PP', and the arcs of great ®P'A', P'B', and PC; then, spherical A P^S and P'A'B' are symmetrical. (§ 620)
But spherical A PAB is isosceles.
.-. A PAB = A P'A'B'. (§ 611)
In like manner,
A PBC = A PB'C and A PCA = A P'C'A'. Then the sum of the areas of A PAB, PBC, and PCA equals the sum of the areas of PA'B, P'B'C, and PCA'. .'. area ABC = area A'B'O.
622. Sch. If P and P' fall without spherical A ABC and A'B'C, we should take the sum of the areas of two isos- celes spherical A, diminished by the area of a third.
623. Cor. I. Two symmetrical spherical triangles are equiv- alent.
624. Cor. II. Spherical pyramids 0-APB, 0-BPC, and 0-CPA are equal, respectively, to spherical pyramids 0-A'PB', O-B'PC, and 0-CPA'. (§ 589)
.-. vol. 0-ABC= vol. 0-A'B'C. Whence, the spherical pyramids corresponding to a pair of vertical triedral angles are equivalent.
EXERCISES.
4. The sura of the angles of a spherical hexagon is greater than 8, and less than 12, right angles. (§ 596.)
5. The sum of the angles of a spherical polygon of n sides is greater than 2 w — 4, and less than 2 n^ right angles.
6. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle point of the base, is perpendicular to the base, and bisects the vertical angle.
352
SOLID GEOMETRY.— BOOK VIII.
Prop. XXXI. Theorem.
625. Two lunes on the same sphere, or equal spheres, are to each other as their angles.
Note. The word ^^lune,'^ in the above statement, signifies the area of the lune.
Case I. When the angles are commensurable.
Given ACBD and AGBE lunes on sphere AB, having their A CAD and CAE commensurable.
To Prove
ACBD Z. CAD
ACBE Z CAE
Proof. Let /. CAa be a common measure of A CAD and CAE, and let it be contained 5 times in Z CAD, and 3 times in Z CAE.
ACAD 5
A CAE 3
(1)
Producing the arcs of division of Z CAD to B, lune ACBD will be divided into 5 parts, and lune ACBE into 3 parts, all of which parts will be equal. (§ 618)
ACBD 5
Prom (1) and (2),
ACBE 3
ACBD ^ Z CAD ACBE A CAE
(2)
(?)
Note. The theorem may be proved in a similar manner when the given lunes are on equal spheres.
THE SPHERE. 353
Case II. When the angles are incommenaurahle.
(Prove as in §§ 189 or 244. Let Z CAD be divided into any number of equal parts, and apply one of these parts to Z CAE as a unit of measure.)
626. Cor. I. The surface of a lune is to the surface of the sphere as the angle of the lune is to four right angles.
For the surface of a sphere may be regarded as a lune whose Z is equal to 4 rt. A.
627. Cor. n. If the unit of measure for angles is the right angle, the area of a lune is equal to twice its angle, multiplied by the area of a tri-rectangidar triangle.
Given L the area of a lune ; A the numerical measure of its Z referred to a rt. Z as the unit of measure ; and T the area of a tri-rectangular A.
To Prove L = 2A x T.
Proof. The area of the surface of the sphere is 8 T.
(§ 609)
.-. -^ = A (§625)
8r 4 ^ ^
.: L = 4x^T=2AxT. 4
628. Sch. I. Let it be required to find the area of a lune whose Z is 50°, on a sphere the area of whose surface is 72.
The Z of the lune referred to a rt. Z as the unit of measure is f ; and T is ^ of 72, or 9.
Then the area of the lune is 2 x 4 X 9, or 10.
354 SOLID GEOMETRY.— BOOK VIII.
629. Def . A tri-rectangular pyramid is a spherical pyra- mid whose base is a tri-rectangular triangle.
630. Sch. n. It may be proved, as in § 625, that
Two spherical tvedges in the same sphere, or equal spheres, are to each other as the angles of the lunes which form their bases.
(The proof is left to the pupil ; see § 619.)
631. Sch. III. It may be proved that
If the unit of measure for angles is the right angle, the volume of a spherical wedge is equal to twice the angle of the lune which forms its base, multiplied by the volume of a tri- rectangular pyramid.
(The proof is left to the pupil ; see §§ 626 and 627.)
632. Def. The sphencal excess of a spherical triangle is the excess of the sum of its angles above 180° (§ 596).
Thus, if the zi of a spherical A are 65°, 80°, and 95°, its spherical excess is 65° + 80° + 95° - 180°, or 60°.
Prop. XXXII. Theorem.
633. If the unit of measure for angles is the right angle, the area of a spherical triangle is equal to its spherical excess, multiplied by the area of a tri-rectangular triangle.
Given A, B, and C the numerical measures of the A of spherical A ABC, referred to a rt. Z as the unit of measure, and T the area of a tri-rectangular A.
To Prove area ABC = (^ + 5 -f O - 2) x T.
THE SPHERE. 355
Proof, Complete circumferences ABA'B'f ACA'C, and BCB'C, and draw diameters AA\ BB', and CC. Then, since ABA'C is a lune whose Z is A, we have
area ABC + area ABC = 2 ^ x T (§ 626). (1)
And since BAB'C is a lune whose Z is J5,
area ^5(7 + area AB'C =2Bx T. (2)
Again, area A'B'C = area ^J5(7. (§ 620)
Adding area ABC to both members, we have
area ABC -\- area AB'C= area of lune CBCA
= 2CxT. (3)
Adding (1), (2), and (3), and observing that the sum of the areas of A ABC, A'BC, AB'C, and A'B'C is equal to the area of the surface of a hemisphere, or 4 T, we have 2 area ABC + 4: T= {2 A + 2 B-\-2 C) x T. .-. area ABC-\-2 T={A + B+C) x T.
.-. area ABC = (A-i- B + C -2) x T.
634. Sch. I. Let it be required to find the area of a spherical A whose A are 105°, 80°, and 95°, on a sphere the area of whose surface is 144.
The spherical excess of the spherical A is 100°, or J^ re- ferred to a rt. Z as the unit of measure ; and the area of a tri-rectangular A is |^ of 144, or 18.
Then the area of the spherical A is y- x 18, or 20.
635. Sch. n. It may be proved, as in § 633, that
If the unit of measure for angles is the right angle, the volume of a triangular spherical pyramid is equal to the spherical excess of its base, multiplied by the volume of a tri-rectangidar pyramid.
(The proof is left to the pupil ; see §§ 624 and 630.)
EXERCISES.
7. What is the volume of a spherical wedge the angle of whose base is 127° 30', if the volume of the sphere is 112 ?
8. In figure of Prop. XVII., prove A' = 180° - a.
356 SOLID GEOMETRY, —BOOK VIII.
Prop. XXXIII. Theorem.
636. If the unit of measure for angles is the right angle, the area of any spherical polygon is equal to the sum of its angles, diminished by as many times two right angles as the figure has sides less two, multiplied by the area of a tri- rectangular triangle.
Given K the area of any spherical polygon, n the number of its sides, s the sum of its A referred to a rt. Z as the unit of measure, and T the area of a tri-rectangular A.
To Prove K= [s - 2 (ii - 2)] x T.
Proof. The spherical polygon can be divided into ri — 2 spherical A by drawing diagonals from any vertex.
Now, if the unit of measure for A is the rt. Z, the area of each spherical A is equal to the sum of its A, less 2 rt. A, multiplied by T. (§ 633)
Hence, if the unit of measure for A is the rt. Z, the sum of the areas of the spherical A is equal to the sum of their A, diminished by n — 2 times 2 rt. A, multiplied by T.
But the sum of the A of the spherical A is equal to the sum of the A of the spherical polygon.
Whence, K=[s-2 (n - 2)] x T.
637. Sch. It may be proved, as in § 636, that If the unit of measure for angles is the right angle, the volume of any spherical pyramid is equal to the sum of the angles of its base, diminished by as many times two right angles as the base has sides less two, multiplied by the volume of a tri-recta7igular pyramid. (The proof is left to the pupil.)
THE SPHERE. 357
EXERCISES.
9. The area of a lune is 28^. If the area of the surface of the sphere is 120, what is the angle of the lune ?
10. Find the area of a spherical triangle whose angles are 103°, 112°, and 127°, on a sphere the area of whose surface is 160.
11. Find the volume of a triangular spherical pyramid the angles of whose base are 02°, 119°, and 134°; the volume of the sphere being 192.
12. What is the ratio of the areas of two spherical triangles on the same sphere whose angles are 94", 135°, and 140°, and 87°, 105°, and 118°, respectively ?
13. The area of a spherical triangle, two of whose angles are 78° and 99°, is 34 1. If the area of the surface of the sphere is 234, what is the other angle ?
14. The volume of a triangular spherical pyramid, the angles of whose base are 105°, 126°, and 147°, is 60^ ; what is the volume of the sphere ?
15. The sides opposite the equal angles of a bi- rectangular triangle are quadrants. (§ 442.)
16. The sides of a spherical triangle, on a sphere
the area of whose surface is 156, are 44°, 63°, and 97°. Find the area of its polar triangle.
17. Find the area of a spherical hexagon whose angles are 120°, 139°, 148°, 155°, 162°, and 167°, on a sphere the area of whose surface is 280.
18. Find the volume of a pentagonal spherical pyramid the angles of whose base are 109°, 128°, 137°, 153°, and 158° ; the volume of the sphere being 180.
19. The volume of a quadrangular spherical pyramid, the angles of whose base are 110°, 122°, 135°, and 146°, is 12f ; what is the volume of the sphere ?
20. The area of a spherical pentagon, four of whose angles are 112°, 131°, 138°, and 168°, is 27. If the area of the surface of the sphere is 120, what is the other angle ?
21. If two straight lines are tangent to a sphere at the same point, their plane is t,angent to the sphere. (§ 400.)
358 SOLID GEOMETRY. —BOOK VIII.
22. The sum of the arcs of great circles drawn from any point within a spherical triangle to the extremities of any side, is less than the sum of the other two sides of the triangle.
(Compare § 48.)
23. How many degrees are there in the polar distance of a circle, whose plane is 5\/2 units from the centre of the sphere, the diameter of the sphere being 20 units ?
(The radius of the O is a leg of a rt. A, whose hypotenuse is the radius of the sphere, and whose other leg is the distance from its centre to the plane of the O.)
24. The chord of the polar distance of a circle of a sphere is 6. If the radius of the sphere is 5, what is the radius of the circle ?
D
25. If side AB oi spherical triangle ABC is a / quadrant, and side BC less than a quadrant, prove Z^ less than 90°.
,^^,
26. The polar distance of a circle of a sphere is 60°. If the diameter of the circle is 6, find the diameter of the sphere, and the distance of the circle from its centre.
(Represent radius of sphere by 2 x.)
27. Any point in the arc of a great circle bisecting a spherical angle is equally distant (§ 573) from the sides of the angle.
(To prove arc FM = arc PiV. Let U be a pole of arc AB, and F of arc BC. Spherical ^"^ A BFE and BFF are symmetrical by § 602, IL, and FE = FF.)
28. A point on the surface of a sphere, equally distant from the sides of a spherical angle, lies in the arc of a great circle bisecting the angle.
(Fig. of Ex. 27. To prove Z ABF = Z CBF. Spherical A BFE and BFF are symmetrical by § 605, 2.)
29. The arcs of great circles bisecting the angles of a spherical triangle meet in a point equally distant from the sides of the triangle. (Exs. 27, 28, p. 358.)
30. A circle may be inscribed in any spherical triangle.
31. State and prove the theorem for spherical triangles analogous to Prop. IX., L, Book I.
THE SPHEllE. 359
32. State and prove the theorem for spherical triaugles analogous to Prop, v., Book I.
33. State and prove the theorem for spherical triangles analogous to Prop. L., Book I. (Ex. 32.)
34. If PA, PB, and PC are three equal arcs of great circles drawn from point P to the circumference of great circle ABC, prove P a pole of ABC.
(PA and PB are quadrants by Ex. 15, p. 357.)
35. The spherical polygons corresponding to a pair of vertical poly- edral angles are symmetrical. (§ 456.)
36. A sphere may be inscribed in, or circumscribed about, any tetraedron. (Ex. 73, Book VII.)
37. What is the locus of points in space at a given distance from a given straight line ?
38. Equal small circles of a sphere are equally distant from the centre.
39. State and prove the converse of Ex. 38.
40. The less of two small circles of a sphere is at the greater dis- tance from the centre.
41. State and prove the converse of Ex, 40.
42. What is the locus of points on the surface of a sphere equally distant from the sides of a spherical angle ?
43. If two spheres are tangent to the same plane at the same point, the straight line joining their centres passes through the point of contact,
44. The distance between the centres of two spheres whose radii are 25 and 17, respectively, is 28. Find the diameter of their circle of intersection, and its distance from the centre of each sphere.
45. If a polyedron be circumscribed about each of two equal spheres, the volumes of the polyedrons are to each other as the areas of their surfaces.
(Find the volume of each polyedron by dividing it into pyramids.)
46. Either angle of a spherical triangle is greater than the differ- ence between 180° and the sum of the other two angles.
(Fig. of Prop. XX, To prove ZA>1S0°-(ZB + ZC), or XZB + ZC) - 180°, according asZ^+Z(7is<or> 180°. In the latter case, A'C 4- A'B'>B'C'; then use § 593.)
Book IX.
MEASUREMENT OF THE CYLINDER, CONE, AND SPHERE.
THE CYLINDER. Definitions.
638. The lateral area of a cylinder is the area of its lateral surface.
A right section of a cylinder is a section made by a plane perpendicular to the elements of its lateral surface.
639. A prism is said to be inscribed in a cylinder when its lateral edges are elements of the cylindrical surface.
In this case, the bases of the prism are inscribed in the bases of the cylinder.
A prism is said to be circumscribed about a cylinder when its lateral faces are tangent to the cylinder, and its bases lie in the same planes with the bases of the cylinder.
In this case, the bases of the prism are circumscribed about the bases of the cylinder.
640. It follows from § 363 that ^^^
If a prism whose base is a regular KTT^ /~/V
polygon be inscribed in, or circum- / /p^zzzJ-^/ItJ
scribed about, a circular cylinder I I I ffj/ I /
(§ 540), and the number of its faces 11 / ' I !
be indefinitely increased, / / //'..v/.-/::^^-./^/ /
1. The lateral area of the prism \['f J i/i^.
approaches the lateral area of the cyl- ^^■-/-^^ii>'*^^^'^^^
inder as a limit. n^-""^
MEASUREMENT OF THE CYLINDER. 361
2. The volume of the prism approaches the volume of the cylinder as a limit.
3. The perimeter of a right section of the prism approaches the perimeter of a right section of the cylinder as a limit*
Prop. I. Theorem.
641. The lateral area of a circular cylinder is equal to the perimeter of a right section multiplied by an element of the lateral surface.
Given S the lateral area, P the perimeter of a rt. section, and E an element of the lateral surface, of a circular cylinder.
To Prove S = P x E.
Proof. Inscribe in the cylinder a prism whose base is a regular polygon, and let S' denote its lateral area, and P the perimeter of a rt. section.
Then, since the lateral edge of the prism is E,
S' = P' X E. (§ 484)
Now let the number of faces of the prism be indefinitely increased.
Then, S' approaches the limit S,
and P xE approaches the limit P x E. (^ 640, 1, 3)
By the Theorem of Limits, these limits are equal. (§ 188) .-. S = PxE.
* For rigorous proofs of these statements, see Appendix, p. 386.
362 SOLID GEOMETRY. — BOOK IX.
642. Cor. I. Tlie lateral area of a cylinder of revolution is equal to the circumference of its base multiplied by its altitude.
643. Cor. II. If S denotes the lateral area, T the total area, H the altitude, and R the radius of the base, of a cylinder of revolution,
S = 27rEH. (§368)
And, T=2 ttRH + 2 ttR^ (§ 371) = 2 irR{H + R),
Prop. II. Theorem.
644. The volume of a circular cylinder is equal to the prod- uct of its base and altitude.
Given V the volume, B the area of the base, and ^ the altitude, of a circular cylinder.
To Prove V=BxH.
Proof. Inscribe in the cylinder a prism whose base is a regular polygon, and let V denote its Volume, and B' the area of its base.
Then, since the altitude of the prism is H,
r = B'xH. (§499)
Now let the number of faces of the prism be indefinitely increased.
Then, V approaches the limit V. (§ 640, 2)
And, B' X H approaches the limit B x H. (§ 363, II) .-. V=BxH. (?)
MEASUREMENT OF THE CYLINDER.
363
645. Cor. If V denotes the volume, H the altitude, and li the radius of the base, of a circular cylinder,
V= ttH^H. (?)
Prop. III. Theorem.
646. The lateral or total areas of two similar cylinders of revolution (§ 550) are to each other as the squares of their altitudes, or as the squares of the radii of their bases ; and their volumes are to each other as the cubes of their altitudes^ or as the cubes of the radii of their bases.
Given S and s the lateral areas, T and t the total areas, V and V the volumes, H and h the altitudes, and R and r the radii of the bases, of two similar cylinders of revolution.
To Prove ^ = ^=^' = ^, and ^=^ = ^-
Proof. Since the generating rectangles are similar,
h r
H-\-R
(§ 253, 2)
h + r
(§ 240)
and -^ =
S^2jRH
s 2 Trrh T^27rR{H+R) t 2Trr{h-\-r)
V ttR'H
V irT^h
W
(§643)=^x5 = f = — r r r^ h^
(§ 643)
R^R
r r
(§645)=^x^
R^
El
' h^'
364 SOLID GEOMETRY.— BOOK IX.
EXERCISES.
1. Find the lateral area, total area, and volume of a cylinder of revolution, the diameter of whose base is 18, and whose altitude is 16.
2. The radii of the bases of two similar cylinders of revolution are 24 and 44, respectively. If the lateral area of the first cylinder is 720, what is the lateral area of the second ?
3. Find the altitude and diameter of the base of a cylinder of revolution, whose lateral area is 168 tt and volume 504 tt.
(Substitute the given values in the formulse of §§ 643 and 646, and solve the resulting equations.)
4. Find the volume of a cylinder of revolution, whose total area is 170 IT and altitude 12.
5. How many cubic feet of metal are there in a hollow cylindrical tube 18 ft. long, whose outer diameter is 8 in., and thickness 1 in.?
(Find the difference of the volumes of two cylinders of revolution. TT =3.1416.)
6. The cross-section of a tunnel, 2^ miles in length, is in the form of a rectangle 6 yd. wide and 4 yd. high, surmounted by a semicircle whose diameter is equal to the width of the rectangle ; how many cu. yd. of material were taken out in its construction ? (t = 3.1416.)
7. The volume of a cylinder of revolution is equal to its lateral area multiplied by one-half the radius of its base.
THE CONE. DEFINITIONS.
647. The lateral area of a cone, or frustum of a cone, is the area of its lateral surface.
The slant height of a cone of revolution is the straight line drawn from the vertex to any point in the circumfer- ence of the base.
The slant height of a frustum of a cone of revolution is that portion of the slant height of the cone included between the bases of the frustum.
648. A pyramid is said to be inscribed in a cone when its lateral edges are elements of the conical surface ; the base of the pyramid is inscribed in the base of the cone, and its vertex coincides with the vertex of the cone.
MEASUREMENT OF THE CONE. 3^5
A pyramid is said to be circmnscribed about a cone when its lateral faces are tangent to the cone, and its base lies in the same plane with the base of the cone ; the base of the pyramid is circumscribed about the base of the cone, and its vertex coincides with the vertex of the cone.
649. A frustum of a pyramid is said to be inscribed in a fi'iistimi of a cone when its lateral edges are elements of the lateral surface of the frustum of the cone.
In this case, the bases of the frustum of the pyramid are inscribed in the bases of the frustum of the cone.
A frustum of a pyramid is said to be circumscribed about a frustum of a cone when its lateral faces are tangent to the frustum of the cone, and its bases lie in the same planes with the bases of the frustum of the cone.
In this case, the bases of the frustum of the pyramid are circumscribed about the bases of the frustum of the cone.
650. It follows from § 363 that
If a pyramid whose base is a regidar polygon be inscribed in, or circumscribed about, a circular cone
(§ 553), and the number of its faces be in- /fnK
definitely increased, /^ lv\
1. The lateral area of the pyramid ap- y/'y M I • • \ proaches the lateral area of the cone as a /^J-if^--h^^^^:\ \ limit. X^/ / I ' A>
2. The volume of the pyramid approaches n^^^^^^^sJ^^*^^ the volume of the cone as a limit*
651 It follows from the above that
If a frustum of a pyramid whose base is a regular polygon be inscribed in, or circumscribed about, a frustum of a circu- lar cone, and the number of its faces be indefinitely increased,
1. The lateral area of the frustum of the pyramid ap- proaches the lateral area of the frustum of the cone as a limit.
2. Tlie volume of the frustum of the p?/r(xmid approaches the volume of the frxistum of the cone as a limit.
* For rigorous proofs of these statements, see Appendix, p. 388.
366 SOLID GEOMETRY. — BOOK IX.
Prop. IV. Theorem.
652. The lateral area of a cone of revolution is eqUal to the circumference of its base, multiplied by one-half its slant height.
Given S the lateral area, (7 the circumference of the base, and L the slant height, of a cone of revolution.
To Prove S = C x \ L.
Proof. Circumscribe about the cone a regular pyramid; let aS' denote its lateral area, and C" the perimeter of its base.
Now the sides of the base of the pyramid are bisected at their points of contact with the base of the cone. (§ 174)
Then, the slant height of the pyramid is the same as the
slant height of the cone. (§ 508)
.-. >S"=C"xiX. (§ 512)
Now let the number of faces of the pyramid be indefi- nitely increased.
Then, /S" approaches the limit S. (§ 650, 1)
And C x^L approaches the limit C x \ L. (§ 363, I) .'.S=Cx\L. (?)
653. Cor. If S denotes the lateral area, T the total area, L the slant height, and R the radius of the base, of a cone of revolution,
S = 2'!rItx\L{?) =-kUL. And, T = ttRL -f ttR'' (?) = 7rR{L-\- R).
Prop. V. Theorem.
654. The volume of a circular cone is equal to the area of its base, multiplied by one-third its altitude.
MEASUREMENT OF THE CONE. 367
Given V tlie volume, B the area of the base, and if the altitude, of a circular cone. To Prove V=BxiH.
(Inscribe a pyramid whose base is a regular polygon.)
655. Cor. If F denotes the volume, H the altitude, and R the radius of the base, of a circular cone,
V=\.R'H. (?)
Prop. VI. Theorem.
656. Tlie lateral or total areas of tiuo similar cones of revo- lution are to each other as the squares of their slant heights, or as the squares of their altitudes, or as the squares of the radii of their bases; and their volumes are to each other as the cubes of their slant heights, or as the cubes of their altitudes, or as cubes of the radii of their bases.
Given S and s the lateral areas, T and t the total areas, V and V the volumes, L and I the slant heights, H and h the altitudes, and R and r the radii of the bases, of two similar cones of revolution (§ 555).
To Prove ^=^=i^ = ^'=^, and l=Il=El=^.
s t I- le 1^ V P ft" r"
368 SOLID GEOMETRY.— BOOK IX.
(The proof is left to the pupil ; compare § 646.)
Prop. VII. Theorem.
657. TJie lateral area of a frusttim of a cone of revolution is equal to the sum of the circumferences of its bases, multiplied by one-half its slant height.
Given S the lateral area, C and c the circumferences of the bases, and L the slant height, of a frustum of a cone of revolution.
To Prove S = (G + c) x\L,
Proof. Circumscribe about the frustum of the cone a frustum of a regular pyramid ; let >iS" denote its lateral area, and O and c' the perimeters of its bases.
Now the sides of the bases of the frustum of the pyra- mid are bisected at their points of contact with the bases of the frustum of the cone. (§ 174)
Then, the slant height of the frustum of the pyramid is the same as the slant height of the frustum of the cone.
(§ 508) .-. ^'=((7'H-c') x^L. (§513)
Now let the number of faces of the frustum of the pyramid be indefinitely increased.
Then, /S" approaches the limit S, (§ 651, 1)
and (C + c') X ^L approaches the limit {C -\-c) x\L.
(§ 363, I) ... S={C + c)x\L. (?)
MEASUREMENT OF THE CONE. 369
658. Cor. I. If S denotes the lateral area, L the slant height, and E and r the radii of the bases, of a frustum of a cone of revolution,
S = (2TrE + 2 Trr) xiL(?) =ir{R+ r)L.
659. Cor. II. We may write the first result of § 668
S = 2'jrK\{R + r)xL.
But, 2Tr X ^{R + r) is the circumference of a section equally distant from the bases. (§ 132)
Whence, the lateral area of a frustum of a coiie of revolu- tion is equal to the circumference of a section equally distant from its bases, multiplied by its slant height.
Prop. VIII. Theorem.
660. Tlie volume of a frustum of a circular cone is equal to the sum of its bases and a mean proportional between its bases, multiplied by one-third its altitude.
Given V the volume, B and b the areas of the bases, and H the altitude, of a frustum of a circular cone.
To Prove F= (jB -f & + VBxb) x\H.
(Inscribe a frustum of a pyramid whose base is a regular polygon. Then apply § 524.)
661. Cor. If V denotes the volume, H the altitude, and R and r the radii of the bases, of a frustum of a circular cone, B = ■7rR\ b = Trr", and Vslib = V^i^^V = irRr. (?) Then, F= (ttR^ + Tr?'^ + TrRr) X \H = ^TriR' -{■ r" + Rr) H.
370 SOLID GEOMETRY.— BOOK IX.
EXERCISES.
8. Find the lateral area, total area, and volume of a cone of revo- lution, the radius of whose base is 7, and whose slant height is 25.
9. Find the lateral area, total area, and volume of a frustum of a cone of revolution, the diameters of whose bases are 16 and 6, and whose altitude is 12.
10. The slant heights of two similar cones of revolution are 9 and 15, respectively. If the volume of the second cone is 625, what is the volume of the first ?
11. Find the volume of a cone of revolution, whose slant height is 29 and lateral area 580 tt.
12. Find the lateral area of a cone of revolution, whose volume is 320 TT and altitude 15.
13. The altitude of a cone of revolution is 27, and the radius of its base is 16. What is the diameter of the base of an equivalent cylinder of revolution, whose altitude is 16 ?
14. The area of the entire surface of a frustum of a cone of revolu- tion is 306 TT, and the radii of its bases are 11 and 5. Find its lateral area and volume.
15. The volume of a frustum of a cone of revolution is 6020 tt, its altitude is 60, and the radius of its lower base is 15. Find the radius of its upper base and its lateral area.
16. Find the altitude and lateral area of a cone of revolution, whose volume is 800 tt, and whose slant height is. to the diameter of its base as 13 to 10.
17. The total areas of two similar cylinders of revolution are 32 and 162, respectively. If the volume of the second cylinder is 1458, what is the volume of the first ?
(Let X and y denote the altitudes of the cylinders.)
18. The volumes of two similar cones of revolution are 343 and 512, respectively. If the lateral area of the first cone is 196, what is the lateral area of the second ?
19. A cubical piece of lead, the area of whose entire surface is 384 sq. in., is melted and formed into a cone of revolution, the radius of whose base is 12 in. Find the altitude of the cone.
20. A tapering hollow iron column, 1 in. thick, is 24 ft. long, 10 in. in outside diameter at one end, and 8 in. in diameter at the other; how many cubic inches of metal were used in its construction ?
(Find the difference of the volumes of the frustums of two cones of revolution, tt = 3.1416.)
MEASUREMENT OF THE SPHERE.
371
21. U the altitude of a cone of revolution is three-fourths the radius of its base, its volume is equal to its lateral area multiplied by one-fifth the radius of its base.
THE SPHERE. DEFINITIONS.
662. A zone is a portion of the surface of a sphere in- cluded between two parallel planes.
Tlie circumferences of the circles which bound the zone are called the bases, and the perpendicular distance between their planes the altitude.
A zone of one base is a zone one of whose bounding planes is tangent to the sphere.
A spherical segment is a portion of a sphere included be- tween two parallel planes.
The circles which bound it are called the bases, and the perpendicular distance between them the altitude.
A spherical segment of one base is a spherical segment one of whose bounding planes is tangent to the sphere.
663. If semicircle ACEB be revolved about diameter AB as an axis, and CD and EF are lines ± AB, arc CE generates a zone whose altitude is DF, figure CEFD a spherical segment whose altitude is DF, arc AC a zone of one base, and figure ACD a spherical segment of one base.
664. If a semicircle be revolved about its diameter as an axis, the solid generated by any sector of the semicircle is called a spherical sector.
Thus, if semicircle ACDB be revolved about diameter AB as an axis, sector OCD generates a spherical sector.
The zone generated by the arc of the sector is called the base of the spherical sector.
372
SOLID GEOMETRY — BOOK IX.
Prop. IX. Theorem.
665. Tlie area of the surface generated by the revolution of a straight line about a straight line in its plane, not parallel to and not intersecting it, as an axis, is equal to its projection on the axis, multiplied by the circumference of a circle, whose radius is the perpendicular erected at the middle point of the line and terminating in the axis.
Given str. line AB revolved about str. line FM in its plane, not II to' and not intersecting it, as an axis; lines AC and BD±FM, and EF the J_ erected at the middle point of AB terminating in FM.
To Prove area AB* = CD x 2 tt EF. (§ § 276, 368)
Proof. Draw line AG J_ BD, and line EH ± CD. The surface generated by AB is the lateral surface of a frustum of a cone of revolution, whose bases are generated by AC and BD.
.'. area AB = ABx2Tr EH. But A ABG and EFH are similar. AB^EF AG eh' .'. AB X EH= AG X EF
= CD X EF.
Substituting, we have
area AB==CD x2 7r EF.
(§ 659) (§ 262)
(?)
(§ 232)
(?)
* The expression " area AB'''' is used to denote the area of the $i(r- face generated by AB.
MEASUREMENT OF THE SPHERE. 373
Prop. X. Theorem.
666. If an isosceles triangle he revolved about a straight line in its plane, not parallel to its base, as an axis, which passes through its veHex ivithout intersecting its surface, the volume of the solid generated is equal to the area of the surface generated by the base, multiplied by one-third the altitude.
^0 Given isosceles A OAB revolved about str. line OF in its plane, not II to base AB, as an axis ; and line OC 1. AB. To Prove vol. OAB * = area AB x \ OC. Proof. Draw lines AD and BE 1. OF; and produce BA to meet OF a^t F.
Now, vol. OBF = vol. OBE + vol. BEF
= \^BE' xOE + \itBE^xEF {^Qb^)
= \ ttBE" X {OE + EF) = i ttBE X be X OF.
But BE X OF = OC X BF, for each expresses twice the
area of AOBF. (?)
.-. vol. OBF = 1 ttBE xOCx BF.
But ttBE X BF is the area of the surface generated by BF.
(§ 653) .-. vol. OBF= area BF x ^ OC. (1)
Similarly, vol. OAF = area AF x ^ OC. (2)
Subtracting (2) from (1), we have
vol. OAB = (area BF- area AF) x\OC = area AB x \ OC.
* The expression " vol, OAB " is used to denote the volume of the solid generated by OAB.
374 SOLID GEOMETRY.— BOOK IX.
Prop. XI. Thp:orem.
667. The area of a zone is equal to its altitude multiplied by the circumference of a great circle.
M
A'
C,
Given arc AB revolved about diameter OM as an axis, lines AA^ and BB' J_ OM, and R the radius of the arc. To Prove area of zone generated by AB = A'B' x 2 ttR. Proof. Divide arc AB into three equal arcs, AC, CD, and DB, and draw chords AC, CD, and DB.
Also, draw lines CC and DD'±OM, and line OE ± AC. .'. area^(7 = .4'(7' x^irOE,
area CD = CD' x 2 irOE, etc. (§ 665)
Adding these equations, we have area of surface generated by broken line ACDB
= (^'C + C'Z)' + etc.) X 2 ttOE = A'B' x 2 ttOE. Now let the subdivisions of arc AB be bisected indefinitely. Then, area of surface generated by broken line ACDB approaches area of surface generated by arc AB as a limit.
(§363,1*) And, A'B' X 2 irOE approaches A'B' x 2 ttJ? as a limit.
(§ 364, 1*)
* The broken line ACDB is called a regular broken line, and is said
to be inscribed in arc AB ; the theorems of §§ 363, I, and 364, 1, are
evidently true when, instead of the perimeter of a regular inscribed
polygon, we have a regular broken line inscribed in an arc.
For a rigorous proof of the statement that the area of the surface generated by ACDB approaches the area of the surface generated by arc AB as a limit, see Appendix, p. 390.
MEASUREMENT OF THE SPHERE. 375
Then, area of zone generated by arc AB = A'B' x 2 ttR.
(§ 188)
668. Sch. The proof of § 667 holds for any zone which lies entirely on the surface of a hemisphere ; for, in that case, no chord is II OM, and § 665 is applicable.
Since a zone which does not lie entirely on the surface of a hemisphere may be considered as the sum of two zones, each of which does lie entirely on the surface of a hemi- sphere, the theorem of § 667 is true for any zone.
669. Cor. I. If S denotes the area of a zone, h its alti- tude, and E the radius of the sphere,
8 = 2^^1.
670. Cor. II. Since the surface of a sphere may be re- garded as a zone whose altitude is a diameter of the sphere, it follows that
The area of the surface of a sphere is equal to its diameter multiplied by the circumference of a great circle.
671. Cor. m. Let S denote the area of the surface of a sphere, M its radius, and D its diameter.
Then, S = 2 R x2 7rR(?) = A irR'.
That is, the area of the surface of a sphere is equal to the square of its radius multiplied by 4 tt.
Again, S = 7rx(2Ry = ttI^.
That is, the area of the surface of a sphere is equal to the square of its diameter multiplied by tt.
672. Cor. IV. The surface of a sphere is equivalent to four great circles.
For ttR^ is the area of a great O. (?)
673. Cor. V. The areas of the su^r faces of two spheres are to each other as the squares of their radii, or as the squares of their diameters.
(The proof is left to the pupil ; compare § 372.)
376
SOLID GEOMETRY.— BOOK IX.
EXERCISES.
22. Find the area of the surface of a sphere whose radius is 12.
23. Find the area of a zone whose altitude is 13, if the radius of the sphere is 16.
24. Find the area of a spherical triangle whose angles are 125°, 133°, and 156°, on a sphere whose radius is 10.
Prop. XII. Theorem.
674. The volume of a spherical sector is equal to the area of the zone which forms its base, multiplied by one-third the radius of the sphere.
Given sector OAB revolved about diameter OM as an axis, and R the radius of the arc.
To Prove volume of spherical sector generated by OAB = area of zone generated by AB x\R.
Proof. Divide arc AB into three equal arcs, AC, CD, and DB, and draw chords AC, CD, and DB.
Also, draw lines OC and OD, and line OE Jl AC. .-. vol. 0^C= area AG x \ OE,
vol. OCD = area CD x ^ OE, etc. (§ 666) Adding these equations, we have
volume of solid generated by polygon OACDB = (area AC + area CD + etc.) x ^ OE = area ACDB x | OE. Now let the subdivisions of arc AB be bisected indefi- nitely.
MEASUREMENT OF THE SPHERE. 377
Then, volume of solid generated by polygon OACDB approaches volume of solid generated by sector OAB as a limit. (§ 363, II *)
And area of surface generated by ACDB x \ OE ap- proaches area of surface generated by arc AB x ^ i2 as a limit. (§§ 363, I, 364, 1 1)
Then, volume of solid generated by sector OAB
= area of zone generated by arc AB x ^ R. (?)
675. Sch. It is evident, as in § 668, that the theorem of § 674 holds for any spherical sector.
676. Cor. I. If V denotes the volume of a spherical sec- tor, h the altitude of the zone which forms its base, and li the radius of the sphere,
V=2'7rEh X ii2(§ 669) = i^I^k.
677. Cor. II. Since a sphere may be regarded as a spherical sector whose base is the surface of the sphere,
The volume of a sphere is equal to the area of its surface multiplied by one-third its radius.
678. Cor. m. Let V denote the volume of a sphere, E
its radius, and D its diameter. •
Then, F = 4 TT i?2 X i i? (§ 671) = 1 7ri2».
That is, the volume of a sphere is equal to the cube of its radius multiplied by | tt.
Again, F = 7rZ>2 X i Z> (§ 671) = \ -kB^.
That is, tlie volume of a sphere is equal to the cube of its diameter multiplied by \ tt.
* The polygon OACDB is called a regular polygonal sector^ and is said to be inscribed in sector OAB ; the theorem of § 363, II, is evi- dently true when, instead of a regular inscribed polygon, we have a regular polygonal sector inscribed in a sector.
For a rigorous proof of the statement that the volume of the solid generated by OA CDB approaches the volume of the solid generated by sector OAB as a limit, see Appendix, p. 391.
t See note foot of p. 374.
378
SOLID GEOMETRY.— BOOK IX.
679. Cor. rV. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diame- ters.
(The proof is left to the pupil.)
680. Cor. V. The volume of a spherical pyramid is equal to the area of its base multiplied by one-third the radius of the sphere.
Given P the volume of a spherical pyramid, K the area of its base, and R the radius of the sphere.
To Prove
F=KxlB.
Proof. Let n denote the number of sides of the base of the spherical pyramid, s the sum of its A referred to a rt. Z as the unit of measure, T the area of a tri-rectangular A, T' the volume of a tri-rectangular pyramid, S the area of the surface of the sphere, and V its volume.
P _[s-2(n-2)-] xT' _T' ir~[s-2(7i-2)] xT~ T'
V ^sr ^r
S ST t'
Then,
Also,
(§§636,637) (§ 609)
P K
V
S'
i-
R^
4.TrR
-,(§§671, 678) = ii^.
P=K x^R.
Prop. XIII. Problem.
681. Given the radii of the bases, and the altitude, of a spherical segment, to find its volume.
MEASUREMENT OF THE SPHERE. 379
Given O the centre of arc ADB, lines AA and BB' 1. to diameter OM, AA' — r\ BB' = r, A'B' = h, and figure ADBB'A' revolved about OM as an axis.
Required to express volume of spherical segment gener- ated by ADBB'A' in terms of r, r', and A.
Solution. Draw lines OA, OB, and AB ; also, line OC ± AB, and line AE 1. BB' ; and denote radius OA by R.
Now, vol. ADBB'A' = vol. ACBD + vol. ABB' A'. (1)
Also, • vol. ACBD = vol. OADB - vol. OAB.
But, vol. OADB = I TrR'h. (§ 676)
And, vol. OAB = area AB x i OC (§ 666)
= h x27rOCxiOC (§665)
.'. vol. ^Ci>i? = |7ri?% - I TT OO'A = |7r(i22- 00")/^. But, E'-OC' = AC' (§ 273)
= (i^y (?)
.-. vol. ACDB = I TT X i Zb" X /i = i ttZb" Jl Now, ^5" = ^:^' + AE" (?)
= (,. _ r'y + ^2 (?)
.-. vol. ACDB = i TT [(r - r^ + ^i^j;^^ Also, vol. ABB' A' = i tt (r^ + r'^ + rr') /i. (§ 661)
Substituting in (1), we have Yol. ADBB'A'
= i TT [(?• - r'f + 7^2] /i + 1 TT (2r2 + 2r'2 + 2rr') /i
= i-
2rr' + r'^ + 7^2 _^ 2 r^ + 2r'2 + 2rr') h
380 SOLID GEOMETRY. —BOOK IX.
682. Cor. If r denotes the radius of the base, and h the altitude, of a spherical segment of one base, its volume is iTrr'h + l'rrh^
EXERCISES.
25. Find the volume of a sphere whose radius is 12.
26. Find the volume of a spherical sector, the altitude of whose base is 12, the diameter of the sphere being 25.
27. Find the volume of a spherical segment, the radii of whose bases are 4 and 5, and whose altitude is 9.
28. Find the radius and volume of a sphere, the area of whose surface is 324 w.
29. Find the diameter and area of the surface of a sphere whose volume is M^ t,
30. The surface of a sphere is equivalent to the lateral surface of its circumscribed cylinder.
31. The volume of a sphere is two-thirds the volume of its circum- scribed cylinder.
32. A spherical cannon-ball 9 in. in diameter is dropped into a cubical box filled with water, whose depth is 9 in. How many cubic inches of water will be left in the box ? (ir = 3.1416.)
33. What is the angle of the base of a spherical wedge whose volume is ^^ tt, if the radius of the sphere is 4 ?
34. Find the volume of a quadrangular spherical pyramid, the angles of whose base are 107°, 118°, 134°, and 146° ; the diameter of the sphere being 12.
35. The surface of a sphere is equivalent to two-thirds the entire surface of its circumscribed cylinder.
36. Prove Prop. IX. when the straight line is parallel to the axis.
37. Find the area of the surface and -the volume of a sphere inscribed in a cube the area of whose surface is 486.
38. How many spherical bullets, each f in. in diameter, can be formed from five pieces of lead, each in the form of a cone of revolu- tion, the radius of whose base is 5 in., and whose altitude is 8 in. ?
39. A cylindrical vessel, 8 in. in diameter, is filled to the brim with water. A ball is immersed in it, displacing water to the depth of 2\ in. Find the diameter of the ball.
MEASUREMENT OF THE SPHERE. 381
40. If a sphere 6 in, in diameter weighs 351 ounces, what is the weight of a sphere of the same material whose diameter is 10 in. ?
41. If a sphere whose radius is 12^ in. weighs 3125 lb., what is the radius of a sphere of the same material whose weight is 819^ lb. ?
42. The altitude of a frustum of a cone of revolution is 3J, and the radii of its bases are 5 and 3 ; what is the diameter of an equivalent sphere ?
43. Find the radius of a sphere whose surface is equivalent to the entire surface of a cylinder of revolution, whose altitude is 10^, and radius of base 3.
44. The volume of a cylinder of revolution is equal to the area of its generating rectangle, multiplied by the circumference of a circle whose radius is the distance to the axis from the centre of the rectangle.
45. The volume of a cone of revolution is equal to its lateral area, multiplied by one-third the perpendicular from the vertex of the right angle to the hypoteniise of the generating triangle.
46. Two zones on the same sphere, or equal spheres, are to each other as their altitudes.
47. The area of a zone of one base is equal to the area of the circle whose radius is the chord of its generating arc. (§ 270, 2.)
48. If the radius of a sphere is B, what is the area of a zone of one base, whose generating arc is 45° ? (Ex. 65, p. 210.)
49. If the altitude of a cone of revolution is 15, and its slant height 17, find the total area of an inscribed cylinder, the radius of whose base is 5.
(Let the cone and cylinder be generated by the revo- lution of rt. A ABC and rect. CDEF about ^C as an axis.)
50. Find the area of the surface and the volume of a sphere cir- cumscribing a cylinder of revolution, the radius of whose base is 9, and whose altitude is 24.
51. An equilateral triangle, whose side is 6, revolves about one of its sides as an axis. Find the area of the entire surface, and the volume, of the solid generated.
52. A cone of revolution is inscribed in a sphere whose diameter is I the altitude of the cone. Prove that its lateral surface and vol- ume are, respectively, | and /^ the surface and volume of the sphere.
382
SOLID GEOMETRY. —BOOK IX.
53. Find the volume of a sphere circumscribing a cube whose volume is 64.
54. A cone of revolution is circumscribed about a sphere whose diameter is two-thirds the altitude of the cone. Prove that its lateral surface and volume are, respectively, three-halves and nine-fourths the surface and volume of the sphere.
55. If the radius of a sphere is 25, find the lateral area and volume of an inscribed cone, the radius of ^f^ whose base is 24.
(Two solutions.)
56. If the volume of a sphere is ^^ ir, find the lateral area and volume of a circumscribed cone whose altitude is 18.
57. Find the volume of a spherical segment of one base whose altitude is 6, the diameter of the sphere being 30.
B
58. A square whose area is A revolves about its diago- nal as an axis. Find the area of the entire surface, an;' c the volume, of the solid generated.
59. The altitude of a cone of revolution is 9. At what distances from the vertex must it be cut by planes parallel to its base, in order that it may be divided into three equivalent parts ? (§ 656.)
(Let V denote the volume of the cone, x the distance from the vertex to the nearer plane, and y the distance to the other.)
60. Given the radius of the base, B, and the total area, T, of a cylinder of revolution, to find its volume.
(Find H from the equation T-2irBH^2 irB^.)
61. Given the diameter of the base, D, and the volume, F, of a cylinder of revolution, to find its lateral area and total area.
62. Given the altitude, H, and the volume, F, of a cone of revo- lution, to find its lateral area.
63. Given the slant height, i, and the lateral area, 8, of a cone of revolution, to find its volume.
MEASUREMENT OF THE SPHERE.
383
64. A circular sector whose central angle is 45'' and radius 12 revolves about a diameter perpendicular to one of its bounding radii. Find the volume of the spherical sector generated.
65. Given the area of the surface of a sphere, S, to find its volume.
66. Given the volume of a sphere, F, to find the area of its surface.
67. A right triangle, whose legs are a and b, revolves about its hypotenuse as an axis. Find the area of the entire surface, and the volume, of the solid generated.
68. The parallel sides of a trapezoid are 12 and 26, respectively, and its non-parallel sides are 13 and 15. Find the volume generated by the revolution of the trapezoid about its longest side as an axis.
(Represent BE by x.)
69. An equilateral triangle, whose altitude is A, revolves about one of its altitudes as an axis. Find the area of the surface, and the volume, of the solids generated by the triangle, and by its inscribed circle. (Ex. 21, p. 151.)
70. Find the lateral area and volume of a cylinder of revolution, whose altitude is equal to the diameter of its base, inscribed in a cone of revolution whose altitude is h, and radius of base r.
(Represent altitude of cylinder by x.)
71. Find the lateral area and volume of a cylinder of revolution, whose altitude is equal to the diameter of its base, inscribed in a sphere whose radius is r.
72. An equilateral triangle, whose side is a, revolves about a straight line drawn through one of its vertices parallel to the opposite side. Find the area of the en- tire surface, and'the volume, of the solid generated.
(The solid generated is the difference of the cylinder generated by BCHG, and the cones generated by ABG and ACH.)
73. The outer diameter of a spherical shell is 9 in., and its thick- ness is 1 in. What is its weight, if a cubic inch of the metal weighs Jib.? (ir = 3.1416.)
384
SOLID GEOMETRY.— BOOK IX.
74. Find the diameter of a sphere in which the area of the sur- face and the volume are expressed by the same numbers.
75. A regular hexagon, whose side is «, revolves about its longest diagonal as an axis. Find the area of the entire surface, and the volume, of the solid generated.
76. The sides AB and BC of rectangle ABCD are 5 and 8, respec- tively. Find the volumes generated by the revolution of triangle ACD about sides AB and BC as axes.
77. The sides of a triangle are 17, 25, and 28. Find the volume generated by the revolution of the triangle about its longest side as an axis. (§ 324.)
78. A frustum of a circular cone is equivalent to three cones, whose common altitude is the altitude of the frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum. (§ 660.)
79. The volume of a cone of revolution is equal to the area of its generating triangle, multiplied by the circumference of a circle whose radius is the distance to the axis from the intersection of the medians of the triangle. (§140.)
80. If the earth be refgarded as a sphere whose radius is B, what is the area of the zone visible from a point whose height above the surface is if? (§ 271, 2.)
81. The sides AB and BC of acute-angled triangle ABC are \/241 and 10, respectively. Find the volume of the solid generated by the revolution of the triangle about an axis in its plane, not intersecting its surface, whose dis- tances from A, B, and C are 2, 17, and 11, respectively.
82. A projectile consists of two hemispheres, connected cylinder of revolution. If the altitude and diameter of the base cylinder are 8 in. and 7 in., respectively, find the number of inches in the projectile, (tt = 3.1416.)
83. A segment of a circle, whose bounding arc is a quadrant, and whose radius is r, revolves about a diameter parallel to its bounding chord. Find the area of the'entire surface, and the volume, of the solid generated.
by a of the cubic
MEASUREMENT OF THE SPHERE.
385
84. If any triangle be revolved about an axis in its plane, not parallel to its base, vsrhich passes through its vertex without intersect- ing its surface, the volume of the solid generated is equal to the area of the surface generated by the base, multiplied by one-third the altitude.
Fig. 2.
Fig. S.
(Compare § 666. Case I., Figs. 1 and 2, when a side coincides with the axis ; there are two cases according as AE falls on BC, or BC produced. Case II. , Fig. 3, when no side coincides with the axis ; prove by Case I.)
85. If any triangle be revolved about an axis which passes through its vertex parallel to its base, the volume of the solid generated is equal to the area of the surface generated by the base, multiplied by one-third the altitude.
(Compare Ex. 72, p. 383. There are two cases according as AD falls on 5C, OT BC produced.)
86. Find the area of the surface of the sphere circumscribing a regular tetraedron, whose edge is 8.
(Draw lines DOE and AOF ± to A ABC and BCD, respectively.)
386 SOLID GEOMETRY.
APPENDIX.
PROOF OF STATEMENT MADE IN ELEVENTH LINE, PAGE 201.
683. Theorem. The circumference of a circle is shorter than the perimeter of any circumscribed polygon. ''r-.^E
Given polygon ABCD circumscribed about a O.
To Prove circumference of O shorter than perimeter ABCD.
Proof. Of the perimeters of the O and of its circumscribed polygons, there must be one perime- A ter such that all the others are of equal or greater length.
But no circumscribed polygon can have this perimeter.
For, if we suppose polygon ABCD to have this perimeter, and draw a tangent to the O, meeting CD and DA at points E and F, respec- tively, then since str. line EF is < broken line EDF., the perimeter of circumscribed polygon ABCEF is < perimeter ABCD.
Hence, the circumference of the O is < the perimeter of any cir- cumscribed polygon.
PROOFS OF THE LIMIT STATEMENTS OF §640.
684. We assume the following :
A portion of a plane is less than any other surface having the same boundaries.
685. Theorem. The total surface of a circular cylinder is less than the total surface of any circumscribed
prism.*
Given prism AC circumscribed about circular cylinder EG.
To Prove total surface EGr < total sur- face AC.
Proof. Of the total surfaces of the cylinder and of its circumscribed prisms, there must be one total surface such that the area of every other is either equal to or > it.
APPENDIX. 387
But no circumscribed prism can have this total surface.
For suppose prism AC to have this total surface; and let BGDFE — E' be a circumscribed prism, whose face EF' intersects faces AB' and AD' in lines EE' and FF', respectively.
Now, face EF' is < sum of faces AE', AF', AEF, and A'E'F'.
(§684)
Whence, total surface of prism BCDFE — E' is < total surface of prism AC.
Then, total surface of cylinder EG is < total surface of any cir- cumscribed prism.
Proofs of the Limit Statements of § 640.
686. Let L denote the lateral edge, H the altitude, S and s the the lateral areas, V and v the volumes, E and e the perimeters of rt. sections, and B and 6 the areas of the bases of the circumscribed and inscribed prisms, respectively ; also, JS' the lateral area of the cylinder, V its volume, E' the perimeter of a rt. section, and B' the area of the
1. We have, 8 + 2 B> S' + 2 B'. (§685)
.-. S+2(B- B>) > S'.
Again, the total surface of the inscribed prism is < the total surface of the cylinder. (§ 684)
.-. S< -\-2B'>s-{-2h, or S'>s^2{h- B').
Then, S+2(B-B')>S'>s + 2(b-B').
Now if the number of faces of the prisms be indefinitely increased, B- B' andb - B' approach the limit 0. (§ 363, II)
Again, the difference between the perimeters of the bases of the prisms approaches the limit 0. (§ 363, I)
Then, the total surface of the circumscribed prism continually de- creases, but never reaches the total surface of the inscribed prism ; and the total surface of the inscribed prism continually increases, but never reaches the total surface of the circumscribed prism. (§ 684)
Then, the difference between S + 2 B and s + 2b can be made less than any assigned value, however small.
Whence, S + 2 B - (s -\- 2 b) , or S - s -\- 2 (B - b), approaches the limit 0.
But B — b approaches the limit 0. (§ 363, II)
Whence, S — s approaches the limit 0.
Then, S' is intermediate in value between two variables, the differ- ence between which approaches the limit 0.
388 SOLID GEOMETRY.
Then, the difference between either variable and S', that is, S + 2(B -B>) - S' and S' -s-2(b- B'), approaches the limit 0.
Whence, S — S' and S' — s approach the limit 0. Hence, S and s approach the limit S'.
2. We have, V= B x H &ndv = b x H. (§ 499) Whence, V-v = BxH-bx H= (B - b) x H.
Now if the number of faces of the prisms be indefinitely increased, B — b, and therefore V— v, approaches the limit 0. (§ 363, II)
But V is evidently > v, and < V. Then, V— V and V — v approach the limit 0. Whence, V and v approach the limit V.
3. We have, S = E x L and s = e x L. (§ 484)
Then, ^ = - and e=-; OT,U-e = ^=-?.
L L L
Now if the number of faces of the prisms be indefinitely increased, 8 — s^ and therefore E — e, approaches the limit 0. (§ 640, 1)
But E', the perimeter of a rt. section of the cylinder, is < JE"; for the theorem of § 683 is evidently true when for the O is taken any closed curve whose tangents do not intersect its surface ; also, E' is > e. (Ax. 4)
Then, E — E' and E' — e approach the limit 0.
Whence, E and e approach the limit E'.
PROOFS OF THE LIMIT STATEMENTS OF §650.
687. Theorem. The total surface of a circular cone is less than the total surface of any circumscribed pyramid.
Given pyramid S-ABCD circumscribed about circular cone S-EF.
To Prove total surface S-EF<C total sur- face S-ABCD.
Proof. Of the total surfaces of the cone and of its circumscribed pyramids, there must be one total surface such that the area of every other is either equal to or > it.
But no circumscribed pyramid can have this J,otal surface.
For suppose pyramid S-ABCD to have this total surface ; and let S-BCDFE be a circumscribed pyramid, whose face SEF intersects SAB and SAD in lines ^S*^ and SF, respectively.
APPENDIX. 389
Now, face SEF is < sum of faces 8AE, SAF, and AEF. (§ 684) Whence, total surface of pyramid S-BCDFE is < total surface of pyramid S-ABCD.
Then, total surface of cone S-EF is < total surface of any circum- scribed pyramid.
Proofs of the Limit Statements of § 650.
688. Let H denote the altitude, S and s the lateral areas, V and V the volumes, and B and h the areas of the bases, of the circum- scribed and inscribed pyramids, respectively ; also, S' the lateral area of the cone, V its volume, and B' the area of its base.
1. We have, S-\- B>8' + B<. (§ 687)
.-. ^4- {B-B')>S'.
Again, the total surface of the inscribed pyramid is < the total surface of the cone. (§ 684)
.-. 8' + B<>s^-h, or S'>s^{h-B').
Then, 8 -\- {B - Bi)> 8' > s -\- {h - B>).
Now if the number of faces of the pyramids be indefinitely in- creased, 5 - 5' and 6 - B' approach the limit 0. (§ 363, II)
Also, the difference between the perimeters of the bases of the pyramids approaches the limit 0. (§ 363, I)
Then, 8 -V B continually decreases, and s -\- h continually increases ; and the difference between them can be made less than any assigned value, however small. (§ 684)
Then, 8- s-\-{B -h) approaches the limit 0.
But 5 - & approaches the limit 0. (§ 363, II)
Whence, 8 — s approaches the limit 0.
Then, 8' is intermediate in value between two variables, the differ- ence between which approaches the limit 0.
Whence, the difference between either variable and 8', that is, 8-\-{B-B')- 8' and 8' -s-(b-B'), approaches the limit 0.
Then, 8—8' and 8' — s approach the limit 0.
Whence, 8 and s approach the limit 8'.
2. We have, V= B x } H Rnd v = h x ^ H. (§ 521) Whence, V - v = {B - b) x ^ H.
Now if the number of faces of the pyramids be indefinitely increased, B -b, and therefore V—v, approaches the limit 0. (§ 363, II)
But, V is evidently > v, and < V. Then, V-V and V - v approach the limit 0. Whence, V and v approach the limit V'.
390 SOLID GEOMETRY.
PROOF OF THE LIMIT STATEMENT IN NOTE FOOT OF PAGE 374.
689. Theorem. If a regular broken liiie, inscribed in an arc. be revolved about a diameter, not intersecting the arc, as an axis, and the subdivisions of the arc be bisected indefinitely, the area of the surface generated by the broken line approaches the area of the surface generated by the arc as a limit.
Given regular broken line ABCD, inscribed in arc AD, revolving about diameter OM as an axis.
To Prove that, if the subdivisions of arc AD be bisected indefinitely, area of surface generated by ABCD approaches area of surface generated by arc AD as a limit.
Proof. Let A'B', B'C, and CD' be tangents || to AB, BC, and CD, respectively, points A', B', C, and D' being in radii OA, OB, OC, and OD, respectively, produced ; and let S, s, and S' denote the areas of the surfaces generated by A'B' CD', and ABCD, and arc AD, respectively.
Of the surfaces generated by arc AD, by ABCD, and by regular inscribed broken lines obtained by bisecting the subdivisions of the arc indefinitely, there must be one surface such that the areas of all the others are either equal to or < it.
But no regular inscribed broken line can generate this surface.
For if this were the case, by bisecting the subdivisions of the arc, a regular inscribed broken line would be obtained having the same pro- jection on the axis ; but the ± from 0 to each line would be greater, and hence the surface generated would be greater.
(§ 665, and Note foot of p. 374.)
Hence, surface generated by arc AD is > surface generated by ABCD ; that is, S' is > s.
Again, of the surfaces generated by arc AD, by A'B' CD', and by regular circumscribed broken lines obtained by bisecting the sub- divisions of the arc indefinitely, there must be one surface such that the areas of all the others are either equal to or > it.
But po regular circumscribed broken line can generate this surface.
For if this were the case, by bisecting the subdivisions of the arc, a regular circumscribed broken line would be obtained in which the _L from 0 to each line would be the same ; but the projection on the axis would be smaller, and hence tihp ;sH.rf.ape generated would be smaller.
APPENDIX. 391
Hence, surface generated by arc AD is < surface generated by A'B'C'D' ; that is, >S" is < S.
Then, S - S' and S' - s are < S - s.
Now if the subdivisions of arc AD be bisected indefinitely, the difference between broken lines A'B'C'D' and ABCD approaches the limit 0. (Note foot p. 374.)
Then, the difference between the projections on OM oi A'B'C'D' and ABCD approaches the limit 0.
Also, the difference between the Js from O to A'B' and AB ap- proaches the limit 0. (Note foot p. 374. )
Then, the difference between the areas of the surfaces generated by A'B'C'D' and ABCD, that is, 8-s approaches the limit 0. (§ 665)
Then, S — S' and S' — s approach the limit 0.
Whence, S and s approach the limit ^S^'.
PROOF OF THE LIMIT STATEMENT IN NOTE FOOT OF PAGE 377.
690. Theorem. If a regular polygonal sector, inscribed in a sector of a circle, be revolved about a diameter, not crossing the sector, as an axis, and the subdivisions of the arc be bisected indefinitely, the volume of the solid generated by the polygonal sector approaches the volume of the solid generated by the sector as a limit.
Given regular polygonal sector OABCD, inscribed in sector OAD, revolved about diameter OM as an axis. (Fig. of § 689. )
To Prove that, if the subdivisions of arc AD be bisected indefi- nitely, volume of solid generated by OABCD approaches volume of solid generated by sector OAD as a limit.
Proof. Let A'B', B'C, and CD' be tangents || to AB, BC, and CD, respectively, points A', B', C, and D' being in radii OA, OB, OC, and OD, respectively, produced ; and let V, v, and V denote the volumes of the solids generated by OA'B'C'D', OABCD, and sector OAD, respectively.
Then, V is evidently > v, and < F.
Whence, V — V and V — v are < F— v.
Now if the subdivisions of arc AD be biseqted indefinitely, the difference between the areas of OA'B'C'D' and OABCD, and there- fore V— v, approaches the limit 0. (Note foot p. 377.)
Then, V — V and V — v approach the limit 0.
Whence, V and v approach the limit V.
SUGGESTIONS FOR THE USE OF COLORED PLATES.
In beginning the study of solid geometry, a new difficulty is encountered, the difficulty of seeing the figures correctly. Through plane geometry, the pupil has acquired the habit of looking at the figures simply as lines making different angles and running in varying directions, but always limited to one plane. To the untrained eye, the line figures in solid geometry do not look essentially different. The teacher sees, the pupil does not, and, worse than all, too frequently the teacher fails to realize that what represents to him a solid figure is, to the pupil, a number of lines similar to those in plane geometry, only hopelessly compli- cated in arrangement.
The first thing necessary, then, is to train a class to visualize correctly, — to see in imagination, not a seeming confusion of lines, but the solids outlined by those lines.
In the hope of accomplishing this, various aids have been offered by text-books in the way of graphic representation, but all of them, while attractive at first, have, when tried, fallen short of expectation in teaching value.
Work with actual models is accurate and helpful, but pho- tographic reproductions of these models make nearly the same demands upon the untrained imagination that the line figures do. Shaded figures have been used, but the simi- larity in tone of grays and blacks is confusing to the unedu- cated eye.
With the color scheme here presented, the confusion
394 SOLID GEOMETRY.
vanishes, and the pupil not only may see, but must see, the planes in their true relations to each other. If his first glimpse of figures for solids is right, he is ready then to look for depth, distance, and three dimensions, in all suc- ceeding figures.
The few colored figures here presented are valuable in the beginning, to show the pupil the kind of thing that he is to look for — what he is expected to see. Take, for instance, the figure on page 236. To the beginner there is little sug- gestion of various planes intersecting, disappearing behind each other, and reappearing, but by Plate I all this is in- stantly revealed. The correct visual impression here gained will then be transferred naturally to the line figure.
Another objection to the aids thus far presented lies in the fact that the text-book does all the work, leaving the pupil only an observer. If the work stops with looking at the figures and studying from them, their greatest teaching value is lost. It is comparatively easy, with a figure that has been carefully drawn and effectively shaded or colored, to grasp for the moment the general idea indicated, but the impression will be neither complete nor lasting.
Purposely only a few suggestive figures are here pre- sented in color, it being the plan that the pupil, from the figures given in the text, and from the accompanying dem- onstration, shall interpret in color the solids indicated. When he is compelled thus to fix the limitations of the planes, he is led to definite knowledge that is otherwise impossible. Here is represented all the vast distance that lies between looking at a picture done by some one else, and reproducing that picture yourself, — all the difference be- tween observing and doing.
The plan here offered is capable of practical application in several ways.
Send the class to the board to draw the figures for the day with colored crayons ; the result will reveal their under- standing or misunderstanding of the proposition under con-
THE USE OF COLORED PLATES. 395
sideration. With the color in their own hands, pupils are compelled to decide where plane intersects plane, where one disappears behind another, and many other things that escaped their observation in studying from the book. Take, for instance, the figure on page 236. It probably never occurred to the pupil in studying to observe into how many planes the argument is carried. When he colors it he must know.
This first work should be done rapidly, with no attempt at finished drawings. Sometimes it is well to have a class draw entirely free-hand, laying in the color rapidly, attempt- ing only to bring out the geometric idea. At first strongly contrasting colors should be used, and, as the work is not permanent, they may be even crude, if only striking.
Following this, certain figures should be put into perma- nent form. Such drawing should be done carefully, with as close mathematical accuracy as board and chalk will allow. Here attention should be given to the color scheme and the result made restful to the eye. There is actual teaching value in having these difficult figures long before the attention.
With the figures thus before the class, it is easy in the few spare moments that occasionally come at the close of a recitation, to give a quicks review that would not be possible if time had to be consumed in drawing.
Blackboard work is strengthened by outlining all planes in strong white lines, just as in the book they are outlined in black.
Another useful expedient in the use of color is the careful preparation of plates outside of class. The most interesting and effective figures should be selected, and all members of the class required to execute a certain number, this number varying according to the ability of different classes. It is usually well to suggest a uniform size of paper ; seven by nine inches, approximately, is desirable. One figure only should be placed on a sheet. As to size of figure, it is bet- ter not to dictate. When the first drawings are brought
393 SOLID GEOMETRY.
together, it will not take a class long to decide which is most effective. From this they will modify their scale, approximating the best, but retaining perfect individuality. . One additional direction should be given, applying equally to board work and to work on paper. Leave only such lines as would be visible if the planes were opaque. A glance at the colored plates here given will show that by omitting the dotted lines the figure is more effective, and the solidarity greatly emphasized. The geometric value of the lines is not lost, for the eye naturally carries them along behind the plane, and joins the parts correctly if they reappear. In outlining the figure at first, these lines, of course, should be drawn, for by them is frequently determined where the planes intersect.
As to the medium used, that is a matter of taste and equipment. Colored pencils are easiest for the untrained hand, and good effects can be obtained with them. If any one in the class handles water colors, he should be encouraged to use them, for they make stronger figures, and the influ- ence of even one or two working in them will elevate the entire standard.
Some or all of these expedients may be used as the condi- tions of individual classes indicate, but let it always be insisted that the class do the work. The most carefully executed drawing of teacher or text-book is worth less than the poorest attempt of the poorest pupil.
Finally, the method here presented is offered only as a practical suggestion for clearer teaching, not as an integral part of geometry, and may be used or not as teachers desire. Like everything else, it is capable of abuse and perversion, and whoever uses it should be ever watchful lest it overstep its proper limitations. Its purpose is not to produce a fine set of drawings, but to assist in teaching, geometry. It is a means, not an end ; an expedient, not a science.
Plate I.
Plate II.
Plate III.
Plate IV.
Plate VI.
Plate VIII.
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/ (M |
/ |
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/<^ |
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Plate IX.
Plate
INDEX TO DEFINITIONS.
Acute angle, § 27. Adjacent angles, § 23.
diedral angles, § 428. Alternate-exterior angles, § 71. -interior angles, § 71. Alternation, § 235. Altitude of a cone, § 553.
of a cylinder, § 540.
of a frustum of a cone,
§ 553. of a frustum of a pyramid,
§606. of a parallelogram, ^ 104. of a prism, § 466. of a pyramid, § 502. of a spherical segment,
§662. of a trapezoid, § 104. of a triangle, § 60. of a zone, § 662. Angle, § 20.
at the centre of a regular
polygon, § 341. between two intersecting
curves, § 583. inscribed in a segment,
§148. of a lune, § 616. Angles of a polygon, § 118.
of a quadrilateral, § 103. of a spherical polygon,
§ 587. of a triangle, § 57. Angular degree, § 29. Antecedents of a proportion, § 229.
Apothem of a regular polygon,
§ 341. Arc of a circle, § 142. Area of a surface, § 302. Axiom, § 15.
Axis of a circle of a sphere, § 567. of a circular cone, § 553. of a circular cylinder, § 546. • of symmetry, § 387.
Base of a cone, § 553.
of a polyedral angle, § 452. of a pyramid, § 502. of a spherical pyramid, § 589. of a spherical sector, § 664. of a spherical wedge, § 617. of a triangle, § 60. Bases of a cylinder, § 540.
of a parallelogram, § 104.
of a prism, § 466.
of a spherical segment, § 662.
of a trapezoid, § 104.
of a truncated prism, § 472.
of a truncated pyramid,
§ 505. of a zone, § 662. Bi- rectangular triangle, § 598. Broken line, § 7.
Central angle, § 148. Centre of a circle, § 142.
of a parallelogram, § 111,
of a regular polygon, § 341
of a sphere, § 561.
of symmetry, § 386.
397
398
INDEX TO DEFINITIONS.
Chord of a circle, § 147. Circle, § 142.
circumscribed about a poly- gon, § 151. inscribed in a polygon, §151. Circles tangent externally, § 150. tangent internally, § 150. Circular cone, § 553.
cylinder, § 540. Circumference, § 142. Commensurable magnitudes, § 181. Common measure, § 181. tangent, § 150. Complement of an angle, § 30. of an arc, § 190. Complementary angles, § 30. Composition, § 237. Concave polygon, § 121. Concentric circles, § 146. Conclusion, § 38. Cone, § 553.
of revolution, § 555. Conical surface, § 553. Consequents of a proportion, § 229. Constant, § 185. Converse of a proposition, § 39. Convex polyedral angle, § 453. polyedron, § 463. polygon, § 121. spherical polygon, § 588. Corollary, § 15. Corresponding angles, § 71. Cube, § 474. Curve, § 7. Curved surface, § 10. Cylinder, § 640.
of revolution, § 550. Cylindrical surface, § 540.
Decagon, § 119. Degree of arc, § 190. Determination of a plane, § 394. of a straight line, §18. Diagonal of a polyedron, § 461.
Diagonal of a polygon, § 118.
of a quadrilateral, § 103. of a spherical polygon, §587. Diameter of a circle, § 142. of a sphere, § 561. Diedral angle, § 428. Dimensions of a rectangle, § 304. of a rectangular paral- lelepiped, § 487. Directrix of a conical surface, § 553. of a cylindrical surface, §540. Distance between two points on the surface of a sphere, §573. of a point from a line,
§47. of a point from a plane, §410. Division, § 238. Dodecaedron, § 462. Dodecagon, § 119.
Edge of a diedral angle, § 428. Edges of a polyedral angle, § 452.
of a polyedron, § 461. Element of a conical surface, § 553. of a cylindrical surface, §540. Enneagon, § 119. Equal angles, § 22.
diedral angles, § 432. figures, § 22. polyedral angles, § 454. Equiangular polygon, § 120.
triangle, § 58. Equilateral polygon, § 120.
spherical triangle,
§587. triangle, § 58. Equivalent solids, § 465.
surfaces, § 303. Exterior angles, § 71.
of a triangle, § 57.
INDEX TO DEFINITIONS.
399
Extremes of a proportion, § 229.
Face angles of a polyedral angle,
§ 452. Faces of a diedral angle, § 428. of a polyedral angle, § 452. of a polyedron, § 461. Figure symmetrical with respect to a centre, § 390. symmetrical with respect to an axis, § 391. Figures symmetrical with respect to a centre, § 388. symmetrical with respect to an axis, § 388. Foot of a line, § 397. Fourth proportional, § 231. Frustum of a cone, § 653.
of a pyramid, § 506. of a pyramid circum- scribed about a frus- tum of a cone, § 649. of a pyramid inscribed in a frustum of a cone, §649.
Greneratrix of a conical surface, §553. of a cylindrical sur- face, § 540.
Geometrical figure, § 12.
Geometry, § 13.
Great circle of a sphere, § 567.
Hendecagon, § 119.
Heptagon, § 119.
Hexaedron, § 462.
Hexagon, § 119.
Homologous, §§ 65, 123.
Hypotenuse of a right triangle,
§59. Hypothesis, § 15.
Icosaedron, § 462. Incommensurable magnitudes, §181.
Indirect method of proof, § 50. Inscribed angle, § 148. Inscriptible polygon, § 151. Interior angles, § 71. Inversion, § 236. Isoperimetric figures, § 378. Isosceles spherical triangle, § 587. triangle, § 58.
Lateral area of a cone, § 647.
of a cylinder, § 638. of a frustum of a cone,
§647. of a prism, § 466. of a pyramid, § 502. Lateral edges of a prism, § 466.
of a pyramid, § 502. Lateral faces of a prism, § 466.
of a pyramid, § 502. Lateral surface of a cone, § 553.
of a cylinder, § 540. Legs of a right triangle, § 59. Limit of a variable, § 186. Line, § 3.
Locus of a series of points, § 141. Lower base of a frustum of a cone, § 553. nappe of a conical surface, §553. Lune, § 616.
Material body, § 1. Mean proportional, § 230. Means of a proportion, § 229. Measure of a magnitude, § 180.
of an angle, § 29. Median of a triangle, § 139. Mutually equiangular polygons, §122. equiangular spherical
polygons, § 599. equilateral polygons,
§ 122. equilateral spherical polygons, § 599.
400
INDEX TO DEFINITIONS.
Numerical measure, § 180.
Oblique angles, § 27.
lines, § 27.
prism, § 470. Obtuse angle, § 27. Octaedron, § 462. Octagon, § 119.
Parallel lines, § 52.
planes, § 397. Parallelogram, § 104. Parallelopiped, § 474. Pentagon, § 119. Perimeter of a polygon, § 118. Perpendicular Ijnes, § 24.
planes, § 436. Plane, § 9.
angle of a diedral angle,
§ 429. figure, § 12. geometry, § 14. tangent to a cone, § 553. tangent to a cylinder, § 540. tangent to a frustum of a
cone, § 553. tangent to a sphere, § 564. Point, § 4.
of contact of a line tangent
to a circle, § 149. of contact of a line tangent
to a sphere, § 564. of contact of a plane tangent to a sphere, § 564. Points symmetrical with respect to a line, § 387. symmetrical with respect to a point, § 386. Polar distance of a circle of a sphere, § 576. '
triangle of a spherical tri- angle, § 590. triangles, § 592. Poles of a circle of a sphere, § 567. Polyedral angle, § 452. Polyedron, § 461.
Polyedron circumscribed about a sphere, § 564. inscribed in a sphere, §564. Polygon, § 118.
circumscribed about a
circle, § 151. inscribed in a circle, § 151. Postulate, § 15. Prism, § 466.
circumscribed about a cylin- der, § 639. inscribed in a cylinder, § 639. Problem, § 15.
Projection of a line on a line, § 276. of a line on a plane,
§ 447. of a point on a line,
§ 275. of a point on a plane, §447. Proportion, § 227. Proposition, § 15. Pyramid, § 502.
circumscribed about a
cone, § 648. inscribed in a cone, § 648.
Quadrangular prism, § 469.
pyramid, § 603. Quadrant, § 146. Quadrilateral, § 103.
Kadius of a circle, § 142.
of a regular polygon, § 341.
of a sphere, § 661. Ratio, § 180.
Reciprocally proportional magni- tudes,' § 281. Rectangle, § lj)5.
Rectangular parallelopiped, § 474 Rectilinear figure, § 12. Re-entrant angle, § 121. Regular polyedron, § 536.
INDEX TO DEFINITIONS.
403
Regular polygon, § 339. prism, § 471. pyramid, § 504. Rhomboid, § 105. Rhombus, § 105. Right angle, § 24.
angled spherical triangle, § 587.
circular cone, § 553.
cylinder, § 540.
diedral angle, § 436.
parallelopiped, § 474.
prism, § 470.
section of a cylinder, § 638.
section of a prism, § 473.
triangle, § 69.
Scalene triangle, § 58. Scholium, § 15. Secant, § 149. Sector of a circle, § 147. Segment of a circle, § 147. Segments of a line by a point, § 250. Semicircle, § 147. Semi-circumference, § 146. Sides of a polygon, § 118.
of a quadrilateral, § 103. of a spherical polygon, § 587. of a triangle, § 57. of an angle, § 20. Similar arcs, § 369.
cones of revolution, § 555. cylinders of revolution,
§550. polyedrons, § 527. polygons, § 252. sectors, § 369. segments, § 369. Slant height of a cone of revolu- tion, § 647. of a frustum of a cone ofrevolution,§647. of a frustum of a regu- lar pyramid, § 511. of a regular pyramid, §508.
Small circle of a sphere, § 567. Solid, § 2.
geometry, § 14. Sphere, § 561.
circumscribed about a
polyedron, § 564. inscribed in a polyedron, §564. Spherical angle, § 583.
excess of a spherical tri- angle, § 632. polygon, § 587. pyramid, § 589. sector, § 664. segment, § 662. segment of one base,
§ 662. triangle, § 587. wedge, § 617. Square, § 105. Straight line, § 7.
divided in extreme and mean ratio externally, § 296. divided in extreme and mean ratio internally, § 296. oblique to a plane,
§397. parallel to a plane,
§ 397. perpendicular to a
plane, § 397. tangent to a circle,
§ 149. tangent to a sphere, § 564. Straight lines divided proportion- ally, § 243. Sublended arc, § 147. Supplement of an angle, § 30. of an arc, § 190. Supplementary-adjacent angles, §33. angles, § 30. Surface, § 2.
402
INDEX TO DEFINITIONS.
Surface of a material body, § 1.
of a solid, § 2. Symmetrical polyedral angles, § 455. spherical polygons, §591.
Tangent circles, § 150. Tetraedron, § 462. Theorem, § 15. Third proportional, § 230. Transversal, § 71. Trapezium, § 104. Trapezoid, § 104. Triangle, § 57. Triangular prism, § 469.
pyramid, § 503. Triedral angle, § 452. Tri-rectangular triangle, § 698. pyramid, § 629. Truncated prism, § 472.
pyramid, § 505.
Unit of measure, § 180.
of surface, § 302.
of volume, § 464. Upper base of a frustum of a cone, §553.
Upper nappe of a conical sutface, §553.
Variable, § 184. Vertex of a cone, § 553.
of a conical surface, § 553. of a polyedral angle, § 452. of a pyramid, § 502. . of a spherical pyramid, § 589. of a triangle, § 60. of an angle, § 20. Vertical angle of a triangle, § 60. angles, § 28. diedral angles, § 428. polyedral angles, § 452. Vertices of a polyedron, § 461. of a polygon, § 118. of a quadrilateral, § 103. of a spherical polygon^
§587. of a triangle, § 57. Volume of a solid, § 464.
Zone, § 662.
of one base, § 662.
ANSWERS
TO
NUMERICAL EXERCISES.
>j»ic
|
Book I. |
|
|
4. 24°. |
5. 63° 30', 26° 30'. 8. 22° 30', 157° 30'. |
|
9. 37°. |
24. ^ = 112°30', jB=C=33°45'. 88. 7. |
Book II.
12. 28°. 13. 44° 30'. 14. 12°. 15. 54° 30'. 16. 178°
17. 112° 30'. 18. 83°, 89° 30', 97°, 90° 30', 74° 30'.
52. Z AED = 14° 30', Z AFB = 10° 30'.
55. 114° 30', 89° 30', 65° 30', 90° 30'.
67. 97°30', 89°30', 82°30', 90°30'.
Book III.
1. 112. 2. 42. 3. If 4. 63. 5. J5C, 3^, 2f ; C^, 4, 3;
AB, 4^, Sj%. 6. BC, llf, 18 j; CA, 20, 28; AB, 35, 40. 7. 19f, 25^. 9. 4 ft. 6 in. 10. 12. 11. 15.
12. 37 ft. 1 in. 13. 47 ft. 6 in. 14. -i/VS.
15. 15V2in. 16. 41. 17. 5^. 18. 21. 19. 24. 25. 18. 28. 48. 29. 10. 30. 131 31. 9V2. 32. 45. 34. 17|. 37. 50. 41. Vl29, 2V21, V20i. 42. i^. 47. 36. 49. 63. 50. 4 and 3 ; -i/ and |. 56. 24.
403
404 NUMERICAL EXERCISES.
57. 17. 58. 21,28. 59. 8V3. 60. BE = 4., ED = 12. 62. 6V3. 67. 14. 70. 21. 74. ffand^; 9 and 5.
Book IV. 1. 30f ft. 2. 8 ft. 9 in. 3. 14, 12. 4. 6 ft. 11 in., 20 ft. 9 in. 5. 6 sq. ft. 60 sq. in. 6. 30 V3. 7. 26 yd. 1 ft. 8. 2 sq. ft. 48 sq. in. 9. 243. 10. 210; 24if, 15, 16|.
11. 73. 12. 117. 16. 2 ft. 10 in. 18. -2/V3. 19. 3V3. 21. 120. 24. 210. 25. 18. 26. li ft. 27. 6. 28. 4V3. 29. 1260. 33. 150. 34. 17. 36. 624. 37. 540 sq. in. 38. 28 ft. 41. -1/. 42. 30,16. 45. it^./47. AD=^^-\/2, AE=llV2. 48. 54. 51. Area ^5Z>=39, area ^OZ>=45. 52. 1010. 53. 336.
Book V.
32." Area, -^I^TT. 33. Circumference, 34 tt. 34. 64:121.
36. 9. 37. 13. 38. |V2. 39. ^tt. 40. -^-^-rr.
41. 9.8268. 42. J^tt. 43. 392. 44. 48 tt. 45. 1.2732.
46. -«/7r. 47. 67r. 48. 167r. 49. 37r,12 7r. SO.Stt, 87rV2.
51.9.06. 52. 416 TT sq.ft. 53.120.99 ft. 54. 57 in. 60. 67.295''+. 61. 2.658+. 62. 5.64+.
APPENDIX TO PLANE GEOMETRY.
58. 10V7. 62. 8. 63. ff. 91. 480.
Book VII. 1. 4:3. 2. 2:5. 4. 42. 5. 1 ft. 9 in. 6. ^^\ cu. in. ; 63f sq. in. 7. 574. 8. 1008. 9. 12, 7. 10. 1944.
12. Volume, 50V3. 14. Volume, ^f^VS. 15. 17. 17. 2400 sq. in. 18. 770. 19. Volume, 48 V5. 20. 144. 21. 512, 384. 22. 1705. 23. 10, 1. 24. 36 sq. in. 25. 12 in. 28. V273, 18V237, 180V3. 29.iVll8,
ANSWERS. 405
3V109, 15. 30. V97, 12V93, 72V3. 31. 4V39, 504V3, 936V3. 32. 6V3, 56V26, 503f 33. 4Vl0, 72V39, 672V3. 34. 150. 35. 320. 36. '84(0 37. 700, 1568. 38. |V57, 640V3. 39. 42V9i, 624VT40. 108, 21V39. 41. 240, J^Vil9. 49. 4V3, |V2. 50. 15. 51. 768, 2340. 59. 438. 63. 9600 1b. 64. 50. 69. 168 V3, 15V219. 76. 3456 cu. in. 77. 6 ft. 78. 4 ft. 6 in. 79. 5^4 in. 80. 960,3072. 81. 128. 82. 12. «3. 6. 86. 36V3.
|
Book VIII. |
|
|
7. 39|. 9. 86° 24'. 10. 36. 11. |
44. 12. 3:2. |
|
13. 108°. 14. 220. 16. 334. 17. |
66f 18. 36i. |
|
19. 60. 20. 153°. 23. 45°. 24. 4f |
26. 4V3, Vs. |
|
44. 30,8,20. |
|
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Book IX. |
1. 288 TT, 450 TT, 1296^. 2. 2420. 3. 14, 12.
4. 300 TT. 5. 2.7489+. 6. 167803.68. 8. 175 tt, 224 7r, 392 7r. 9..143 7r, 216 TT, 388 tt. 10. 135.
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37r
24. 130 TT. 25. 2304 TT. 26. 1250^7. 27. 306 tt.
28. Volume, 972 tt. 29. Area of surface, 225 tt.
32. 347.2956. ' 33. 56° 15'. 34. 58 tt. 37. 81 tt,
^TT. 38. 8192. 39. 6 in. 40. 1625 oz.
41. 8 in. 42. 7. 43. 4i. 48. 7rR'(2-V2).
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53. 32 7rV3. 55. 720 tt, 3456 tt; 960 tt, 6144 tt.
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406 NUMERICAL EXERCISES.
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