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ELEMENTS

or

PLANE AND SOLID GEOMETRY.

BY

G. A. WENT WORTH, A. M.,

PROFESSOR OF MATHEMATICS IN PHILLIPS EXETER ACADEMY.

BOSTON: PUBLISHED BY GINN, HEATH, & CO.

18.85. .

*-<LdX<

Copyright, 1877. By GINN AND HEATH.

j. s. cushing,

Superintendent of Printino, ioi Pearl St., Boston.

PREFACE

Most persons do not possess, and do not easily acquire, the power of abstraction requisite for apprehending the Geometri- cal conceptions, and for keeping in mind the successive steps of a continuous argument. Hence, with a very large proportion of beginners in Geometry, it depends mainly upon the form in which the subject is presented whether they pursue the study with indifference, not to say aversion, or with increasing interest and pleasure.

In compiling the present treatise, this fact has been kept con- stantly in view. All unnecessary discussions and scholia hav« been avoided ; and such methods have been adopted as experi- ence and attentive observation, combined with repeated trials, have shown to be most readily comprehended. No attempt has been made to render more intelligible the simple notions of position, magnitude, and direction, which every child derives from observation ; but it is believed that these notions have been limited and defined with mathematical precision.

A few symbols, which stand for Avords and not fur operations, have been used, but these are of so great utility in giving style and perspicuity to the demonstrations that no apology seems necessary for their introduction.

Great pains have been taken to make the page attractive. The figures are large and distinct, and are placed in the middle of the page, so that they fall directly under the eye in imme- diate connection with the corresponding text. The given lines

M306198

IV PREFACE.

of the figures are full lines, the lines employed as aids in the demonstrations are short-dotted, and the resulting lines are long- dotted.

In each proposition a concise statement of what is given is printed in one kind of type, of what is required in another, and the demonstration in still another. The reason for each step is indicated in small type between that step and the one follow- ing, thus preventing the necessity of interrupting the process of the argument by referring to a previous section. The number of the section, however, on which the reason depends is placed at the side of the page. The constituent parts of the propo- sitions are carefully marked. Moreover, each distinct assertion in the demonstrations, and each particular direction in the construc- tions of the figures, begins a new line ; and in no case is it neces- sary to turn the page in reading a demonstration.

This arrangement presents obvious advantages. The pupil perceives at once what is given and what is required, readily refers to the figure at every step, becomes perfectly familiar with the language of Geometry, acquires facility in simple and accu- rate expression, rapidly learns to reason, and lays a foundation for the complete establishing of the science.

A few propositions have been given that might properly bo considered as corollaries. The reason for this is the great diffi- culty of convincing the average student that any importance should be attached to a corollary. Original exercises, however, have been given, not too numerous or too difficult to discourage the beginner, but well adapted to afford an effectual test of the degree in which he is mastering the subjects of his reading. Some of these exercises have been placed in the early part of the work in order that the student may discover, at the outset, that to commit to memory a number of theorems and to repro- duce them in an examination is a useless and pernicious labor ; but to learn their uses and applications, and to acquire a readi- ness in exemplifying their utility, is to derive the full benefit of that mathematical training which looks not so much to the

PREFACE.

attainment of information as to the discipline of the mental fac- ulties.

It only remains to express my sense of obligation to Dr. D. F. Wells for valuable assistance, and to the University Press for the elegance with which the book has been printed ; and also to give assurance that any suggestions relating to the work will be thankfully received.

G. A. WENTWORTH.

Phillips Exeter Academy, January, 1878.

NOTE TO THIRD EDITION.

In this edition I have endeavored to present a more rigor- ous, but not less simple, treatment of Parallels, Ratio, and Limits. The changes are not sufficient to prevent the simulta neous use of the old and new editions in the class ; still they arc very important, and have been made after the most careful and prolonged consideration.

I have to express my thanks for valuable suggestions received from many correspondents ; and a special acknowledgment is due from me to Professor C. H. Judson, of Furman University, Greenville, South Carolina, to whom I am indebted for assist- ance in effecting many improvements in this edition.

TO THE TEACHER.

When the pupil is reading each Book for the first time, it will be well to let him write his proofs on the blackboard in his own lan- guage ; care being taken that his language be the simplest possible, that the arrangement of work be vertical (without side work), and that the figures be accurately constructed.

This method will furnish a valuable exercise as a language lesson, will cultivate the habit of neat and orderly arrangement of work, and will allow a brief interval for deliberating on each step.

After a Book has been read in this way the pupil should review the Book, and should be required to draw the figures free-hand. He

Vi PREFACE.

should state and prove the propositions orally, using a pointer to indicate on the figure every line and angle named. He should be encouraged, in reviewing each Book, to do the original exercises ; to state the converse of propositions ; to determine from the statement, if possible, whether the converse be true or false, and if the converse be true to demonstrate it ; and also to give well-considered answers to questions which may be asked him on many propositions.

The Teacher is strongly advised to illustrate, geometrically and arithmetically, the principles of limits. Thus a rectangle with a constant base 6, and a variable altitude x, will afford an obvious illustration of the axiomatic truth contained in [4], page 88. If x increase and approach the altitude a as a limit, the area of the rec- tangle increases and approaches the area of the rectangle a b as a limit ; if, however, x decrease and approach zero as a limit, the area of the rectangle decreases and approaches zero for a limit. An arith- metical illustration of this truth would be given by multiplying a constant into the approximate values of any repetend. If, for exam- ple, we take the constant 60 and the repetend .3333, etc., the approxi- mate values of the repetend will be fa ffa iWr» l^A% etc> anc^ these values multiplied by 60 give the series 18, 19.8, 19.98, 19.998, etc., which evidently approach 20 as a limit ; but the product of 60 into £ (the limit of the repetend .333, etc.) is also 20.

Again, if we multiply 60 into the different values of the decreasing series, fa -gfa, sfaj, 3iri5inr> e*c-i which approaches zero as a limit, we shall get the decreasing series, 2, \, fa, jfo, etc. ; and this series evidently approaches zero as a limit.

In this way the pupil may easily be led to a complete comprehen- sion of the whole subject of limits.

The Teacher is likewise advised to give frequent written examina- tions. These should not be too difficult, and sufficient time should be allowed for accurately constructing the figures, for choosing the best language, and for determining the best arrangement.

The time necessary for the reading of examination-books will be diminished by more than one-half, if the use of the symbols employed in this book be permitted.

G. A. W.

Phillips Exeter Academy, January, 1879.

CONTENTS.

PLANE GEOMETEY.

BOOK I. Rectilinear Figures.	Pagr
Introductory Remarks	. 3
Definitions	4
Straight Lines	6
Plane Angles	7
Angular Magnitude	9
Superposition	10
Mathematical Terms	. 11
Axioms and Postulates	12
Symbols and Abbreviations ....	. 13
Perpendicular and Oblique Links .	14
Parallel Lines	. 24
Triangles	37
Quadrilaterals	. 58
Polygons in General	63
BOOK II. Circles.
Definitions	73
Straight Lines and Circles ....	. 75
Measurement	86
Theory of Limits	. 87
Supplementary Propositions ....	100
Constructions	. 103

BOOK III. Proportional Lines and Similar Polygons.

Theory of Proportion .128

Proportional Lines 139

Similar Polygons 143

Constructions 164

CONTENTS.

BOOK IV. Comparison and Measurement of the Sur- faces of Polygons.

Comparison and Measurement of Polygons . . .174

Constructions 194

BOOK V. Regular Polygons and Circles.

Regular Polygons and Circles 210

Constructions 224

Isoperimetrical Polygons. Supplementary . . . 237

Symmetry. Supplementary 245

SOLID GEOMETRY.

BOOK VI. Planes and Solid Angles.

Lines and Planes 251

Dihedral Angles , 268

Supplementary Propositions 275

Polyhedral Angles 277

BOOK VII. Polyhedrons, Cylinders, and Cones.

Prisms 286

Pyramids 302

Similar Polyhedrons 317

PvEGUlar Polyhedrons . . . . . . . 322

Supplementary Propositions 326

Cylinders 328

Cones 339

BOOK VIII. The Sphere.

Sections and Tangents 349

Distances on the Surface of the Sphere . . . 356

Spherical Angles . . . . . . . . . . 363

Spherical Polygons and Pyramids 365

Comparison and Measurement of Spherical Surfaces . 383

Volume or the Sphere 396

ELEMENTS OF GEOMETRY.

BOOK I.

RECTILINEAR FIGURES.

Introductory Remarks.

A rough block of marble, under the stone-cutter's hammer, may be made to assume regularity of form.

If a block be cut in the shape repre- ^ y

sented in this diagram,

It will have six flat faces.

Each face of the block is called a Sur- face.

If these surfaces be made smooth by pol- ishing, so that, when a straight-edge is applied to any one of them, the straight-edge in every part will touch the surface, the surfaces are called Plane Surfaces.

The sharp edge in which any two of these surfaces meet is called a Line.

The place at which any three of these lines meet is called a Point.

If now the block be removed, we may think of the place occupied by the block as being of precisely the same shape and size as the block itself; also, as having surfaces or boundaries which separate it from surrounding space. We may likewise think of these surfaces as having lines for their boundaries or limits ; and of these lines as having points for their extremities or limits.

A Solid, as the term is used in Geometry, is a limited por- tion of space.

After we acquire a clear notion of surfaces as boundaries of solids, we can easily conceive of surfaces apart from solids, and

GEOMETRY. BOOK I.

suppose them of unlimited extent. Likewise we can conceive of lines apart from surfaces, and suppose them of unlimited length; of points apart from lines as having position, but no extent.

Definitions.

1. Def. Space or Extension has three Dimensions, called Length, Breadth, and Thickness.

2. Def. A Point has position without extension.

3. Def. A Line has only one of the dimensions of exten- sion, namely, length.

The lines which we draw are only imperfect representations of the true lines of Geometry.

A line may be conceived as traced or generated by a point in motion.

4. Def. A Surface has only two of the dimensions of ex- tension, length and breadth.

A surface may be conceived as generated by a line in motion.

5. Def. A Solid has the three dimensions of extension, length, breadth, and thickness. Hence a solid extends in all direc- tions.

A solid may be conceived as generated by a surface in motion.

Thus, in the diagram, let the upright surface A B CD move to the right to the position E F H K. The points A, B, C, and D will generate the lines AE, BF, CK, and D H respectively. C *

And the lines A B, B D, DC, and A C will generate the sur- faces A F, B H, D K, and A K respectively. And the surface ABC D will generate the solid A H.

The relative situation of the two points A and H involves three, and only three, independent elements. To pass from A to H it is necessary to move East (if we suppose the direction A E to

! I

]H

DEFINITIONS.

be due East) a distance equal to A E, North a distance equal to E F, and down a distance equal to F H.

These three dimensions we designate for convenience length, breadth, and thickness.

6. The limits (extremities) of lines are points. The limits (boundaries) of surfaces are lines. The limits (boundaries) of solids are surfaces.

7. Def. Extension is also called Magnitude.

When reference is had to extent, lines, surfaces, and solids are called magnitudes.

8. Def. A Straight line is a line which has the same direction throughout its whole extent.

9. Def. A Curved line is a line which changes its direction at every point.

10. Def. A Broken line is a series of con- nected straight lines.

When the word line is used a straight line is meant; and when the word curve is used a curved line is meant.

1 1. Def. A Plane Surface, or a Plane, is a surface in which, if any two points be taken, the straight line joining these points will lie wholly in the surface.

12. Def. A Curved Surface is a surface no part of which is plane.

1 3. Figure or form depends upon the relative position of points. Thus, the figure or form of a line (straight or curved) depends upon the relative position of points in that line ; the figure or form of a surface depends upon the relative position of points in that surface.

When reference is had to form or shape, lines, surfaces, and solids are called figures.

6 GEOMETRY. BOOK I.

1 4. Def. A Plane Figure is a figure, all points of which are in the same plane.

15. Def. Geometry is the science which treats of position, magnitude, and form.

Points, lines, surfaces, and solids, with their relations, arc the geometrical conceptions, and constitute the subject-matter of Geometry.

16. Plane Geometry treats of plane figures.

Plane figures are either rectilinear, curvilinear, or mixtilinear.

Plane figures formed by straight lines are called rectilinear figures ; those formed by curved lines are called curvilinear fig- ures ; and those formed by straight and curved lines are called mixtilinear figures.

17. Def. Figures which have the same form are called Similar Figures. Figures which have the same extent are called Equivalent Figures. Figures which have the same form and extent are called Equal Figures.

On Straight Lines.

18. If the direction of a straight line and a point in the line be known, the position of the line is known; that is, a straight line is determined in position if its direction and one of its points be known.

Hence, all straight lines which pass through the same point in the same direction coincide.

Between two points one, and but one, straight line can be drawn ; that is, a straight line is determined in position if two of its points be known.

Of all lines between two points, the shortest is the straight iine ; and the straight line is called the distance between the two points.

DEFINITIONS.

The point from which a line is drawn is called its origin.

19. If a line, as G B, ± % £, be produced through C,

the portions GB and G A may be regarded as different lines having opposite directions from the point G.

Hence, every straight line, as A B, £ £ , has two opposite

directions, namely from A toward B, which is expressed by say- ing line A B, and from B toward A, which is expressed by saying line B A.

20. If a straight line change its magnitude, it must become

longer or shorter. Thus by prolonging A B to C, A f ?,

AC = AB + BC; and conversely, BG = AG~AB.

If a line increase so that it is prolonged by its own magnitude several times in succession, the line is multiplied, and the result- ing line is called a multiple of the given line. Thus, if A B =

BG = G D, etc., £_* % £_!, then A C= 2 A B, AB =

3AB, etc.

It must also be possible to divide a given straight line into an assigned number of equal parts. For, assumed that the »th part of a given line were not attainable, then the double, triple, quadruple, of the nth part would not be attainable. Among these multiples, however, we should reach the nth multiple of this nth part, that is, the line itself. Hence, the line itself would not be attainable ; which contradicts the hypothesis that we have the given line before us.

Therefore, it is always possible to add, subtract, multiply, and divide lines of given length.

21. Since every straight line has the property of direction, it must be true that two straight lines have either the same direction or different directions.

Two straight lines which have the same direction, without coin- ciding, can never meet ; for if they could meet, then we should have two straight lines passing through the same point in the same direction. Such lines, however, coincide. § 18

8 GEOMETRY. BOOK I.

22. Two straight lines which lie in the same plane and have different directions must meet if sufficiently prolonged ; and must have one, and but one, point in common.

Conversely : Two straight lines lying in the same plane ivhich do not meet have the same direction; for if they had different directions they would meet, which is contrary to the hypothesis that they do not meet.

Two straight lines which meet have different directions; for if they had the same direction they would never meet (§ 21), which is contrary to the hypothesis that they do meet.

On Plane Angles.

23. Def. An Angle is the difference in direction of two lines. The point in which the lines (prolonged if necessary) meet is called the Vertex, and the lines are called the Sides of the angle.

An angle is designated by placing a letter at its vertex, and one at each of its sides. In reading, we name the three let- ters, putting the letter at the vertex between the other two. When the point is the vertex of but one angle we usually name the letter at the vertex only ; thus, in Fig. 1, we read the angle by

calling it angle A. But in Fig. 2, H is the common vertex of two angles, so that if we were to say the angle H, it would not be known whether we meant the angle marked 3 or that marked 4. We avoid all ambiguity by reading the former as the angle E H D, and the latter as the angle E H F.

DEFINITIONS.

9

angles

D

The magnitude of an angle depends wholly upon the extent of opening of its sides, and not upon their length. Thus if the sides of the angle B AC, namely, A B and A C, be prolonged, their extent of opening will not be altered, and the size of the angle, consequently, will not be changed.

24. Def. Adjacent Angles are angles having a common vertex and a common side between them. Thus the C D E and CDF are adjacent angles.

25. Def. A Right Angle is an angle included between two straight lines which meet each other so that the two adjacent angles formed by producing one of the lines

through the vertex are equal. Thus if the straight line A B meet the straight line C D so that the adjacent angles A BC and ABD are equal to one another, each of these an- gles is called a right angle.

26. Def. Perp>endicular Lines are lines which make a right angle with each other.

27. Def. An Acute Angle is an angle less than a right angle ; as the angle B A C.

28. Def. An Obtuse Angle is an angle greater than a right angle ; as the angle DEF.

29. Def. Acute and obtuse angles, in distinction from right angles, are called ob- lique angles ; and intersecting lines which are not perpendicular to each other are called oblique lines.

30. Def. The Complement of an angle is the difference between a right angle and the given angle. Thus A B D is the complement

B

D

D

of the angle BBC; also D B C is, the com- plement of the angle ABD.

10

GEOMETRY. BOOK I.

31. Def. The Supplement of an angle is the difference between two right angles and the given angle. Thus A CD is the supplement of the angle DC B\ also D C B is the supplement of the angle AC D.

32. Def. Vertical Angles are angles which have the same vertex, and their sides extending in opposite directions. Thus the angles A OB and COB are vertical angles, as also the angles A 0 C sm&DOB.

1>

B

On Angular Magnitude.

,1'

0

B>

C A

33. Let the lines B B' and A A' be in b

the same plane, and let B B' be perpen- dicular to A A' at the point 0.

Suppose the straight line 0 C to move in this plane from coincidence with 0 A, about the point 0 as a pivot, to the po- sition 0 C ; then the line 0 C describes or generates the angle A 0 C.

The amount of rotation of the line, from the position 0 A to the position 0 C, is the Angular Magnitude A 0 C.

If the rotating line move from the position 0 A to the po- sition 0 B, perpendicular to 0 A, it generates a right angle ; to the position 0 A' it generates two right angles ; to the position OB', as indicated by the dotted line, it generates three right angles; and if it continue its rotation to the position 0 A, whence it started, it generates four right angles.

Hence the whole angular magnitude about a point in a plane is equal to four right angles, and the angular magnitude about a point on one side of a straight line drawn through that point is equal to two right angles.

DEFINITIONS.

11

0

//

o

Fig. 2.

34. Now since the augular magnitude about the point 0 is neither increased nor diminished by the number of lines which radiate from that point, the sum of all the angles about a point in a plane, as AOB+BOC+COD, etc., in Fig. 1, is equal to four right angles ; and the sum of all the angles about a point on one side of a straight line drawn through that point, as AOB+BOC+COD, etc., Fig. 2, is equal to two right angles.

Hence two adjacent angles, OCA and OGB, jy

formed by two straight lines, of which one is produced from the point of meeting in both di- rections, are supplements of eacli other, and may J be called supplementary adjacent angles.

On the Method of Superposition.

35. The test of the equality of two geometrical magnitudes is that they coincide point for point.

Thus, two straight lines are equal, if they can be so placed that the points at their extremities coincide. Two angles arc equal, if they can be so placed that their vertices coincide in position and their sides in direction.

In applying this test of equality, we assume that a line may be moved from one place to another without altering its length ; that an angle may be taken up, turned over, and put down, without altering the difference in direction of its sides.

12

GEOMETRY. BOOK I.

This method enables us to com- pare unequal magnitudes of the same kind. Suppose we have two angles, ABC and A' B' C. Let the side B C be placed on the side B' C, so that the vertex B shall fall on B', then if the side B A fall on B' A1, the angle ABC equals the angle A' B' C ; if the side B A fall between B' C and B' A' in the direction B' D, the angle A B C is less than A' B' C" ; but if the side B A fall in the direction B'E, the angle A B C is greater than A' B' C.

This method of superposition en- B q

ables us to add magnitudes of the

same kind. Thus, if we have two c D

straight lines A B and CD, by A B

placing the point C on B, and keeping C D in the same direc- tion with A B, we shall have one continuous straight line A D equal to the sura of the lines A B and C D.

Again : if we have the angles ABC and D E F, by placing the vertex B on E and the side BC in the direction of ED, the angle ABC will take the position A ED, and the angles D E F and ABC will together equal the an- gle AEF.

Mathematical Terms.

36. Def. A Demonstration is a course of reasoning by which the truth or falsity of a particular statement is logically established.

37. Def. A Theorem is a truth to be demonstrated.

38. Def. A Construction is a graphical representation of a geometrical conception.

39. Def. A Problem is a construction to be effected, or a question to be investigated.

DEFIXITIONS. 13

40. Def. An Axiom is a truth which is admitted without demonstration.

41. Def. A Postulate is a problem which is admitted to be possible.

42. Def. A Proposition is either a theorem or a problem.

43. Def. A Corollary is a truth easily deduced from the proposition to which it is attached.

44. Def. A Scholium is a remark upon some particular fea- ture of a proposition.

45. Def. An Hypothesis is a supposition made in the enunciation of a proposition, or in the course of a demonstration.

46. Axioms.

1. Tilings which are equal to the same thing are equal to each

other.

2. When equals are added to equals the sums are equal.

3. When equals are taken from equals the remainders are equal.

4. When equals are added to unequals the sums are unequal.

5. When equals are taken from unequals the remainders are

unequal.

6. Things which are double the same tiling, or equal things,

are equal to each other.

7. Things which are halves of the same thing, or of equal

things, are equal to each other.

8. The whole is greater than any of its parts.

9. The whole is equal to all its parts taken together.

47. Postulates.

Let it be granted —

1. That a straight line can be drawn from any one point to any

other point.

2. That a straight line can be produced to any distance, or can

be terminated at any point.

3. That the circumference of a circle can be described about any

centre, at any distance from that centre.

14

GEOMETRY. BOOK I.

48. Symbols and Abbreviations.

.'. therefore.

= is (or are) equal to.

Z angle.

A angles.

A triangle.

A triangles.

II parallel. O parallelogram HJ parallelograms. _L perpendicular. Jl perpendiculars, rt. Z right angle, rt. A right angles.

> is (or are) greater than. < is (or are) less than, rt. A right triangle, rt. A right triangles.

O circle.

(D circles.

+ increased by.

— diminished by.

X multiplied by.

-r- divided by.

Post, postulate.

Def. definition.

Ax. axiom.

Hyp. hypothesis.

Cor. corollary.

Q. E. D. quod erat demonstran- dum.

Q. E. F. quod erat faciendum.

Adj. adjacent.

Ext.- int. exterior-interior.

Alt. -int. alternate-interior.

Iden. identical.

Cons, construction.

Sup. supplementary.

Sup. adj. supplementary-adja- cent.

Ex. exercise.

111. illustration.

perpendicular and oblique lines. 15

On Perpendicular and Oblique Lines.

Proposition I. Theorem.

49. When one straight line crosses another straight line the vertical angles are equal.

P

Let line 0 P cross A B at C.

We are to prove Z OCB = Z A C P.

ZOCA + ZOCB = 2 rt. A, § 34

{fn ing siqy.-adj. A).

ZOCA + ZACP=2vt. A, § 34

(being sup. - adj. A).

.'.ZOCA + ZOCB = ZOCA + ZACP. Ax. 1.

Take away from each of these equals the common Z 0 C A. Then ZOCB = Z AC P.

In like manner we may prove

Z ACO = Z PCB.

Q. E. D.

50. Corollary. If two straight lines cut one another, the four angles which they make at the point of intersection are together equal to four right angles.

L6 GEOMETRY. BOOK I.

Proposition II. Theorem.

51. When the sum of two adjacent angles is equal to two right angles, their exterior sides form one and the same straight line.

0

--F

Let the adjacent angles Z OCA + Z 0 C B = 2 rt. A. We are to prove A C and C B in the same straight line. Suppose C F to be in the same straight line with A C.

Then ZOCA + ZOCF=2 it. A. §34

(being sup. -adj. A).

But ZOCA + ZOCB = 2Tt A. Hyp.

.•.ZOCArt-ZOCF=ZOCA + ZOCB. Ax. 1.

Take away from each of these equals the common Z 0 C A.

Then Z OCF=Z OGB.

.'. C B and C F coincide, and cannot form two lines as rep- resented in the figure.

.'.AC and C B are in the same straight line.

Q. E. D.

PERPENDICULAR AND OBLIQUE LINES.

17

Proposition III. Theorem. 52. A perpendicular measures the shortest distance from a point to a straight line.

Let A B be the given straight line, C the given point, and CO the perpendicular.

We are to prove C 0 < any other line drawn from C to A B, as C F.

Produce CO to E, making 0 E = C 0.

Draw EF. On A B as an axis, fold over 0 C F until it comes into the plane of OFF.

The line 0 C will take the direction of 0 E, (since ZCOF=ZEOFf each being a rt. Z. ).

The point C will fall upon the point E, (since 0 0— 0 E by cons. ). .-.line C F= line F E} § 18

(having their extremities in tlie same points).

.'. CF+ FE=2 CF, and CO +OE=2 CO. Cons.

But CO+OE<CF+FE, §18

(a straight line is tlie shortest distance between two points). Substitute 2 C 0 for C 0 + 0 E,

and 2 CF for CF+ F E ; then we have 2 C0<2CF.

.'. CO < CF.

Q. E. D.

18

GEOMETRY. BOOK I.

Proposition IV. Theorem.

53. Two oblique lines drawn from a point in a perpen- dicular, cutting off equal distances from the foot of the per- pendicular, are equal.

Let F C be the perpendicular, and C A and C 0 two oblique lines cutting off equal distances from F.

We are to prove C A = C 0.

Fold over C FA, on C F as an axis, until it comes into the plane of C F 0.

FA will take the direction of FO, (since Z.CFA = ZCFO, each being art. /.).

Point A will fall upon point 0, (FA = FO, by hyp.).

.-.line C A = line CO, (their extremities being the same points).

§18

Q. E. D.

PERPENDICULAR AND OBLIQUE LINES. 19

Proposition V. Theorem.

54. The sum of two lines drawn from a point to the ex- tremities of a straight line is greater than the sum of two other lines similarly drawn, but included by them.

A B

Let C A and C B be two lines drawn from the point C to the extremities of the straight line A B. Let 0 A and 0 B be two lines similarly drawn, but included byCA andCB.

We are to prove CA + CB>OA + OB.

Produce A 0 to meet the line C B at E.

Then AC+ CE>AO+ OB, §18

(a straight line is the shortest distance between tico 2>oints),

and BE+ OE> BO. § 18

Add these inequalities, and we have

CA + CE+BE+OE>OA + OE+OB.

Substitute for CE + BE its equal C B,

and take away 0 E from each side of the inequality.

We have CA + CB > 0 A + 0 B.

Q. E. D

20

GEOMETRY.

BOOK I.

Proposition VI. Theorem.

55. Of two oblique lines drawn from the same point in a perpendicular, cutting off unequal distances from the foot of the perpendicular } the more remote is the greater. C

Let G F be perpendicular to A B, and C K and C H two oblique lines cutting off unequal distances from F.

We are to prove C H > C K.

Produce C F to E, making FE=CF. Draw EK and EH.

CH=HE,?m&CK= KE, §53

(two oblique lines drawn from the same point in a J_, cutting off equal dis- tances from the foot of the _L, are equal).

But CH+HE>CK+KE, §54

(The sum of two oblique lines drawn from a point to the extremities of a straight line is greater than tht sum of two other lines similarly drawn, but included by them);

.\2 CII>2CK;

.\CH>CK.

Q. E. D.

56. Corollary. Only two equal straight lines can be drawn from a point to a straight line ; and of two unequal lines, the greater cuts off the greater distance from the foot of the perpen- dicular.

PERPENDICULAR AND OBLIQUE LINES.

21

Proposition VII. Theorem.

57. Two equal oblique lines, drawn from the same point in a perpendicular, cut off equal distances from the foot of the perpendicular.

. C

Let C F be the perpendicular, and C E and C K be two equal oblique lines drawn from the point C.

We are to prove

FE=FK.<

Fold over C FA on C F as an axis, until it comes into the plane of CFB.

The line FE will take the direction FK, (Z CFE = ZCFK, each being a rt. Z ).

Then the point E must fall upon the point K ;

otherwise one of these oblique lines must be more remote from the _L,

and .'. greater than the other; which is contrary to the hypothesis. § 55

.'.FE = FK.

Q. E. D.

GEOMETRY. — BOOK I.

Proposition VIII. Theorem.

58. If at the middle point of a straight line a perpen- dicular be erected,

I. Any point in the perpendicular is at equal distances from the extremities of the straight line.

II. Any point without the perpendicular is at unequal distances from the extremities of the straight line.

Let P R be a perpendicular erected at the middle 01 the straight line A B, 0 any point in PR, and C any point without P R.

I. Draw OA and OB. We are to prove 0 A = 0 B. Since P A = P B,

OA = OB, § 53

(two oblique lines drawn from tlie same jmint in a ±, cutting off equal dis- tances from tJw, foot of the ±, are equal).

II. Draw CA and C B.

We are to prove C A and C B unequal.

One of these lines, as CA, will intersect the _L. From D, the point of intersection, draw D B.

PERPENDICULAR AND OBLIQUE LINES. 23

DB = DA, §53

(five oblique lines drawn from the same point in a ±, cutting off equal dis- tances from the foot of the ±, are equal).

CB< CD + I)B, § 18

(a straigld line is Uie shortest distance between two points).

Substitute for D B its equal D A, then

CB< CD + DA.

But CD + DA = CA, Ax. 9.

,'.CB< CA.

Q. E. D.

59. The Locus of a point is a line, straight or curved, con- taining all the points which possess a common property.

Thus, the perpendicular erected at the middle of a straight line is the locus of all points equally distant from the extremi- ties of that straight line.

60. Scholium. Since two points determine the position of a straight line, two points equally distant from the extremities of a straight line determine the perpendicular at the middle point of that line.

Ex. 1. If an angle be a right angle, what is its complement?

2. If an angle be a right angle, what is its supplement 1

3. If an angle be # of a right angle, what is its complement 1

4. If an angle be £ of a right angle, what is its supplement 1

5. Show that the bisectors of two vertical angles form one and the same straight line.

6. Show that the two straight lines which bisect the two pairs of vertical angles are perpendicular to each other.

24

GEOMETRY. — BOOK I.

Proposition IX. Theorem.

61. At a point in a straight line only one perpendicular to that line can be drawn ; and from a point without a straight line only one perpendicular to that line can be drawn.

AE

A F

B

Fig. 1.

EB

Fig. 2.

Let B A {fig. 1) be perpendicular to C D at the point B.

We are to prove B A the only p>erpendicular to G D at the point B.

If it be possible, let B E be another line _L to G D at B.

Then Z EBD is a rt. Z. §26

But Z ABD is art. Z. § 26

.'.Z EBD = Z ABD. Ax. 1.

That is, a part is equal to the whole ; which is impossible.

In like manner it may be shown that no other line but B A is _L to GD at B.

Let AB {fig. 2) be perpendicular to G D from the point A. We are to prove A B the only _L to G D from the point A.

If it be possible, let A E be another line drawn from A 1_ to GD.

Conceive Z A E B to be moved to the right until the ver- tex E falls on B, the side E B continuing in the line G D.

Then the line E A will take the position B F.

Now if A E be J_ to C D, B F is JL to C D, and there will be two J» to C D at the point B ; which is impossible.

In like manner, it may be shown that no other line but A B is _L to GD from A. q. e. d.

62. Corollary. Two lines in the same plane perpendicular to the same straight line have the same direction ; otherwise they would meet (§ 22), and we should have two perpendicular lines drawn from their point of meeting to the same line ; which is impossible.

PARALLEL LINES. 25

On Parallel Lines.

63. Parallel Lines are straight lines which lie in the same plane and have the same direction, or opposite directions.

Parallel lines lie in the same direction, when they are on the same side of the straight line joining their origins.

Parallel lines lie in opposite directions, when they are on opposite sides of the straight line joining their origins.

64. Two parallel lines cannot meet. § 21 65.' Two lines in the same plane perpendicular to a given

line have the same direction (§ 62), and are therefore parallel.

66. Through a given point only one line can be drawn par- allel to a given line. § 18

If a straight line EF cut two other straight lines A B and C D, it makes with those lines eight angles, to which par- ticular names are given.

The angles 1, 4, 6, 7 are called Interior angles.

The angles 2, 3, 5, 8 are called Exterior angles.

The pairs of angles 1 and 7, 4 and 6 are called Alternate- interior angles.

The pairs of angles 2 and 8, 3 and 5 are called Alternate- exterior angles.

The pairs of angles 1 and 5, 2 and 6, 4 and 8, 3 and 7 are called Exterior-iyiterior angles.

GEOMETRY. BOOK I.

Proposition X. Theorem.

67. If a straight line be perpendicular to one of two parallel lines } it is perpendicular to the other.

M>

E-

~~-X

Let A B and E F be two parallel lines, and let H K be p erp en die ular to A B.

We are to prove H K _L to E F.

Through C draw UN JL to UK.

Then ' MN is II to A B. § 65

{Two lines in the same plane JL to a given line arc parallel).

But EFi&WtoAB, By p.

.'. E F coincides with M N. § G6

{Through the same point only one line can be drawn \\ to a given line).

.'. E F is ± to HK, that is UK is -L to EF.

Q. E. D.

PARALLEL LINES. 27

Proposition XI. Theorem.

68. If two parallel straight lines he cut by a third straight line the alternate-interior angles are equal.

A B F

c b H

Let E F and Gil be two parallel straight lines cut by the line BO.

We are to prove Z B = /.0.

Through 0, the middle point of B C, draw A D ±.to G H.

Then A D is likewise _L to E F, § 67

(a straight line ± to one of two lis is ± to the oilier),

that is, C D and B A are both _L to A D. Apply figure C 0 D to figure B 0 A so that 0 D shall fall on OA.

Then 0 C will fall on OB,

(since Z CO D = /.BOA, being vertical A) ;

and point C will fall upon B,

(siiice 0 C — 0 B by construction).

Then _L CD will coincide with ± B A, § 61

(/rem a point without a straig/U line only one ± to tliat line can be drawn).

.'. Z. 0 G D coincides with Z 0 B A, and is equal to it.

Q. E. D.

Scholium. By the converse of a proposition is meant a proposition which has the hypothesis of the first as conclusion and the conclusion of the first as hypothesis. The converse of a truth is not necessarily true. Tims, parallel lines never meet ; its converse, lines which never meet are parallel, is not true unless the lines lie in the same plane.

Note. — The converse of many propositions will be omitted, but their statement and demonstration should be required as an important exercise for the student.

28 GEOMETRY. — BOOK I.

Proposition XII. Theorem.

69. Conversely : When two straight lines are cut by a third straight line, if the alternate-interior angles be equal, the two straight lines are parallel.

Let E F cut the straight lines A B and C D in the points H and K, and let the Z A HK = Z HKD.

We are to prove A B II to C D. Through the point H draw M N II to CD ;

then Z MHK = Z H K D, § 68

(being alt. -int. A ).

But Z A HK = Z HKD, Hyp.

.'.Z MHK= Z AHK. Ax. 1.

.'.the lines M N and A B coincide. But J/iVis II to CD; Cons.

.'. AB, which coincides with M N, is II to CD.

Q. E. o.

PARALLEL LINES. 29

Proposition XIIT. Theorem.

70. If 'two parallel lines be cut by a third straight line, the exterior-interior angles are equal.

E

Let AB and C D be two parallel lines cut by the straight line E F, in the points II and K.

We are to prove Z EHB = Z H K D.

ZEHB = /.AIIK, §49

{being vertical A).

But ZAIIK = ZIIKD, § 08

(being alt. -int. A).

.-.Z EIIB = Z HKD. Ax. 1

In like manner we may prove

ZEIIA = ZHKC.

Q. E. D.

71. Corollary. The alternate-exterior angles, EHB and C K F, and also A II E and D K F} are equal.

30 GEOMETRY. — BOOK I.

Proposition XIV. Theorem.

72. Conversely : When two straight lines are cut by a third straight line, if the exterior-interior angles be equal, these two straight lines are parallel.

Let EF cut the straight lines AB and CD in the points II and K, and let the Z EHB = Z HKD.

We are to prove A B II to C D . Through the point H draw the straight line M N II to CD.

Then ZEHN=ZHKD, §70

(being ext. -int. A ).

But Z EHB = Z HKD. Hyp.

.-. Z EHB = Z EHN. Ax. 1.

.*. the lines M N and A B coincide. But MNh II to CD, Cons.

,\ AB, which coincides with M N, is II to CD.

Q. E. D.

PARALLEL LINES. 31

Proposition XV. Theorem.

73. If two parallel lines be cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles.

E

Let AB and C D be two parallel lines cut by the straight line EF in the points II and K.

We are to prove A B II K + Z II K D = two rt. A.

Z EIIB + Z BIIK=2 rt. A, § 34

(being sup. -adj. A ).

But Z EIIB = A II KB, §70

(being exl.-int. A ).

Substitute Z HKD for Z EIIB in the first equality; then Z BHK + Z HKD = 2 rt. A.

Q. E. d.

32 GEOMETRY. — BOOK I.

Proposition XVI. Theorem.

74. Conversely : When two straight lines are cut hy a third straight line, if the two interior angles on the same side of the secant line he together equal to two right angles, then the two straight lines are parallel.

E

Let EF cut the straight lines AB and CD in the points H and K, and let the Z. B II K + A H K D equal two right angles.

We are to prove A B II to CD.

Through the point IT draw MN II to CD.

Then Z. NHK + Z HKD = 2 rt. A, § 73

(being two interior A on the same side of the secant line).

But ZBHK+ZHKD = 2 rt. A. Hyp.

.'./.NHK+AHKD = ABHK+/.HKD. Ax. 1. Take away from each of these equals the common /.HKD, then ANHK=ABHK.

.'. the lines A B and M N coincide. But MN\% II to CD; Cons.

.', AB. which coincides with M iV, is II to CD.

Q. E D.

PARALLEL LINES.

83

Proposition XVII. Theorem.

75. Two straight lines which are parallel to a third straight line are parallel to each other.

H

i

A' Let AB and CD be parallel to E F. We are to prove A B II to C D.

Draw H K± to EF.

Since C D and EF are II, HK is J_ to C D, § 67

(if a straigld line be _L to one of tiro lis, it is _L to the other also).

Since A B and EF are II, UK is also _L to A B, § 67

.\ZHOB = Z II PD,

(each being art. /.).

.'. ABisW to CI), § 72

(ivhen two straight lines are cut by a third straight line, if the ext. -int. A be equal, the two Hues are II ).

Q. E. D.

34

GEOMETRY.

BOOK I.

Proposition XVIII. Theorem.

76. Two parallel lines are everywhere equally distant from each other.

E M H

I)

F P K

Let A B and CD be two parallel lines, and from any two points in A B, as E and II, let EF and II K be drawn perpendicular to A B. We are to prove E F = UK

Now EF and UK are J_ to C D, § 67

(a line ± to one of two lis is ± to the other also).

Let M be the middle point of E H. * Draw MP ± to A B. On MP as an axis, fold over the portion of the figure on the right of MP until it comes into the plane of the figure on the left.

MB will fall on MA, (for ZPMH=APME, each being art. Z ) ;

the point H will fall on E, {for M H= ME, by hyp.) ;

HK will fall on EF,

(for ZMHK= ZMEF, each being art. Z ) ;

and the point K will fall on E F, or E F produced.

Also, PD will fall on P C,

(Z MPK= Z MPF', each being a rt. Z) ;

and the point K will fall on P C.

Since the point K falls in both the lines EF and P C,

it must fall at their point of intersection F.

.\HK= EF, § 18

(their extremities being the same points).

Q. E. D.

PARALLEL LINES.

35

Proposition XIX. Theorem.

77. Two angles whose sides are parallel, two and two, and lie in the same direction, or opposite directions, from their vertices, are equal. A D

D>

Fig. 1. Fig. 2.

Let A B and E (Fig. 1) have their sides B A and E D, and BC and EF respectively, parallel and lying in the same direction from their vertices.

We are to prove the Z B = Z E.

Produce (if necessary) two sides which are not II until they

intersect, as at H ;

then Z B = Z DHC, ' § 70

(being i.r/.-inf. A ),

and ZE = ZDHC, §70

.'.ZB = ZE. Ax. 1

Let A B' and E> (Fig. 2) have B' A1 and W D', and B' C and E' F' respectively, parallel and lying in oppo- site directions from their vertices.

We are to prove the Z B' = Z E1. Produce (if necessary) two sides which are not intersect, as at H1.

Then Z B' = Z E IT C,

[I" in'/ i xt. -inf. A),

and

Z E' = Z E> H' C

(beiiuj alt. -int. A ) ;

.'. Z B' = Z E>

until they §70 §68

Ax. 1.

Q. E. D.

36 GEOMETRY. BOOK I.

Proposition XX. Theorem.

78. If two angles have two sides parallel and lying in the same direction from their vertices, while the other two sides are parallel and lie in opposite directions, then the two angles are supplements of each other.

Let A BC and D E F be two angles having B C and ED parallel and lying in the same direction from their vertices, while E F and B A are parallel and lie in opposite directions.

We are to prove /.ABC and Z D E F supplements of each other.

Produce (if necessary) two sides which are not II until they intersect as at H.

ZABC = ZBHD, §70

(being ext.-int. A ).

ZDEF==ZBHE, §68

(being alt. -int. A ).

But Z B II D and Z B HE are supplements of each other, § 34

sup. -adj. A ).

.'. Z ABC and Z D E F, the equals of Z BED and Z B H E, are supplements of each other.

Q. E. D.

TRIANGLES. 37

On Triangles.

79. Def. A Triangle is a plane figure bounded by three straight lines.

A triangle has six parts, three sides and three angles.

80. When the six parts of one triangle are equal to the six parts of another triangle, each to each, the triangles are said to be equal in all respects.

81. Def. In two equal triangles, the equal angles are called Homologous angles, and the equal sides are called Homologous sides.

82. In equal triangles the equal sides are opposite the equal angles.

ISOSCELES. EQUILATERAL.

83. Def. A Sealene triangle is one of which no two sides are equal.

84. Def. An Isosceles triangle is one of which two sides <ire equal.

85. Def. An Equilateral triangle is one of which the three sides are equal.

86. Def. The Base of a triangle is the side on which the triangle is supposed to stand.

In an isosceles triangle, the side which is not one of the equal sides is considered the base.

38

GEOMETRY.

BOOK I.

87. Def. A Right triangle is one which has one of the angles a right angle.

88. Def. The side opposite the right angle is called the Hypotenuse.

89. Def. An Obtuse triangle is one which has one of the angles an obtuse angle.

90. Def. An Acute triangle is one which has all the angles acute.

EQUIANGULAR.

91. Def. An Equiangular triangle is one which has all the angles equal.

92. Def. In any triangle, the angle opposite the base is called the Vertical angle, and its vertex is called the Vertex of the triangle.

93. Def. The Altitude of a triangle is the perpendicular distance from the vertex to the base, or the base produced.

94. Def. The Exterior angle of a triangle is the angle in- cluded between a side and an adjacent side produced, as /. CBD.

95. Def. The two angles of a triangle which are opposite the exterior angle, are called the two opposite interior angles, as A A and G.

-in ANGLES. 39

96. Any side of a triangle is less than the sum of the other two sides.

Since a straight line is the shortest distance between two

points,

AC<AB+BC

97. Am/ tide of a triangle is greater than the difference of the other two sides.

In the inequality A C < A B + B C,

take away A B from eacli side of the inequality.

Then AC~AB<BC; or

BO AG- AB.

Ex. 1. Show that the sum of the distances of any point in a triangle from the vertices of three angles of the triangle is greater than half the sum of the sides of the triangle.

2. Show that the locus of all the points at a given distance from a given straight line A B consists of two parallel lines, drawn on opposite sides of A B, and at the given distance from it.

3. Show that the two equal straight lines drawn from a point to a straight line make equal acute angles with that line.

4. Show that, if two angles have their sides perpendicular, each to each, they are either equal or supplementary.

40 GEOMETRY. BOOK I.

Proposition XXI. Theorem.

98. The sum of the three angles of a triangle is equal to two right angles.

C '

Let A B G be a triangle.

We are to prove AB-\rABCA + AA = two rt. A.

Draw C E II to A B, and prolong A C.

ThenZ ECF+ZECB+ZBCA = 2 rt. A, § 34 (the sum of all the A about a point on tlu same side of a straight line

= 2rt. A ).

But ZA=ZECF, §70

(being ext. -int. A ),

and Z B = Z. BCE, §68

( being alt. -int. A ) .

Substitute for Z E CF&nd Z B C E their equal A, A and B. Then Z A + Z B + Z B OA = 2 rt. A.

Q. E. D.

99. Corollary 1. If the sum of two angles of a triangle be known, the third angle can be found by taking this sum from two right angles.

100. Cor. 2. If two triangles have two angles of the one equal to two angles of the other, the third angles will be equal.

TRIANGLES. 41

101. Cor. 3. If two right triangles have an acute angle of the one equal to an acute angle of the other, the other acute angles will be equal.

102. Cor. 4. In a triangle there can be but one right angle, or one obtuse angle.

103. Cor. 5. In a right triangle the two acute angles are complements of each other.

104. Cor. 6. In an equiangular triangle, each angle is one third of two right angles, or two thirds of one right angle.

Proposition XXII. Theorem.

105. The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Let BC II be an exterior angle of the triangle ABC. We are to prove Z B Gil = Z A + Z B.

Z BCII + Z ACB=2vt A, § 34

(beivcf sup.-mlj. A).

Z A + Z B+ Z ACB = 2 rt. A, § 98

{three AofaA = two rt. A ).

.'.Z BCII+ Z ACB = Z A + Z B + Z A C B. Ax. 1.

Take away from each of these equals the common Z A C B ;

then ZBCII = ZA + ZB.

Q. E- D.

42

GEOMETRY. BOOK I.

Proposition XXIII. Theorem.

106. Two triangles are equal in all respects when two sides and the included angle of the one are equal respectively to two sides and the included angle of the other.

B A'

In the triangles ABO and A' B' C, let AB = A'h',

A $=A' C',Z A=Z A'.

We are to prove A A B C = A A1 B' C.

Take up the A A B C and place it upon the A A' B' C so that A B shall coincide with A' B'.

Then A G will take the direction of A' C, (for Z A = Z A', by hyjy.),

the point C will fall upon the point C, (forAC=A'Ci,byhyp.);

.\CB = C B',

(their extremities being the same points).

,'. the two A coincide, and are equal in all respects.

IS

Q. E. D,

TRIANGLES. 43

Proposition XXIV. Theorem.

107. Two triangles are equal in all respects when a side and two adjacent angles of the one are equal respectively to a side and two adjacent angles of the other.

C C

B A'

In the triangles A BC and A' B' C, let A B = A' B', Z A = Z A', Z B = Z B'.

We are to prove A A B C = A A' B' C.

Take up A A BC and place it upon A A' B' C, so that A B shall coincide with A' B'.

A C will take the direction of A' C, (for Z A = ZA', by. hyp.) ;

the point C, the extremity of A C, will fall upon A' C or A' C produced.

B 0 will take the direction of B1 C, (for ZB = ZB', by hyp.);

the point C, the extremity of B C, will fall upon B' C or B1 C produced.

.*. the point C, falling upon both the lines A' C and B' C, must fall upon a point common to the two lines, namely, C.

.*. the two A coincide, and are equal in all respects.

Q. E. D.

44

GEOMETRY.

BOOK I.

Proposition XXV. Theorem.

108. Two triangles are equal when the three sides of the one are equal respectively to the three sides of the other. B B>

V

B> In the triangles ABC and A' B' C, let A B = A' B'} A C = A' C, BG = B'C. We are to prove A A B G = A A' B' C. Place A A' B' C in the position A B' C, having its greatest side A' C in coincidence with its equal A C, and its vertex at B', opposite B.

Draw B B' intersecting A C at H.

Since AB = AB', Hyp.

point A is at equal distances from B and B'.

Since B C = B> C, Hyp.

point C is at equal distances from B and B'.

.*. A C is JL to BB' at its middle point, § 60

{two points at equal distances from the extremities of a straight line deter- mine the _l_ at the middle of that line).

Now if A A B' C be folded over on A G as an axis until it comes into the plane of A ABC,

II B' will fall on H B, (for /.AHB = ZAHB', each being a rt. Z),

and point B' will fall on B, (for HW = HB).

.'. the two A coincide, and are equal in all respects.

Q. E. D.

TRIANGLES.

45

Proposition XXVI. Theorem.

109. Two right triangles are equal when a side and the hypotenuse of the one are equal respectively to a side and the hypotenuse of the other.

C B*

In the right triangles ABC and A' B' C", let AB = A' B', and AC = A'C.

We are to prove A A B C = A A' B' C.

Take up the A A B C and place it upon A A' B' C", so that A B will coincide with A' B'.

Then B C will fall upon B' C,

{for ZABC=ZA'B'C, each being a rt. Z ),

and point C will fall upon C ;

otherwise the equal oblique lines A C and A' C would cut off unequal distances from the foot of the _L, which is im- possible, § 57 (two equal oblique lines from a point in a JL cut off equal distances from the foot of the. _L).

,\ the two A coincide, and are equal in all respects.

Q. E. D.

46

GEOMETRY.

BOOK I.

Proposition XXVII. Theorem.

110. Two right triangles are equal token the hypotenuse and an acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other.

In the right triangles ABC and A' B' C, let AG = A' C, and Z A= Z A'.

We are to prove	AABC = AA'B'C.
	AC = A' C,	Hyp.
	z a=z a;	Hyp.

then Z C = Z C, \ § 101

(if two rt. A have an acute Z of the one equal to an acute A of the other, then the other acute A are equal).

.'.AABC = AA'B'C, § 107

(two A are equal when a side and two adj. A of the one are equal respectively to a side and two adj. A of the other).

Q. E. D.

111. Corollary. Two right triangles are equal when a side and an acute angle of the one are equal respectively to an homologous side and acute angle of the other.

TRIANGLES. 47

Proposition XXYIII. Theorem.

112. In an isosceles triangle the angles opposite the equal sides are equal.

C

Let ABC be an isosceles triangle, having the sides AC and CB equal.

We are to prove Z A = Z B.

From C draw the straight line CE so as to bisect the Z A CB.

Id the A ACE and BCE,

AC=BCf Hyp.

CE=C E, Iden.

ZACE=ZBCE; Cons.

.'.AACE = ABCE, §106

(two & are equal wlicn tin, sides and the included Z. of the one are equal respectively to two sides and the included Z. of the other).

.'.ZA=ZB,

(being liomologous A of equal A ).

Q. E. D.

Ex. If the equal sides of an isosceles triangle be produced, show that the angles formed with the base by the sides produced are equal.

48

GEOMETRY. BOOK I.

Proposition XXIX. Theorem.

113. A straight line which bisects the angle at the vertex af an isosceles triangle divides the triangle into two equal triangles, is perpendicular to the base, and bisects the base.

C

Let the line C E bisect the A AC B of the isosceles AACB.

We are to prove I. AACE = ABCE; II. UneCE ±to AB; III. AE = BE.

I. In the A ACE and B C E,

AC=BC, Hyp.

CE=CE, Iden.

AACE= ZBCE. Cons.

.'.A ACE = A BCE, §106

(having two sides and the included A of the one equal respectively to two sides and the included A of the other).

Also, II. A CEA = Z CEB,

(being homologous A of equal A ).

.*. CEis±to AB, (a straight line meeting another, making the adjacent A equal, is JL to

that line).

Also, III. AE=EB,

(being homologous sides of equal & ).

Q. E. D.

TRIANGLES.

49

Proposition XXX. Theorem.

114. If two angles of a triangle be equal, the sides op- posite the equal angles are equal, and the triangle is isosceles.

In the triangle ABC, let the Z B = Z C. We are to prove AB = AC.

Draw4Z)J_to BC.

In the rt. A A DB and A D C,

AD = AD,

ZB = ZC,

.'. rt. A A D B = rt. A A D C,

Iden.

§ HI

(having a side and an acute Z of the one equal respectively to a side and an acute Z of the other).

.\AB = AC, (being homologous sides of equal &).

Q. E. D.

Ex. Show that an equiangular triangle is also equilateral.

50 GEOMETRY. — BOOK I.

Proposition XXXI. Theorem.

115. If two triangles have two sides of the one equal respectively to two sides of the other, but the included a?igle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

B B

E

In the A ABC and ABE, let A B = A B, BG^BE; but Z ABO Z ABE.

We are to prove A G > A E.

Place the A so that A B of the one shall coincide with A B of the other.

Draw B F so as to bisect Z EBG. Draw EF. In the A EBF&nd GBF

EB = BC, Hyp.

BF=BF, Iden.

ZEBF=Z GBF, Cons.

.\ the A EBFtrnd CBF&ie equal, § 106

(having two sides and the included Z of one equal respectively to two sides and the included Z. of the other). .'.EF=FC, (being homologous sides of equal & ). Now AF+ FE> AE, § 96

(the sum of two sides of a A is greater than the third side). Substitute for FE its equal FG. Then AF+ FG>AE; or, A C> A E.

Q. E. D.

TRIANGLES. 51

Proposition XXXII. Theorem.

116. Conversely: If two sides of a triangle be equal respectively to two sides of another, but the third side of the first triangle be greater than the third side of the second, then the angle opposite the third side of the first triangle is greater than the angle opposite the third side of the second.

In the A ABC and A' B' C, let AB = A'B', AC = A'C') but BOB' C.

We are to prove Z A > Z A'. If Z A = Z A',

then would AABC = AA'B'C, § 106

(having two sides and the included Z. of the one equal respectively to two sides and the included A of the other),

and BC = B C,

(being homologous sides of equal A ).

And if A < A',

then would BC<B'C; § 115

(if two sides of a Abe equal respectively to two sides of another A, but the included Z of the first be greater than the included A of the second, the third side of the first will be greater than the third side of the second.)

But both these conclusions are contrary to the hypothesis ;

.'. Z A does not equal Z A', and is not less than Z A'.

.'.ZA>Z A'.

Q. E. D

52 GEOMETRY. — BOOK I.

Proposition XXXIII. Theorem.

117. Of two sides of a triangle, that is the greater which is opposite the greater angle.

In the triangle ABC let angle AG B be greater than angle B.

We are to prove A B > AG.

Draw C E so as to make A B G E = Z5.

Then EC = EB, §114

(being sides opposite equal A ).

Now AE+EOAG, §96

(the sum of two sides of a A is greater than the third side).

Substitute for EG its e^ual E B. Then

AE+ EB> AG, or

A B > A G.

Q. E. D.

Ex. ABG and ABB are two triangles on the same base A B, and on the same side of it, the vertex of each triangle being without the other. If A G equal A D, show that B C cannot equal B D.

TRIANGLES. 53

Proposition XXXIV. Theorem.

118. Of two angles of a triangle, that is the greater which is opposite the greater side.

D

C

In the triangle ABC let A B be greater than A C.

We are to prove Z AC B > Z /?.

Take A E equal to AC)

Draw^C.

Z AEC = £ ACE, §112

(being A opposite equal sides).

But ZAEOZB, §105

(an exterior Z of a A is greater than either opposite interior Z ),

and ZACB>ZACE.

Substitute for Z A C E its equal Z A EC, then

ZACB>ZAEC.

Much more is Z A CB > Z B.

Q. E. D.

Ex. If the angles ABC and AC B, at the base of an isosceles triangle, be bisected by the straight lines B .0, CD, show that BBC will be an isosceles triangle.

54 GEOMETRY. BOOK I.

Proposition XXXV. Theorem. 119. The three bisectors of the three angles of a triangle meet in a point.

Let the two bisectors of the angles A and C meet at 0, and 0 B be drawn.

We are to prove B 0 bisects the Z B.

Draw the Jfc OK, OP, and OH.

Inthert. A 0 C K and OOP,

OC=OC, Iden.

Z00K = Z00P, Cons.

.-.A OCK=A OOP, § 110

(having the hypotenuse and an acute Z of the one equal respectively to the hypotenuse and an acute Z of tlie other).

.'. OP=OK,

(homologous sides of equal & ).

In the rt. A OA P and OAH,

OA = OA, Iden.

ZOAP = ZOAH, Cons.

.'.AOAP = AOAH, §110

{having the hypotenuse and an acute Z of the one equal respectively to the hypotenuse and an acute Z of the other).

.'.OP=OH,

(being homologous sides of equal & ). But we have already shown 0 P *= 0 K,

.\OII= OK, Ax. 1

Now in rt. A 0 HB and 0 KB

TRIANGLES. 55

OH = OK, and 0 B = 0 B,

.'.AOHB = A OKB, § 109

{having the hypotenuse and a side of the one equal respectively to the hypote- nuse and a side of tJie other).

../. OBH = Z OBKt

(being homologous A of equal &. ).

Q. E. D.

Proposition XXXVI. Theorem. 120. The three perpendiculars erected at the middle joints of the three sides of a triangle meet in a point. A

F

Let DD', EE', FF', be three perpendiculars erected at D, E, F, the middle points of A B, A C, and B C.

We are to prove they meet in some point, as 0.

The two Ji D D' and E E' meet, otherwise they would be parallel, and A B and A C, being _l§ to these lines from the same point A, would be in the same straight line;

but this is impossible, since they are sides of a A.

Let 0 be the point at which they meet.

Then, since 0 is in D L1, which is _L to A B at its middle point, it is equally distant from A and B. § 59

Also, since 0 is in E E', _L to A C at its middle point, it is equally distant from A and C.

.'. 0 is equally distant from B and C ;

.'. 0 is in FF _L to B C at its middle point, § 59

(the locus of all points equally distant from the extremities of a straight line

is the ± erected at the middle of that line).

Q. E. D.

56 GEOMETRY. BOOK I.

Proposition XXXVII. Theorem.

121. The three perpendiculars from the vertices of a tri- angle to the opposite sides meet in a point.

In the triangle ABC, let B P, AH, G K, be the per- pendiculars from the vertices to the opposite sides.

We are to prove they meet in some point, as 0. Through the vertices A, B, C, draw

A'B' II to BO, A' C II to A C, B' C II to A B. In the A A B A1 and ABC, we have

AB = AB, Iden.

ZABA' = ZBAC, §68

(being alternate interior A ),

ZBAA' = ZABC. §68

.-. A ABA' = A ABC, § 107

{having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

.'.A'B = AC,

(being homologous sides of egual A ).

TRIANGLES. 57

In the A C B C and A B C,

BC = BC, Iden.

ZCBC' = Z.BCAt §68

{being alternate interior A ).

ZBCC' = ZOBA. §68

.\ACBC' = AABC, § 107

(having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

.\BG' = AC1

(being homologous sides of equal A ).

But we have already shown A1 B = A 0,

.\A'B = BC, Ax. 1.

.\ B is the middle point of A' C

Since B P is -L to A C, Hyp.

it is J_ to A' C, § 67

(a straight line which is ± to one of two lis is _L to the other also).

But B is the middle point of A' C ;

.'. BP is J_ to A' C at its middle point.

In like manner we may prove that

A H is J_ to A' B' at its middle point,

and C K _L to B' C at its middle point.

.'. B P, A H, and C K are Js erected at the middle points of the sides of the A A' B' C

.'. these J§ meet in a point. § 120

(the three J§ erected at the middle points of the sides of a A meet in a point).

Q. E. D.

58

GEOMETRY. BOOK I.

On Quadrilaterals.

122. Def. A Quadrilateral is a plane figure bounded by- four straight lines.

123. Def. A Trapezium is a quadrilateral which has no two sides parallel.

124. Def. A Trapezoid is a quadrilateral which has two sides, and only two sides, parallel.

125. Def. A Parallelogram is a quadrilateral which has its opposite sides parallel.

TRAPEZIUM.

PARALLELOGRAM.

126. Def. A Rectangle is a parallelogram which has its angles right angles.

127. Def. A Square is a parallelogram which has its angles right angles, and its sides equal.

128. Def. A Rhombus is a parallelogram which has its sides equal, but its angles oblique angles.

129. Def. A Rhomboid is a parallelogram which has its angles oblique angles.

The figure marked parallelogram is also a rhomboid.

RECTANGLE.

QUADRILATERALS. 59

130. Def. The side upon which a parallelogram stands, and the opposite side, are called its lower and upper bases ; and the parallel sides of a trapezoid are called its bases.

131. Def. The Altitude of a parallelogram or trapezoid is the perpendicular distance between its bases.

132. Def. The Diagonal of a quadrilateral is a straight line joining any two opposite vertices.

Proposition XXXVIII. Theorem.

133. The diagonal of a parallelogram divides the figure into two equal triangles.

A E

Let ABC E be a parallelogram, and A C its diagonal. We are to prove AABC=AA EC. In the A ABC and A EC

AC = AC, Iden.

ZACB = ZCAE, §68

( being alt. -int. A ). ZCAB = ZACE, §68

.\AABC = AAEC, § 107

(having a side and two adj. A of the one equal respectively to a side and tiro

adj. A of the other).

Q. E. D.

60 GEOMETRY. BOOK I.

Proposition XXXIX. Theorem.

134. In a parallelogram the opposite sides are equal, and the opposite angles are equal.

B G

A E

Let the figure ABC E be a parallelogram.

We are to prove B C = A E, and AB = EC,

also, ZB = ZE,andZBAE = ZBCE.

Draw A C.

AABC = AAEC, §133

(the diagonal of a O divides the figure into two equal & ).

.-.BC = AE1

and AB = CE,

(being homologous sides of equal A ).

ZB = Z E,

(being homologous A of equal A ).

Z BAC = Z ACEf

and ZEAC = ZACB,

{being homologous A of equal A).

Add these last two equalities, and we have

ZBAC + Z EAC = ZACE+ZACB; or, ZBAE = ZBCE.

Q. E. D.

1 35. Corollary. Parallel lines comprehended between par- allel lines are equal.

QUADRILATERALS. 61

Proposition XL. Theorem.

136. If a quadrilateral have two sides equal and par- allel, then the other two sides are equal and parallel, and the figure is a parallelogram.

B C

Let the figure ABC E be a quadrilateral, having the side A E equal and parallel to B C.

We are to prove A B equal and II to E C.

Draw A G. In the A A B C and A E C

BC = AE, Hyp.

AC = AC, Iden.

ZBCA=ZCAE, §68

(being alt. -int. A).

.\AABC = AACE, §106

(kceoing two sides and the included Z. of the one equal respectively to two sides and the included Z. of the other).

.'.AB = EC,

(being homologous sides of equal A ).

Also, ZBAC = ZACE,

(being homologous A of equal A ) ;

.'.A Bis II to EG, §69

(when two straight lines are cut by a third straight line, if the alt. -int. A be equal the lines are parallel).

.*. the figure ABC E is a O, § 125

(the opposite sides being parallel).

Q. E. D.

62 GEOMETRY. — BOOK I.

Proposition XLL Theorem.

137. If in a quadrilateral the opposite sides be equal, the figure is a parallelogram.

B C

A

Let the figure A B G E be a quadrilateral having BG = AE and AB = EC.

We are to prove figure ABC E a E3.

Draw A C.

In the A ABC &nd A EG

BG = AEf Hyp.

AB = GE, Hyp.

A C = A G, Iden.

.\AABG = AAEC, § 108

(having three sides of the one equal respectively to three sides of the other).

.'.ZACB = Z GAE,

and ZBAC = ZAGE,

(being homologous A of equal A ).

.'.BC is II toAE,

and A Bis II to EC, §69

(when two straight lines lying in the same plane are cut by a third straight line, if the alt.-int. A be equal, the lines are parallel).

.'. the figure A B G E is a O, § 125

(having its opposite sides parallel).

QUADRILATERALS.

63

Proposition XLII. Theorem. 138. The diagonals of a parallelogram bisect each other. B

Let the figure ABCE be a parallelogram, and let the diagonals A C and BE cut each other at 0.

We are to prove AO = OC, and B 0 = 0 E.

In the A A 0 E and B 0 C

AE=BC,	§ 134
(being opposite sides of a CD '),
Z OAE = Z OCB,	§ 68
(being alt. -int. A ),
Z OEA=Z OBC;	§ 68
.'.AAOE = ABOC,	§ 107

(having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

.'.AO = OC,

and BO = 0 E.

(being homologous sides of equal A ).

Q. E. D.

64 GEOMETRY. — BOOK I.

Proposition XLIII. Theorem.

139. The diagonals of a rhombus bisect each other at

right angles.

A E

B C

Let the figure A B C E be a rhombus, having the diagonals AC and BE bisecting each other at 0.

We are to prove Z A 0 E and Z A 0 B rt. A.

In the A A 0 E and A 0 B,

AE = AB, §128

(being sides of a rhombus) ;

OE=OB, §138

(the diagonals of a EJ bisect each other) ;

AO = AO, Iden.

.'.AAOE=AAOB, § 108

(having three sides of the one equal respectively to three sides of the other) ;

.'.ZAOE = ZAOB,

(being homologous A of equal A ) ; .'. Z A 0 E and Z A 0 B are rt. A. § 25

( When one straight line meets another straight line so as to make the adj. A I, each Z. is art. Z).

Q. E. D,

QUADRILATERALS. 65

Proposition XLIV. Theorem.

140. Two parallelograms, having two sides and the in- cluded angle of the one equal respectively to Uoo sides and the included angle of the other, are equal in all respects.

B C B a

A'

In the parallelograms A B C I) and A' B' C D't le AB = A'B'9 AD = A'D't and A A = A A'.

We are to prove that the UD are equal.

Apply O A BCD to O A' B' CD', so that A D will fall on and coincide with A' L'.

Then A B will fall on A' B', {for ZA = ZA',bU hi/p.),

and the point B will fall on B'y {for AB= A' B\ by ifjp.).

Now, B C and B' C are "both II to A' D' and are drawn

through point B' ';

.'. the lines B C and B' C coincide, § 66

and C falls on B' C or B' C produced.

In like manner D C and D' C are II to A' B' and are drawn through the point D'.

.'. D C and D' C coincide ; § 66

.*. the point C falls on D' C, or B' C produced ;

.'. C falls on both B' C and D' O ';

.'. C must fall on a point common to both, namely, C.

.'.the two UJ coincide, and are equal in all respects.

Q. E. D.

141. Corollary. Two rectangles having the same base and

altitude are equal ; for they may be applied to each other and

will coincide.

66 GEOMETRY. BOOK I.

Proposition XLV. Theorem.

142. The straight line which connects the middle points of the non-parallel sides of a trapezoid is parallel to the par- allel sides j and is equal to half their sum.

A F E

Let SO be the straight line joining the middle points of the non-parallel sides of the trapezoid ABCE.

We are to prove SO II to A E and B 0 ; also SO = ^(AE + BC).

Through the point 0 draw FH II to A B,

and produce B 0 to meet FO H at H.

In the A FOE and COH

OE=00, Cons.

ZOEF=ZOCH, §68

(being alt. -int. A ),

ZFOE = ZOOfft §49

(being vertical A ).

.\AFOE = ACOH, §107

(having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

QUADRILATERALS. 67

.'.FE=C H,

and OF=OH,

(being homologous sides of equal A ).

Now FH = AB, § 135

( II lines comprehended between II lines are equal) ;

.'.FO = AS. Ax. 7.

.'. the figure AFOSis&CJ, § 136

(having two opposite sides equal and parallel).

.'.SO is II to AF, § 125

(being opposite sides of a Of).

£0 is also II to BC, (a straight line II to one of two II lines is II to the other also).

Now	SO = AF,	§125
	(being opposite sides of a CJ)t
and	SO = BH.	§125
But	AF=AE-FEf
and	BH=BC+ CH.

Substitute for A F and BH their equals, A E — FE and BC+ CH,

and add, observing that CH= FE;

then 2SO = AE+BC.

.-.SO = b(AE+ BC).

Q. E. D.

68

GEOMETRY.

BOOK I.

On Polygons in General.

143. Def. A Polygon is a plane figure bounded by straight lines.

144. Def. The bounding lines are the sides of the polygon, and their sum, as A B + B C + CD, etc., is the Perimeter of the polygon.

The angles which the adjacent sides make with each other are the angles of the polygon.

145. Def. A Diagonal of a polygon is a line joining the vertices of two angles not adjacent.

B B<

C A'

D F'

e m

146. Def. An Equilateral polygon is one which has all its sides equal.

147. Def. An Equiangular polygon is one which has all its angles equal.

148. Def. A Convex polygon is one of which no side, when produced, will enter the surface bounded by the perimeter.

149. Def. Each angle of such a polygon is called a Salient angle, and is less than two right angles.

150. Def. A Concave polygon is one of which two or more sides, when produced, will enter the surface bounded by the perimeter.

151. Def. The angle ED E is called a Re-entrant angle. When the term polygon is used, a convex polygon is meant. The number of sides of a polygon is evidently equal to the

number of its angles.

By drawing diagonals from any vertex of a polygon, the fig- ure may be divided into as many triangles as it has sides less two.

POLYGONS. 69

152. Def. Two polygons are Equal, when they can be divided by diagonals into the same number of triangles, equal each to each, and similarly placed; for the polygons can be applied to each other, and the corresponding triangles will evi- dently coincide. Therefore the polygons will coincide, and be equal in all respects.

153. Def. Two polygons are Mutually Equiangular, if the angles of the one be equal to the angles of the other, each to each, when taken in the same order; as the polygons ABC DE F, and A1 B< C D' E> F, in which Z A = Z A', Z B = Z B', ZC = ZC, etc.

154. Def. The equal angles in mutually equiangular poly- gons are called Homologous angles ; and the sides which lie between equal angles are called Homologous sides.

155. Def. Two polygons are Mutually Equilateral, if the sides of the one be equal to the sides of the other, each to each, when taken in the same order.

Fig. 1. Fig. 2. Fig. 3.

Two polygons may be mutually equiangular without being mutually equilateral ; as Figs. 1 and 2.

And, except in the case of triangles, two polygons may be mutually equilateral without being mutually equiangular; as Figs. 3 and 4.

If two polygons be mutually equilateral and equiangular, they are equal, for they may be applied the one to the other so as to coincide.

156. Def. A polygon of three sides is a Trigon or Tri- angle; one of four sides is a Tetragon or Quadrilateral ; one of five sides is a Pentagon ; one of six sides is a Hexagon ; one of seven sides is a Heptagon ; one of eight sides is an Octagon ; one of ten sides is a Decagon ; one of twelve sides is a Dodecagon.

70

GEOMETRY. BOOK I.

Proposition XL VI. Theorem.

157. The sum of the interior angles of a polygon is equal to two right angles, taken as many times less two as the figure has sides.

A B

Let the figure A BO DBF be a polygon having n sides.

We are to prove

ZA + ZB+ZC, etc., = 2 rt. A (n — 2). From the vertex A draw the diagonals AC, A D, and A E.

The sum of the A of the A = the sum of the angles of the polygon.

Now there are (n — 2) A,

and the sum of the A of each A = 2 rt. A.

98

/.the sum of the A of the A, that is, the sum of the A of the polygon = 2 rt. A (?i — 2).

Q. E. D.

158. Corollary. The sum of the angles of a quadrilateral equals two right angles taken (4 — 2) times, i. e. equals 4 right angles ; and if the angles be all equal, each angle is a right angle. In general, each angle of an equiangular polygon of n

sides is equal to — I- 1 right angles.

POLYGONS. 71

Proposition XLVII. Theorem.

159. The exterior angles of a polygon, made by produ- cing each of its sides in succession, are together equal to four right angles.

Let the figure ABCDE be a polygon, having its sides produced in succession.

We are to prove the sum of the ext. A = 4 rt. A.

Denote the int. A of the polygon by A, B,C,D, E ;

and the ext. A by a, b, c, d, e.

ZA + Za=2vtA, §34

(being siq). -adj. A ).

Z B + A b = 2 rt. A. § 34

In like manner each pair of adj. A = 2 rt. A ;

.'.the sum of the interior and exterior A = 2 rt. A taken as many times as the figure has sides,

or, 2 n rt. A.

But the interior A = 2 rt. A taken as many times as the figure has sides less two, = 2 rt. A (n — 2),

or, 2 n rt. A — 4 rt. A.

.'. the exterior A = 4 rt. A.

Q. E. D.

72 GEOMETRY. BOOK I.

Exercises.

1. Show that the sum of the interior angles of a hexagon is equal to eight right angles.

2. Show that each angle of an equiangular pentagon is f of a right angle.

3. How many sides has an equiangular polygon, four of whose angles are together equal to seven right angles?

4. How many sides has the polygon the sum of whose in- terior angles is equal to the sum of its exterior angles 1

5. How many sides has the polygon the sum of whose in- terior angles is double that of its exterior angles 1

6. How many sides has the polygon the sum of whose exterior angles is double that of its interior angles 1

7. Every point in the bisector of an angle is equally distant from the sides of the angle ; and every point not in the bisector, but within the angle, is unequally distant from the sides of the angle.

8. B A C is a triangle having the angle B double the angle A. If B D bisect the angle B, and meet A C in D, show that B D is equal to A D.

9. If a straight line drawn parallel to the base of a triangle bisect one of the sides, show that it bisects the other also ; and that the portion of it intercepted between the two sides is equal to one half the base.

10. ABC D is a parallelogram, E and F the middle points of A D and B C respectively j show that B E and D F will trisect the diagonal A C.

11. If from any point in the base of an isosceles triangle parallels to the equal sides be drawn, show that a parallelogram is formed whose perimeter is equal to the sum of the equal sides of the triangle.

12. If from the diagonal BD of a square AB CD, BE be cut off equal to B C, and E F be drawn perpendicular to B D, show that D E is equal to E F, and also to F C.

13. Show that the three lines drawn from the vertices of a triangle to the middle points of the opposite sides meet in a point.

BOOK II.

CIRCLES.

Definitions.

160. Def. A Circle is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the Centre.

161. Def. The Circumference of a circle is the line which bounds the circle.

162. Def. A Radius of a circle is any straight line drawn from the centre to the circumference, as 0 A, Fig. 1.

163. Def. A Diameter of a circle is any straight line pass- ing through the centre and having its extremities in the circum- ference, as A B, Fig. 2.

By the definition of a circle, all its radii are equal. Hence, all its diameters are equal, since the diameter is equal to twice the radius.

M M

Fig. 1.

164. Def. An Arc of a circle is any portion of the circum- ference, as A M B, Fig. 3.

165. Def. A Semi-circumference is an arc equal to one half the circumference, as A M B, Fig. 2.

166. Def. A Chord of a circle is any straight line having its extremities in the circumference, as A B, Fig. 3.

Every chord subtends two arcs whose sum is the cir- cumference. Thus the chord A B, (Fig. 3), subtends the arc A MB and the arc A D B. Whenever a chord and its arc are spoken of, the less arc is meant unless it be otherwise stated.

74 GEOMETRY. — BOOK II.

167. Def. A Segment of a circle is a portion of a circle enclosed by an arc and its chord, as A M B, Fig. 1.

168. Def. A Semicircle is a segment equal to one half the circle, as A D C, Fig. 1.

169. Def. A Sector of a circle is a portion of the circle enclosed by two radii and the arc which they intercept, as A C P Fig. 2.

170. Def. A Tangent is a straight line which touches the circumference but does not intersect it, however far produced. The point in which the tangent touches the circumference is called the Point of Contact, or Point of Tangency.

171. Def. Two Circumferences are tangent to each other when they are tangent to a straight line at the same point.

172. Def. A Secant is a straight line which intersects the circumference in two points, as A D, Fig. 3.

173. Def. A straight line is Inscribed in a circle when its extremities lie in the circumference of the circle, as A B, Fig. 1.

An angle is inscribed in a circle when its vertex is in the circumference and its sides are chords of that circumference, as ZABdTig. 1.

A polygon is inscribed in a circle when its sides are chords of the circle, as A A B C, Fig. 1.

A circle is inscribed in a polygon when the circumference touches the sides of the polygon but does not intersect them, as in Fig. 4.

174. Def. A polygon is Circumscribed about a circle when all the sides of the polygon are tangents to the circle, as in Fig. 4.

A circle is circumscribed about a polygon when the circumfer- ence passes through all the vertices of the polygon, as in Fig. 1.

STRAIGHT LINES AND CIRCLES.

75

175. Dep. Equal circles are circles which have equal radii For if one circle be applied to the other so that their centres coincide their circumferences will coincide, since all the points of both are at the same distance from the centre.

176. Every diameter bisects the circle and its circumference. For if we fold over the segment A MB on A B as an axis until it comes into the plane of A P B, the arc A MB will coincide with the arc AP B\ because every point in each is equally dis- tant from the centre 0.

Proposition I. Theorem.

177. The diameter of a circle is greater than any other chord.

Let A B be the diameter of the circle A MB, and A E any other chord.

We are to prove A B > A E.

From C, the centre of the O, draw C E. CE=CB,

(being radii of the same circle).

But AC+CE>AE,

§96

{the sum of two sides ofaA> tlie third side). Substitute for 0 E, in the above inequality, its equal CB. Then A C + CB > A E, or

AB> AE. '

Q. E. D.

76 GEOMETRY. — BOOK II.

Proposition II. Theorem.

178. A straight line cannot intersect the circumference of a circle in more than two points.

M

P

Let HK be any line cutting the circumference A MP.

We are to prove that HK can intersect the circumference in only two points.

If it be possible, let HK intersect the circumference in three points, H, P, and K.

From 0, the centre of the O, draw the radii OH, OP, and OK.

Then OH, OP, and 0 K are equal, § 1 63

(being radii of the same circle).

.'.if HK could intersect the circumference in three points, we should have three equal straight lines OH, OP, and 0 K drawn from the same point to a given straight line, which is impossible, § 56

(only two equal straight lines can be drawn from a point to a straight line).

.*. a straight line can intersect the circumference in only two points.

Q. E. D.

STRAIGHT LINES AND CIRCLES. 77

Proposition III. Theorem.

179. In the same circle, or equal circles, equal angles at the centre intercept equal arcs on the circumference.

P F

In the equal circles ABP and A'B'P' let ZO=ZO'.

We are to prove arc R S = arc R' S'.

Apply Oi^PtoO A'B'P1,

so that Z 0 shall coincide with Z. 0'.

The point R will fall upon 7?', § 176

(for 0 R= Of R', being radii of equal (D),

and the point £ will fall upon S', § 176

(for 0S= Of S', being radii of equal (D).

Then the arc R S must coincide with the arc R'S'. For, otherwise, there would be some points in the circumference unequally distant from the centre, which is contrary to the definition of a circle. § 160

Q. E. D.

78 GEOMETRY. BOOK II.

Proposition IV. Theorem.

180. Conversely : In the same circle, or equal circles, equal arcs subtend equal angles at the centre.

In the equal circles ABP and A' B' P' let arc MS = arc R'S'.

We are to prove A ROS=Z R' 0' S'.

Apply Q) ABP to Q A' B i»,

so that the radius 0 R shall fall upon 0' R'.

Then S, the extremity of arc PS,

will fall upon S', the extremity of arc R' Sf, (for RS=R>S'ibyKyp.).

.'. 0 S will coincide with 0' S', § 18

(their extremities being the same points).

.'. Z. RO S will coincide with, and be equal to, Z. R' 0' 8'.

Q. E. D.

STRAIGHT LINES AND CIRCLES.

79

Proposition V. Theorem.

181. In the same circle, or equal circles, equal arcs are subtended by equal chords.

In the equal circles ABP and A' B' P' let arc RS = arc R'S'.

We are to prove chord R S = chord R' S'.

Draw the radii 0 R, 0 S, 0' R', and 0' S'.

In the A R OS and R' 0' S'

OR=0'R',

(being radii of equal ©),

OS=0'S',

ZO = ZO',

(equal arcs in equal © subtend equal A at the centre).

§176

§176 § 180

§ 106

.'.A ROS = A R'O'S',

(two sides and the included Z of the one being equal respectively to two sides

and the included A of the other).

.'. chord RS = chord R'S', (being homologous sides of equal A ).

Q. E. D.

80 GEOMETRY. BOOK II.

Proposition VI. Theorem.

182. Conversely : In the same circle, or equal circles, equal chords subtend equal arcs.

In the equal circles ABP and A' B' P', let chord RS = chord R'S'.

We are to prove arc R S = arc R' S'.

Draw the radii 0 R, 0 S, 0' R', and 0' S'.

In the AROSsui&R'O'S'

RS=R'S', Hyp.

OR = 0'R'} §176

(being radii of equal CD),

OS=0'S'; §176

.\&ROS = AR'0'S', §108

(three sides of the one being equal to three sides of the other).

.\Z 0=Z 0',

(being homologous A of equal &).

.'.arc RS = &rc R'S', § 179

(in the same O, or equal (D, equal A at the centre intercept equal arcs on the

circumference).

Q. E. D.

STRAIGHT LINES AND CIRCLES.

81

Proposition VII. Theorem.

183. The radius perpendicular to a chord bisect* the chord and the arc subtended bj/ it.

Let AB be the chord, and let the radius C S be per- pendicular to A B at the point if.

We are to prove A M = BM, and arc A S = arc B S.

Draw CA ami C B.

CA = CB,

(being radii of tlie same O) ;

§ 84

.'.A AC B is isosceles, (the opposite sides being equal) ;

.*. k- 08 bisects the base A B and the Z. C, § 113

(the ± drawn from the vertex to the base of an isosceles A bisects tlie base and the Z at the vertex).

.'.AM=BM.

Also, since ZACS = ZBGS,

arc A £■— arc SB, §175

(equal A at tlie centre intercept equal arcs on the circumference).

Q. E. D.

184. Corollary. The perpendicular erected at the middle of a chord passes through the centre of the circle, and bisects the arc of the chord.

82 GEOMETRY. BOOK II.

Proposition VIII. Theorem.

185. In the same circle, or equal circles, equal chords are equally distant from the centre ; and of two unequal chords the less is at the greater distance from the centre.

E F

In the circle A B EC let the chord A B equal the chord C F, and the chord C E be less than the chord G F. Let OP, OH, and 0 K be J§ drawn to these chords from the centre 0.

We are to prove OP = Oil, and OH< OK.

Join OA and OC. In the rt. A A OF and CO II

OA=OC,

(being radii of the same O) ;

AP=CH, § 183

(being halves of equal chords) ;

.'.AAOF = ACOH. §109

.\OF=OH. Again, since CE< CF,

the ZCOF<COF, §116

and the arc CE< the arc CF.

.'.A. OiT will intersect CF in some point, as m. Now OK>Om. Ax. 8

But Om>OII, §52

(a _L is the shortest distance from a point to a straight line).

.'.much more is OK> OH.

Q. E. D.

STRAIGHT LINES AND CIRCLES.

83

Proposition IX. Theorem.

186. A straight line perpendicular to a radius at its extremity is a tangent to the circle.

Let BA be the radius, and MO the straight line perpendicular to BA at A.

We are to prove M 0 tangent to the circle.

From B draw any other line to M 0, as B C H.

BH>BA, §52

(a _1_ measures the shortest distance from a point to a straigJU line).

.'. point ^T is without the circumference.

But B H is any other line than B A,

.'. every point of the line MO is without the circumference, except A.

.'. MO is a tangent to the circle at A. § 171

Q. E. D.

187. Corollary. When a straight line is tangent to a circle, it is perpendicular to the radius drawn to the point of contact, and therefore a perpendicular to a tangent at the point of contact passes through the centre of the circle.

84

GEOMETRY.

BOOK II.

Proposition X. Theorem.

188. When two circumferences intersect each other, the line which joins their centres is perpendicular to their common chord at its middle point.

Let C and C be the centres of two circumferences which intersect at A and B. Let A B be their common chord, and 00' join their centres.

We are to prove C C _L to A B at its middle point.

A _L drawn through the middle of the chord A B passes through the centres C and C, § 184

(a _L erected at the middle of a chord passes through the centre of the O). .*. the line C C, having two points in common with this J_, must coincide with it.

.*. C C is _L to A B at its middle point.

Q. E. D.

Ex. 1. Show that, of all straight lines drawn from a point without a circle to the circumference, the least is that which, when produced, passes through the centre.

Ex. 2. Show that, of all straight lines drawn from a point within or without a circle to the circumference, the greatest is that which meets the circumference after passing through the centre.

STRAIGHT LINES AND CIRCLES.

85

Proposition XL Theorem.

189. When two circumferences are tangent to each other their point of contact is in the straight line joining their centres.

Let the two circumferences, whose centres are C and C, touch each other at 0, in the straight line A B, and let CC be the straight line joining their cen- tres.

We are to prove 0 is in the straight line C C

A _L to A B. drawn through the point 0, passes through the centres C and C, , § 187

(a A. to a tangent at the point of contact passes through tJie centre of tJie O).

.'. the line C C, having two points in common with this _L, must coincide with it.

'. 0 is in the straight line C C.

Q. E. D.

Ex. A B, a chord of a circle, is the base of an isosceles triangle whose vertex C is without the circle, and whose equal sides meet the circle in D and E. Show that CD is equal to CE.

GEOMETRY. — BOOK II.

On Measurement.

190. Def. To measure a quantity of any kind is to find how many times it contains another known quantity of the same kind. Thus, to measure a line is to find how many times it con- tains another known line, called the linear unit.

191. Def. The number which expresses how many times a quantity contains the unit, prefixed to the name of the unit, is called the numerical measure of that quantity ; as 5 yards, etc.

192. Def. Two quantities are commensurable if there be some third quantity of the same kind which is contained an exact number of times in each. This third quantity is called the common measure of these quantities, and each of the given quantities is called a multiple of this common measure.

193. Def. Two quantities are incommensurable if they have no common measure.

194. Def. The magnitude of a quantity is always relative to the magnitude of another quantity of the same kind. No quantity is great or small except by comparison. This relative magnitude is called their Ratio, and this ratio is always an ab- stract number.

When two quantities of the same kind are measured by the same unit, their ratio is the ratio of their numerical measures.

195. The ratio of a to b is written -, or a : b, and by this is meant : *

How many times b is contained in a; a

or, what part a is of b. b

I. If b be contained an exact number of times in a their ratio is a whole number.

If b be not contained an exact number of times in a, but

if there be a common measure which is contained m times in a

m and n times in b, their ratio is the fraction — .

II. If a and b be incommensurable, their ratio cannot be exactly expressed in figures. But if b be divided into n equal parts, and one of these parts be contained m times in a with

a remainder less than - part of b, then — is an approximate n n

value of the ratio — , correct within - . 6 n

THEORY OF LIMITS. 87

Again, if each of these equal parts of b be divided into n equal parts ; that is, if b be divided into n2 equal parts, and if one of these parts be contained m' times in a with a remainder

1 m!

less than — part of b, then — ^ is a nearer approximate value

of the ratio - , correct within —^ . b n

By continuing this process, a series of variable values,

— > — o > — ? j e^c-> w^u be obtained, which will differ less and n nl ir

less from the exact value of - . We may thus find a fraction

which shall differ from this exact value by as little as we please, that is, by less than any assigned quantity.

Hence, an incommensurable ratio is the limit toward which its successive approximate values are constantly tending.

On the Theory of Limits.

196. Dep. When a quantity is regarded as having a fixed value, it is called a Constant ; but, when it is regarded, under the conditions imposed upon it, as having an indefinite number of different values, it is called a Variable.

197. Def. When it can be shown that the value of a vari- able, measured at a series of definite intervals, can by indefinite continuation of the series be made to differ from a given con- stant by less than any assigned quantity, however small, but cannot be made absolutely equal to the constant, that constant is called the Limit of the variable, and the variable is said to approach indefinitely to its limit.

If the variable be increasing, its limit is called a superior limit ; if decreasing, an inferior limit.

198. Suppose a point ± * u[ M" B

to move from A toward B, under the conditions that the first sec- ond it shall move one-half the distance from A to B, that is, to M; the next second, one-half the remaining distance, that is, to M' ; the next second, one-half the •remaining distance, that is, to M", and so on indefinitely.

Then it is evident that the moving point may approach as near to B as we please, but will never arrive at B. For, however

88 GEOMETRY. — BOOK II.

near it may be to B at any instant, the next second it will pass over one-half the interval still remaining ; it must, therefore, approach nearer to B, since half the interval still remaining is some distance, but will not reach B, since half the interval still remaining is not the whole distance.

Hence, the distance from A to the moving point is an in- creasing variable, which indefinitely approaches the constant A B as its limit; and the distance from the moving point to B is a decreasing variable, which indefinitely approaches the con- stant zero as its limit.

If the length of A B be two inches, and the variable be denoted by x, and the difference between the variable and its limit, by v :

after one second, x = l, v = l

after two seconds, # = 1 -f £, v = l

after three seconds, a; = 1 -J- $ + £, v = i

after four seconds, # = l + £ + £-f-&, v = i and so on indefinitely. Now the sum of the series 1 + \ r -f- \ + & etc., is evidently less than 2 ; but by taking a great number -of terms, the sum can be made to differ from 2 by as little as we please. Hence 2 is the limit of the sum of the series, when the number of the terms is increased indefinitely ; and 0 is the limit of the vari- able difference between this variable sum and 2. Urn. wilt be used as an abbreviation for limit. 199. [1] The difference between a variable and its limit is a variable whose limit is zero.

[2] If two or more variables, v, v\ vn, etc., have zero for a limit, their sum, v -f- v'-\-v", etc., will have zero for a limit.

[3] If the limit of a variable, v, be zero, the limit of a ± v will be the constant a, and the limit of a X v will be zero.

[4] The product of a constant and a variable is also a va- riable, and the limit of the product of a constant and a variable is the product of the constant and the limit of the variable.

[5] The sum or product of two variables, both of which are either increasing or decreasing, is also a variable.

THEORY OF LIMITS. 89

Proposition I. [6] If two variables be always equal, their limits are equal.

Let the two variables A M and A N be always equal, and let A C and A B be their respective limits.

We are to prove A C = A B.

Suppose A C > A B. Then we may diminish A C to some value A C such t\mtAC'=AB. p

Since A M approaches indefinitely to Ck A C, we may suppose that it has reached a value A P greater than A C.

Let A Q be the corresponding value of A N.

Then AP=AQ.

Now A C = A B.

But both of these equations cannot be true, for A P > A C, and A Q < A B. .'.AC cannot be greater than A B.

Again, suppose AC < A B. Then we may diminish A B to some value A B' such that A C = A B'.

Since A X approaches indefinitely to A B we may suppose that it lias reached a value A Q greater than A B'.

Let A P be the corresponding value of A M.

Then AP=AQ.

Now A C = A B'.

But both of these equations cannot be true, for A P < A C, and A Q> A B'. .'.AC cannot be less than A B.

Since A C cannot be greater or less than A B, it must be equal to A B. Q- E- D-

[7] Corollary 1. If two variables be in a constant ratio, their limits are in the same ratio. For, let x and y be two variables

x

having the constant ratio r, then - = r, or, x = r yf therefore

lim. ( x) Urn. (x) = lim. (r y) = rX lim. (y), therefore ,, / [ = r.

lim. (y)

[8] Cor. 2. Since an incommensurable ratio is the limit of

a

its successive approximate values, two incommensurable ratios -r

a' and — are equal if they always have the same approximate values

when expressed ivitliin the same measure of precision.

90 GEOMETRY. BOOK II.

Proposition II.

[9] The limit of the algebraic sum of tivo or more variables is the algebraic sum of their limits.

Let x, y, z, be variables, a, b, and c, a - H~

their respective limits, and v, v', and v",

the variable differences between x, y, z, b + —

and a, b, c, respectively.

We are to prove Urn. (x + y + z) = a + b + c. c *-—

Now, x = a — v, y = b — v/, z = c — v". Then, x + y + z = a — v + b — v' + c — v".

.'. lim.(x~\- y-\-z)=lim.(a — v+b — v' + c — v"). [6]

But, Urn. (a — v+b — v' + c — v") = a+b + c. [3]

.*. Urn. (x + y + z) = a + b + c.

Q. E. D.

Proposition III.

[10] The limit of the product of two or more variables is the product of their limits.

Let x, y, z, be variables, a, b, c, their respective limits, and v, v', v", the variable differences between x, y, z, and a, b, c, respectively.

We are to prove Urn. (x y z) — ab c. Now, x = a — v, y — b — v', z = c — v". Multiply these equations together.

Then, xy z = ab c=f= terms which contain one or more of the factors v, v', v", and hence have zero for a limit. [3]

.*. Urn. (xyz) = Urn. (abc^f terms whose limits are zero). [6] But Urn. (a b c =f terms whose limits are zero) = ab c. . * . Urn. (xyz) = a b c.

Q. E. D.

For decreasing variables the proofs are similar.

Note. — In the application of the principles of limits, refer- ence to this section (§ 199) will always include the fundamental truth of limits contained in Proposition I. ; and it will be left as an exercise for the student to determine in each case what other truths of this section, if any, are included in the reference.

MEASUREMENT OP ANGLES. 91

Proposition XII. Theorem.

200. In the same circle, or equal circles, two commen- surable arcs have the same ratio as the angles which thej/ subtend at the centre.

P

In the circle A PC let the two arcs be A B and A C, and AO B and AOC the A which they subtend.

m . arc AB ZAOB

We are to prove = -_ •

F arc A C Z AOC

Let H K be a common measure of A B and A C. Suppose // K to be contained in A B three times, and in A C five times.

arc AB 3

Then 77^ = 7 •

arc A C 5

At the several points of division on A B and A 0 draw radii.

These radii will divide Z. AOC into five equal parts, of

which ZAOB will contain three, § 180

(in the same O, or equal ©, equal arcs subtend equal A at tlie centre).

ZAOB 3

But

' Z AOC 5
arc AB 3
arc AC 5
arc A B ZAOB	Q	Ax. 1.
arc AC Z AOC	E. D.

92

GEOMETRY.

BOOK II.

Proposition XIII. Theorem.

201. In the same circle, or in equal circles, incom- mensurable arcs have the same ratio as the angles which they subtend at the centre.

pt P

In the two equal © ABB and A'B'P1 let AB and A' B' be two incommensurable arcs, and C, C the A which they subtend at the centre.

We are to prove = .

7 arc AB Z 0

Let A B be divided into any number of equal parts, and let one of these parts be applied to A' B' as often as it will be contained in A'B'.

Since AB and A' B' are incommensurable, a certain num- ber of these parts will extend from A1 to some point, as D, leaving a remainder D B' less than one of these parts. Draw CD. Since A B and A'B are commensurable,

arc_^/> = ZA'C'D c 20Q

arcAB ~~ ZACB'

(two commensurable arcs have the same ratio as the A which they subtend at

the centre).

Now suppose the number of parts into which A B is divided to be continually increased ; then the length of each part will become less and less, and the point J) will approach nearer and nearer to B', that is, the arc A' D will approach the arc A' B' as its limit, and the Z A' C D the Z A' OB' as its limit.

MEASUREMENT OP ANGLES. 93

Then the limit of ™a A' D will he alc A' B' , arc A B arc A B

and tlie limit of 4A'C' D will be Z A'C'%. ZACB ZACB

Moreover, the corresponding values of the two variables, namely,

arcA'D , Z A' C D

and 1

arc 4 if ZACB

are equal, however near these variables approach their limits.

. . their limits and are equal. 6 199

sltcAB ZACB X *

Q. E. D.

202. Scholium. An angle at the centre is said to be meas- ured by its intercepted arc. This expression means that an angle at the centre is such part of the angular magnitude about that point (four right angles) as its intercepted arc is of the whole circumference.

A circumference is divided into 360 equal arcs, and each arc is called a degree, denoted by the symbol (°).

The angle at the centre which one of these equal arcs sub- tends is also called a degree.

A quadrant (one-fourth a circumference) contains there- fore 90° ; and a right angle, subtended by a quadrant, con- tains 90°.

Hence an angle of 30° is £ of a right angle, an angle of 45° is £ of a right angle, an angle of 135° is f of a right angle.

Thus we get a definite idea of an angle if we know the number of degrees it contains.

A degree is subdivided into sixty equal parts called min- utes, denoted by the symbol (').

A minute is subdivided into sixty equal parts called sec- onds, denoted by the symbol (").

94

GEOMETRY.

BOOK II.

Proposition XIV. Theorem.

203. An inscribed angle is measured by one-half of the arc intercepted between its sides.

Case I.

In the circle PAB {Fig. 1), let the centre C be in one of the sides of the inscribed angle B.

We are to prove Z B is measured by J arc PA.

Draw CA.

CA = GB,

(being radii of the same O).

.\ZB = ZA, §112

(being opposite equal sides).

ZPCA=ZB+ZA. §105

(the exterior Z of a A is equal to the sum of the two opposite interior A).

Substitute in the above equality Z B for its equal Z A.

Then we have ZPCA = 2ZB.

But Z P C A is measured by A P, § 202

(the A at the centre is measured by the intercepted arc).

.'. 2 Z B is measured by A P.

. ' . Z B is measured by \ A P.

MEASUREMENT OF ANGLES. 95

Case II.

In the circle BAE {Fig. 2), let the centre C fall within the angle E B A.

We are to prove Z E B A is measured by \ arc E A.

Draw the diameter B C P.

Z P B A is measured by \ arc PA, (Case I.)

Z P B E is measured by \ arc P E, (Case I.)

.'. Z PBA + Z PB E is measured by \ (arc P4 + arc P E).

.*. Z E B A is measured by J arc .#.4.

Case III.

In the circle BFP (Fig. 3), let the centre C fall with- out the angle A B F.

We are to prove Z A B F is measured by J arc A F.

Draw the diameter B C P.

Z P B F is measured by \ arc P F, (Case I.)

Z PBA is measured by \ arc PA, (Case I.)

.-. Z PBF— Z PBA is measured by J (arc PP — arc PA).

.'. Z ,4 2? jF is measured by A arc ^1 i^.

Q. E. D.

204. Corollary 1. An angle inscribed in a semicircle is a right angle, for it is measured by one-half a semi-circumfer- ence, or by 90°.

205. Cor. 2. An angle inscribed in a segment greater than a semicircle is an acute angle ; for it is measured by an arc less than one-half a semi-circumference ; i. e. by an arc less than 90°.

206. Cor. 3. An angle inscribed in a segment less than a semicircle is an obtuse angle, for it is measured by an arc greater than one-half a semi-circumference ; i. e. by an arc greater than 90°.

207. Cor. 4. All angles inscribed in the same segment are equal, for they are measured by one-half the same arc.

96 GEOMETRY. BOOK II.

Proposition XV. Theorem.

208. An angle formed by two chords, and whose vertex lies between the centre and the circumference, is measured by one- half the intercepted arc plus one-half the arc intercepted

by its sides produced.

Let the Z AOC be formed by the chords A B and CD.

We are to prove

Z A 0 C is measured by J arc A C + \ arc B D.

Draw A D.

Z COA=Z D + ZA, §105

(the exterior Z of a A is equal to the sum of the two opposite interior A ).

But Z D is measured by \ arc A C, § 203

(an inscribed Z is measured by ? the intercepted arc) ;

and Z A is measured by \ arc B D, § 203

.*. Z C ' 0 A is measured by J arc A C + J arc B D.

Q. E. D.

Ex. Show that the least chord that can be drawn through a given point in a circle is perpendicular to the diameter drawn through the point.

MEASUREMENT OF ANGLES.

97

Proposition XVI. Theorem.

209. An angle formed by a tangent and a chord is measured by one-half the intercepted arc.

Let HAM be the angle formed by the tangent OM and chord AH.

We are to prove

Z HA M is measured by £ arc A EH.

Draw the diameter AC F.

Z FAMisa.it. Z, §186

(the radius drawn to a tangent at the point of contact is A. to it).

Z FA M, being a rt. Z, is measured by J the semi-circum- ference A EF.

Z FA H is measured by \ arc FH, (an inscribed Z is measured by £ the intercepted arc)

§203

.'. Z FAM — A FA H is measured by J (arc A EF— arc HF). .'. Z HA M is measured by J arc A EH.

Q. E. D.

98

GEOMETRY. — BOOK II.

Proposition XYII. Theorem.

210. An angle formed by two secants, two tangents, or a tangent and a secant, and which has its vertex without the circumference, is measured by one-half the concave arc, minus one-half the convex arc.

M D

Fig. 1. Fig. 2. Fig. 3.

Case I. Let the angle 0 {Fig. 1) be formed by the two secants OA and OB.

We are to prove

Z 0 is measured by J arc AB — J arc E C. Draw CB.

ZACB = ZO + ZB, § 105

(the exterior A of a A is equal to the sum of the two opposite interior A ).

By transposing,

ZO = ZACB-ZB,

But Z A CB is measured by J arc A B, § 203

(an inscribed Z is measured by £ the intercepted arc).

and Z B is measured by \ arc C E, * § 203

,\ Z 0 is measured by J arc A B — J arc CE,

MEASUREMENT OF ANGIiES. 99

Case II.

Let the angle 0 {Fig. 2) be formed by the two tan- gents OA and OB.

We are to prove

Z 0 is measured by £ arc A MB — £ arc A SB. Draw A B. ZABC = ZO + ZOAB} §105

(the exterior Z of a A is equal to tfie sum oftJie two opposite interior A ).

By transposing,

Z 0=ZABC-Z OAB.

But Z ABC is measured by \ arc A MB, § 209

(an Z formed by a tangent and a chord is measured by £ the intercepted arc)y

and Z 0 A B is measured by £ arc A SB. § 209

.'. Z 0 is measured by £ arc A MB — £ arc A SB.

Case III.

Let the angle 0 {Fig. 3) be formed by the tangent OB and the secant 0 A.

We are to prove

Z 0 is measured by J arc A D S — J arc C E S. Draw CS.

ZACS = ZO + Z CSO, § 105

(the exterior Zofak is equal to the sum of the two opposite interior A). By transposing,

Z 0 = ZACS~Z CSO.

But Z A CS is measured by J arc A D S, § 203

(being an inscribcdZ).

and Z CSO is measured by £ arc C US, § 209

{being an Z formed by a tangent and a chord).

.'. Z 0 is measured by \ arc A D S — J arc C E S.

Q. E. D.

100

GEOMETRY.

BOOK II.

Supplementary Propositions. Proposition XVIII. Theorem.

211. Two parallel lines intercept upon the circum- ference equal arcs.

A

Let the two parallel lines C A and B F (Fig. 1), inter- cept the arcs C B and A F.

We are to prove arc C B = arc A F.

Draw A B.

£A=£B,

(being alt. -int. A ).

But the arc C B is double the measure of Z. A. and the arc A F is double the measure of A B.

68

.*. arc C B = arc A F.

Ax. (>

Q. E. D.

212. Scholium. Since two parallel lines intercept on the circumference equal arcs, the two parallel tangents M N and 0 P (Fig. 2) divide the circumference in two semi-circumferences AC B and AQ B, and the line A B joining the points of contact of the two tangents is a diameter of the circle.

SUPPLEMENTARY PROPOSITIONS. 101

Proposition XIX. Theorem.

213. If the sum of two arcs be less than a circum- ference the greater arc is subtended by the greater chord ; and conversely, the greater chord subtends the greater arc.

B

P In the circle A CP let the two arcs A B and BC to- gether be less than the circumference, and let AB be the greater.

We are to prove chord A B > chord B C.

Draw A C.

In the A A B C

Z C, measured by J the greater arc AB, § 203

is greater than Z A, measured by \ the less are B C.

.'. the side A B > the side B C, § 117

(in a A the greater Z. has the greater side opposite to it).

Conversely : If the chord A B be greater than the

chord B C.

We are to prove arc A B > arc B C.

In the A A B C,-

AB>BC, Hyp.

.-.ZOA, §118

(in a A the greater side has the greater Z opposite to it).

.'.urcAB, double the measure of the greater Z C, is greater than the arc B C, double the measure of the less Z A.

Q. E. D.

102 GEOMETRY. BOOK II.

Proposition XX. Theorem.

214. If the sum of two arcs be greater than a circum- ference, the greater arc is subtended by the less chord ; and, conversely, the less chord subtends the greater arc.

B

E In the circle BCE let the arcs AECB and BAEC together be greater than the circumference, and let arc AECB be greater than arc B AEG.

We are to prove chord AB < chord B C.

From the given arcs take the common arc AEC ;

we have left two arcs, CB and A B, less than a circumference,

of which CB is the greater.

.'. chord C B > chord A B, § 213

(when the sum of two arcs is less than a circumference, the greater arc is subtended by the greater chord).

.*. the chord A B, which subtends the greater arc AECB, is less than the chord B C, which subtends the less arc BAE C.

Conversely : If the chord A B be less than chord B C.

We are to prove arc AEC B > arc BAEC.

Arc AB + b,tcAECB = the circumference.

Arc BC + arc B A E C = the circumference. .*. arc A B + arc A EC B = arc B C + arc BA EC.

But arc A B < arc B C, § 213

(being subtended by the less chord).

.'. &tcAECB>slicBAEC.

Q. E. D.

CONSTRUCTIONS. 103

On Constructions.

Proposition XXI. Problem.

215. To find a point in a plane, having given its dis- tances from two known points.

X

Let A and B be the two known points; n the dis- tance of the required point from At o its distance from B.

It is required to find a point at the given distances from A and B.

From A as a centre, with a radius equal to n, describe an arc.

From B as a centre, with a radius equal to o, describe an arc intersecting the former arc at C.

G is the required point.

Q. E. F.

216. Corollary 1. By continuing these arcs, another point below the points A and B will be found, which will fulfil the conditions.

217. Cor. 2. When the sum of the given distances is equal to the distance between the two given points, then the two arcs described will be tangent to each other, and the point of tan- gency will be the point required.

104 GEOMETRY. BOOK II.

Let the distance from A to B equal n + o.

From A as a centre, with a \j

radius equal to n, describe an arc ; A- (p -B

and from B as a centre, with A

a radius equal to o, describe an arc. *

These arcs will touch each ~

other at 0, and will not intersect.

.*. G is the only point which can be found.

218. Scholium 1. The problem is impossible when the distance between the two known points is greater than the sum of the distances of the required point from the two given points.

Let the distance from A to B be greater than n + o.

Then from A as a centre, with a radius equal to ?i, de- A' '&

scribe an arc;

and from B as a centre, with a

radius equal to o, describe an arc.

These arcs will neither touch

o nor intersect each other ;

hence they can have no point in common.

219. Scho. 2. The problem is impossible when the distance between the two given points is less than the difference of the distances of the required point from the two given points.

Let the distance from A to B be less than n — o.

From A as a centre, with a radius ^**~~" "^ equal to n, describe a circle ; / \

and from B as a centre, with a j / \ \

radius equal to o, describe a circle. j / \ \

The circle described from B as a \ \ I /

centre will fall wholly within the circle V \v y /

described from A as a centre : o \ /

hence they can have no point in n > •-*

common.

n

CONSTRUCTIONS. 105

Proposition XXII. Problem.

220. To bisect a given straight line. C

X

E

Let AB be the given straight line. It is required to bisect the line A B.

From A and B as centres, with equal radii, describe arcs intersecting at C and E.

Join OE. Then the line C E bisects A B. For, C and E, being two points at equal distances from the extremities A and By determine the position of a J_ to the mid- dle point of A B. § 60

Q. E. F.

Proposition XXIII. Problem.

221. At a given point in a straight line, to erect a perpendicular to that line. R

xix

A

HO" Let 0 be the given point in the straight line AB. It is required to erect a J_ to the line A B at the point 0.

TakeOH=OB. From B and H as centres, with equal radii, describe two arcs intersecting at R.

Then the line joining R 0 is the _L required. For, 0 and R are two points at equal distances from B and H, and .*. determine the position of a JL to the line H B at its middle point 0. § 60

Q. E. F,

106 GEOMETRY. — BOOK II.

Proposition XXIV. Problem.

222. From a point without a straight line, to let fall a perpendicular upon that line.

C

X

V

/

-^—

H \ m ^'K

Let AB be a given straight line, and G a given point without the line.

It is required to let fall a A. to the line A B from the point G.

From G as a centre, with a radius sufficiently great,

describe an arc cutting A B at the points H and K.

From H and K as centres, with equal radii,

describe two arcs intersecting at 0.

Draw G 0,

and produce it to meet A B at m.

G mid- the _L required.

For, G and 0, being two points at equal distances from H and K, determine the position of a ± to the line H K at its middle point. § 60

Q. E. F.

CONSTRUCTIONS.

107

Proposition XXV. Problem.

223. To construct an arc equal to a given arc whose centre is a given point.

F — ^

X

/f\

a<

B B>

Let C be the centre of the given arc A B.

It is required to construct an arc equal to arc A B.

Draw CB, CA, and A B.

From C as a centre, with a radius equal to CB,

describe an indefinite arc B' F.

From B' as a centre, with a radius equal to chord A B,

describe an arc intersecting the indefinite arc at A'.

Then arc A' B' = arc A B.

draw chord A' B'.

For,

and

The (D are equal, (being described with equal radii),

chord A' B' = chord A B ;

.'. arc A' B' = arc A B, (in equal <D equal chords subtend equal arcs).

Cons. § 182

Q. E. F.

108 GEOMETRY. — BOOK II.

Proposition XXVI. Problem.

224. At a given point in a given straight line to con- struct an angle equal to a given angle.

F

\

a

B B'

Let C be the given point in the given line C B4, and C the given angle.

It is required to construct an Z at C equal to the Z C.

Prom C as a centre, with any radius as C B,

describe the arc A B, terminating in the sides of the Z.

Draw chord A B.

From C as a centre, with a radius equal to C B,

describe the indefinite arc B' F.

Prom B' as a centre, with a radius equal to A B,

describe an arc intersecting the indefinite arc at A'.

Draw A' C. Then Z.G' = Za For, iomA'B'.

The (D to which belong arcs A B and A' B' are equal, (being described with equal radii).

and chord A' B' = chord A B ; Cons.

.*.arc^/^/ = arc^j5, § 182

(in equal © equal chords subtend equal arcs).

.-.ZC' = ZC, §180

(in equal (D equal arcs subtend equal A at the centre).

Q. E. F-

CONSTRUCTIONS.

109

Proposition XXVII. Problem. 225. To bisect a given arc.

7*V

Let A OB be the given arc.

It is required to bisect the arc A 0 B.

Draw the chord A B.

From A and B as centres, with equal radii,

describe arcs intersecting at E and C.

Draw EC.

E C bisects the arc A OB.

For, E and C, being two points at equal distances from A and B, determine the position of the J_ erected at the middle of chord A B ; § 60

and a _L erected at the middle of a chord passes through the centre of the O, and bisects the arc of the chord. § 184

Q. E. F.

110

GEOMETRY. BOOK II.

Proposition XXVIII. Problem. 226. To bisect a given angle.

< C

Let A EB he the given angle.

It is required to bisect Z A E B.

From E as a centre, with any radius, as E A,

describe the arc A 0 B, terminating in the sides of the Z.

Draw the chord A B.

From A and B as centres, with equal radii,

describe two arcs intersecting at C.

Join EC.

E 0 bisects the Z E.

For, E and (7, being two points at equal distances from A and B, determine the position of the -L erected at the middle of AB. § 60

And the _L erected at the middle of a chord passes through the centre of the O, and bisects the arc of the chord. § 184

.'. arc A 0 = arc 0 B,

.'.Z AEC = Z BEG, § 180

(in the same circle equal arcs subtend equal A at the centre).

Q. E. F.

CONSTRUCTIONS. Ill

Proposition XXIX. Problem.

227. Through a given point to draw a straight line parallel to a given straight line.

E -^ H

B

Let AB be the given line, and H the given point.

t It is required to draw through the point H a line II to the

line A B.

Draw HA, making the Z H AB.

At the point H construct ZAHE = ZHAB.

Then the line HE is II to A B.

For, ZEHA=ZHAB; Cons.

.'. HE is II to A B, § 69

(whtn two straight lines, lying in the same plane, are cut by a third straight line, if the alt. -int. A be equal, tlie lines are parallel).

Q. E. F.

Ex. 1. Find the locus of the centre of a circumference which passes through two given points.

2. Find the locus of the centre of the circumference of a given radius, tangent externally or internally to a given cir- cumference.

3. A straight line is drawn through a given point A, inter- secting a given circumference at B and C. Find the locus of the middle point P of the intercepted chord B Q,

II 52 GEOMETRY. BOOK 11.

Proposition XXX. Problem.

228. Two angles of a triangle being given to find the

third.

R

8^

v /

E

Let A and B be two given angles of a triangle.

It is required to find the third Z of the A.

Take any straight line, as E F, and at any point, as H.

construct Z R H F equal to Z B,

and Z. SHF equal to Z A.

Then Z RH S is the Z required.

For, the sum of the three A of a A = 2 rt. A, § 98

and the sum of the three A about the point H, on the same side ofFF=2 rt. A. § 34

Two A of the A being equal to two A about the point ff, Cons.

the thhd Z of the A must be equal to the third Z about the point H.

Q. E. F.

CONSTRUCTIONS. 113

Proposition XXXI. Problem.

229. Two sides and the included angle of a triangle being given, to construct the triangle.

K ^r--- A H

E

A

Let the two sides of the triangle be E and Fy and the included angle A.

It is required to construct a A liaving two sides equal to E and F respectively \ and their included Z. — Z. A.

Take HK equal to the side F.

At the point H draw the line H Mt

making the Z K H M = Z A.

On H M take HC equal to E.

Draw C K.

Then A C HK is the A required.

Q.E. F

114 GEOMETRY. — BOOK II.

Proposition XXXII. Problem.

230. A side and two adjacent angles of a triangle being given, to construct the triangle.

	0
	/*»n
	/ \
	/
E	/ /		\	.^>».

Let C E be the given side, A and B the given angles.

It is required to construct a A having a side equal to C E, and two A adjacent to that side equal to A A and B resjyectively.

At point C construct an A equal to A A.

At point E construct an A equal to A B.

Produce the sides until they meet at 0.

Then A C 0 E is the A required.

Q. E. F.

231. Scholium. The problem is impossible when the two given angles are together equal to, or greater than, two right angles.

CONSTRUCTIONS. 115

Proposition XXXIII. Problem.

232. The three sides of a triangle being given, to con- struct the triangle.

\C m

AS--

B o

Let the three sides be m, n, and o.

It is required to construct a A having three sides respectively, equal to m, n, and o.

Draw A B equal to n.

From A as a centre, with a radius equal to o,

describe an arc ;

and from B as a centre, with a radius equal to m,

describe an arc intersecting the former arc at C. Draw CA and C B. Then A C A B is the A required.

Q. E. F.

233. Scholium. The problem is impossible when one side is equal to or greater than the sum of the other two.

116

GEOMETRY. BOOK II.

Proposition XXXIV. Problem.

234. The hypotenuse and one side of a right triangle being given, to construct the triangle.

X
c /	\ \ \	\ \ \
			\
1	i

B Let m be the given side, and o the hypotenuse.

It is required to construct a rt. A having the hypotenuse equal o and one side equal m.

Take A B equal to m.

At A erect a _L, A X.

From B as a centre, with a radius equal to o,

describe an arc cutting A X at G.

Draw CB.

Then A C A B is the A required.

Q. E. F

CONSTRUCTIONS.

117

Proposition XXXV. Problem.

235. The base, the altitude, and an angle at the base, of a triangle being given, to construct the triangle.

Xr

,^-R

™«

Let o equal the base, m the altitude, and C the angle at the base.

It is required to construct a A having the base equal to o> the altitude equal to m, and an Z. at tlie base equal to 0.

Take A B equal to o.

At the point A, draw the indefinite line A R,

making the Z BA R = Z C.

At the point A, erect a A. A X equal to m.

From X draw XS II to A By

and meeting the line A R at S.

Draw SB.

Then A A SB is the A required.

Q. E. F.

118 GEOMETRY. BOOK II.

Proposition XXXVI. Problem.

236. Two sides of a triangle and the angle opposite one of them being given, to construct the triangle.

Case I.

When the given angle is acute, and the side opposite to it is less than the other given side.

D

A

//

x /

\ \

\ y

A^~-—i-—~-^r—^

Let c be the longer and a the shorter given side, and A A the given angle.

It is required to construct a A having two sides equal to a and c respectively, and the Z. opposite a equal to given Z. A. Construct /.DAE equal to the given A A. On AD take A B = c. From JB as a centre, with a radius equal to a, describe an arc intersecting the side A E at C and C".

Draw B C and B C". Then both the A A B C and A B C" fulfil the conditions, and hence we have two constructions.

When the given side a is exactly equal to the _L B C, there will be but one construction, namely, the right triangle ABC.

When the given side a is less than B C, the arc described from B will not intersect A E, and hence the problem is im- possible.

CONSTRUCTIONS. 119

Case II. When the given angle is acute, right, or obtuse, and the side opposite

Koiiis greater than the otJier given side.

D

\

A

B

£

/ \

/

■' \\ n

/ \\ / i \

\ \

\ \ / \

\/ Y Vg ^4 1 A*:

Fig. 1. Fig. 2.

\

s

When the given angle is obtuse. Construct tho Z DAE (Fig. 1) equal to the given Z S.

Take A B equal to a. From B as a centre, with a radius equal to c, describe an arc cutting E A at C, and E A produced at C. Join BCzjidB C.

Then the A A B C is the A required, and there is only one construction ; for the A A BC will not contain the given Z S.

When the given angle is acute, as angle B A C!. There is only one construction, namely, the A B AC (Fig. 1).

When the given Z. is a right angle.

There are two constructions, the equal ABAC and BAC (Fig. 2). Q. E. F.

The problem is impossible when the given angle is right or obtuse, if the given side opposite the angle be less than the other given side. § 117

120 GEOMETRY. BOOK II.

Proposition XXXVII. Problem.

237. Two sides and an included angle of a parallelo- gram being given, to construct the parallelogram. R

i

/

Let m and o be the two sides, and C the included angle.

It is required to construct a O having two adjacent sides equal to m and o respectively, and their included /. equal to Z. C. Draw A B equal to o. From A draw the indefinite line A R, making the Z. A equal to Z C. On A R take A H equal to m. From H as a centre, with a radius equal to o, describe an arc.

From B as a centre, with a radius equal to my

describe an arc, intersecting the former arc at E. Draw EH and E B. The quadrilateral A B E His the O required. For, AB = HE, Cons.

AH = BEt Cons.

.*. the figure A B E H is a O, § 136

(a quadrilateral, which has its opposite sides equal, is a O ).

Q. E. F.

CONSTRUCTIONS. 121

Proposition XXXVIII. Problem. 238. To describe a circumference through three points not in the same straight line.

/ / / / / / 1 1 1	0	\ \ \ \ \ \ \ \ 1 \
.-■	•' '• \	"■*•.. '
\ .•'"''	• !	V(7
*\\		..--' /

Let the three points be A, B, and C.

It is required to describe a circumference through tJie three points Ay By and C.

Draw A B and B C.

Bisect A B and B C.

At the points of bisection, E and F, erect J§ intersect- ing at 0.

From 0 as a centre, with a radius equal to 0 A, describe a circle.

Q ABC is the O required.

For, the point 0, being in the _L E 0 erected at the middle of the line A B, is at equal distances from A and B ;

and also, being in the J_ F 0 erected at the middle of the line C B, is at equal distances from B and C, § 58

{every point in the _L erected at the middle of a straight line is at equal distances from the extremities of that line).

.'. the point 0 is at equal distances from A, B, and C, and a O described from 0 as a centre, with a radius equal to 0 A, will pass through the points A, B, and C.

Q. E. F.

239. Scholium. The same construction serves to describe a circumference which shall pass through the three vertices of a triangle, that is, to circumscribe a circle about a given triangle.

122

GEOMETRY.

BOOK II.

Proposition XXXIX. Problem.

240. Through a given point to draw a tangent to a given circle.

>\E

Case 1 . — When tlie given point is on the circumference.

Let ABC (Fig. 1) be a given circle, and G the given point on the circumference.

It is required to draw a tangent to the circle at C.

From the centre 0, draw the radius 0 C.

At the extremity of the radius, C, draw C M J_ to 0 C.

Then C M is the tangent required, § 186

(a straight line A. to a radius at its extremity is ta.ngent to the O).

Case 2. — When the given point is without the circumference.

Let ABC (Fig. 2) be the given circle, 0 its centre, E the given point without the circumference. It is required to draw a tangent to the circle ABC from

the point E.

Join 0 E.

On 0 E as a diameter, describe a circumference intersecting the given circumference at the points M and H.

Draw 0 M and 0 II, EM and EH.

Now Z OMEisa.xt. Z, §204

(being inscribed in a semicircle).

.'. E M is _L to 0 M at the point M;

.'.EM is tangent to the O, § 186

(a straight line ± to a radius at its extremity is tangent to the O). In like manner we may prove HE tangent to the given O.

Q. E. F.

241. Corollary. Two tangents drawn from the same point to a circle are equal.

CONSTRUCTIONS. 123

Proposition XL. Problem.

242. To inscribe a circle in a given triangle.

fi M

II

Let ABG be the given triangle.

It is required to inscribe a O in the A A B G.

Draw the line A E, bisecting Z A,

and draw the line G E, bisecting Z. G.

Draw EH A. to the line A C.

From E, with radius EH, describe the O K M H.

The O KHM is the O required.

For, draw EK ±to A B,

&ndEM±toBG.

In the rt. A A KE and A HE

AE = AE, Iden.

ZEAK = Z.EAH, Cons.

.\AAKE = A AHE, § 110

(Two rt. A are equal if the hypotenuse and an acute Z of the one be equal respectively to the hypotenuse and an acute Z of the other).

.'. EK=EH, (being homologous sides of equal A).

In like manner it may be shown E M= EH. .'.EKy EH, and E M are all equal. .*. a O described from E as a centre, with a radius equal to EH,

will touch the sides of the A at points H, K, and M, and be inscribed in the A. § 174

Q. E. F.

124

GEOMETRY.

BOOK II.

Proposition XLI. Problem.

243. Upon a given straight line, to describe a segment which shall contain a given angle.

7

M

Let AB be the given line, and M the given angle.

It is required to describe a segment upon the line A B, which shall contain Z M.

At the point B construct Z ABE equal to Z M. Bisect the line A B by the ± F H. Prom the point B, draw B 0 _L to EB. ■ Prom 0, the point of intersection of F H and B 0, as a centre, with a radius equal to 0 B, describe a circumference.

Now the point 0, being in al erected at the middle of A B, is at equal distances from A and B, § 58

(every point in a J_ erected at the middle of a straight line is at equal dis- tances from the extremities of that line) ;

.\ th« circumference will pass through A.

Now B E is ± to OB, Cons.

.'.BE is tangent to the O, § 186

(a straight line A. to a radius at its extremity is tangent to the G).

.'. Z A B E is measured by \ arc A B, § 209

(being an Z formed by a tangent and a chord).

Also any Z inscribed in the segment A II B, as for instance Z A KB, is measured by J arc A B, § 203

(being an inscribed Z ).

CONSTRUCTIONS.

125

.\Z AKB = ZABE,

(being both measured by \ the same arc) ; .'./.AKB = Z M. segment A H B is the segment required.

Q. E. F.

Proposition XLII. Problem.

244. To find the ratio of two commensurable straight

lines.

E H

A » LJL-B

K

v. 1 r

F Let AB and C D be two straight lines.

It is required to find the greatest common measure of A B and C D, so as to express tlieir ratio in figures.

Apply C D to A B as many times as possible.

Suppose twice with a remainder E B.

Then apply E B to C D as many times as possible.

Suppose three times with a remainder F D. Then apply F D to E B as many times as possible.

Suppose once with a remainder H B. Then apply H B to F D as many times as possible.

Suppose once with a remainder K D. Then apply K D to H B as many times as possible. Suppose K D is contained just twice in H B. The measure of each line, referred to K D as a unit, will then be as follows : —

HB =2KD;

FD = IIB+ KD = 3KD EB = FD + HB = 5 KD CD =3EB+ FD = 18KD AB=20D+EB = 41 KD.

. AJB = 41 KD ,

" CD ISKD'

/.the ratio of -—= — .

6 ° 18 Q. E. F.

126 geometry. book h.

Exercises.

1. If the sides of a pentagon, no two sides of which are parallel, be produced till they meet ; show that the sum of all the angles at their points of intersection will be equal to two right angles.

2. Show that two chords which are equally distant from the centre of a circle are equal to each other ; and of two chords, that which is nearer the centre is greater than the one more remote.

3. If through the vertices of an isosceles triangle which has each of the angles at the base double of the third angle, and is inscribed in a circle, straight lines be drawn touching the circle ; show that an isosceles triangle will be formed which has each of the angles at the base one-third of the angle at the vertex.

4. A D B is a semicircle of which the centre is 0 ; and AEG is another semicircle on the diameter AC', A T is a common tangent to the two semicircles at the point A. Show that if from any point F, in the circumference of the first, a straight line FG be drawn to G, the part FK, cut off by the second semicircle, is equal to the perpendicular FH to the tangent A T.

5. Show that the bisectors of the angles contained by the opposite sides (produced) of an inscribed quadrilateral intersect at right angles.

6. If a triangle A B 0 be formed by the intersection of three tangents to a circumference whose centre is 0, two of which, A M and A N, are fixed, while the third, B G, touches the cir- cumference at a variable point P ; show that the perimeter of the triangle A B G is constant, and equal to A M + A N, or 2 AM. Also show that the angle B 0 G is constant.

7. A B is any chord and A G is tangent to a circle at A, G D E a line cutting the circumference in D and E and parallel to A B ; show that the triangle A G D is equiangular to the triangle E A B.

CONSTRUCTIONS. 127

Constructions.

1. Draw two concentric circles, such that the chords of the outer circle which touch the inner may be equal to the diameter of the inner circle.

2. Given the base of a triangle, the vertical angle, and the length of the line drawn from the vertex to the middle point of the base : construct the triangle.

3. Given a side of a triangle, its vertical angle, and the radius of the circumscribing circle : construct the triangle.

4. Given the base, vertical angle, and the perpendicular from the extremity of the base to the opposite side : construct the triangle.

5. Describe a circle cutting the sides of a given square, so that its circumference may be divided at the points of inter- section into eight equal arcs.

6. Construct an angle of 60°, one of 30°, one of 120°, one of 150°, one of 45°, and one of 135°.

7. In a given triangle ABC, draw Q D E parallel to the base B C and meeting the sides of the triangle at D and E, so that D E shall be equal to DB + EC.

8. Given two perpendiculars, A B and CD, intersecting in 0, and a straight line intersecting these perpendiculars in E and F ; to construct a square, one of whose angles shall coincide with one of the right angles at O, and the vertex of the opposite angle of the square shall lie in E F. (Two solutions.)

9. In a given rhombus to inscribe a square.

10. If the base and vertical angle of a triangle be given ; find the locus of the vertex.

11. If a ladder, whose foot rests on a horizontal plane and top against a vertical wall, slip down ; find the locus of its middle point.

BOOK III.

PROPORTIONAL LINES AND SIMILAR POLYGONS.

Ox the Theory of Proportion.

245. Def. The Terms of a ratio are the quantities com- pared.

246. Def. The Antecedent of a ratio is its first term.

247. Def. The Consequent of a ratio is its second term.

248. Def. A Proportion is an expression of equality be- tween two equal ratios.

A proportion may he expressed in any one of the following forms : —

1. a : b : : c : d

2. a : b = c : d

3. a-=c-. b d

Form 1 is read, a is to b as c is to d.

Form 2 is read, the ratio of a to 6 equals .the ratio of c to d.

Form 3 is read, a divided by b equals c divided by d.

The Terms of a proportion are the four quantities com- pared.

The first and third terms in a proportion are the ante- cedents, the second and fourth terms are the consequents.

249. The Extremes in a proportion are the first and fourth terms.

250. The Means in a proportion are the second and third terms.

THEORY OF PROPORTION. 129

251. Def. Ill the proportion a : b : : c : d ; d is a Fourth Proportional to a, b, and c.

252. Def. In the proportion a : b : : b : c ; c is a Third Proportional to a and b.

253. Def. In the proportion a : b : : b : c; b is a .Afea/i Proportional between a and c.

254. Def. Four quantities are Reciprocally Proportional when the first is to the second as the reciprocal of the third is to the reciprocal of the fourth.

Thus a : b : : - : - .

c d

If we have two quantities a and b, and the reciprocals of

these quantities - and - ; these four quantities form a recipro- a b

cal proportion, the first being to the second as the reciprocal of the second is to the reciprocal of the first.

As a : b : : I : - .

b a

255. Def. A proportion is taken by Alternation, when the means, or the extremes, are made to exchange places.

Thus in the proportion

a : b : : o : d, we have either

a : c : : b : d, or, d : b : : c : a.

256. Def. A proportion is taken by Inversion, when the means and extremes are made to exchange places.

Thus in the proportion

a : b : : c : d, by inversion we have

b : a : : d : c.

257. Def. A proportion is taken by Composition, when the sum of the first and second is to the second as the sum of

130 GEOMETRY. BOOK III.

the third and fourth is to the fourth ; or when the sum of the first and second is to the first as the sum of the third and fourth is to the third.

Thus if a : b :	: c : d,
we have by composition,
a + b : b :	: c + d	:d,
or, a + b : a :	: c + d	: c.

258. Def. A proportion is taken by Division, when the difference of the first and second is to the second as the dif- ference of the third and fourth is to the fourth ; or when the difference of the first and second is to the first as the difference of the third and fourth is to the third.

Thus if a : b : : c : d,

we have by division

a — b : b :: c — d : dt

or, a — b : a : : c — d : c,

Proposition I.

259. In every proportion the product of the extremes is equal to the product of the weans.

Let a : b : : c : d.

We are to prove ad = be.

Now		a b =	c d'
whence,	by multiplyi	ngby	bd,
		ad =	--be.

Q. E. D

THEORY OF PROPORTION. 131

In the treatment of proportion, it is assumed that fractions may be found which will represent the ratios. It is evident that a ratio may be represented by a fraction when the two quanti- ties compared can be expressed in integers in terms of any common unit. Thus the ratio of a line 2£ inches long to a line 3 \ inches long may be represented by the fraction §§ when both lines are expressed in terms of a unit TV of an inch long.

But it often happens that no unit exists in terms of which both the quantities can be expressed in integers. In such cases, however, it is possible to find a fraction that will represent the ratio to any required degree of accuracy.

Thus, if a and b denote two incommensurable lines, and b be divided into any integral number (n) of equal parts, if one of these parts be contained in a more than m times, but less than

m + 1 times, then - > — but < — — — ; so that the error b n n

in taking either of these values for -is < — Since n can

b n

be increased at pleasure, - can be made less than any assigned n

value whatever. Propositions, therefore, that are true for — and

n

i — — — , however little these fractions differ from each other, are n

true for - ; and - may be taken to represent the value of — b n b

Proposition II.

260. A mean proportional between two quantities is equal to the square root of their product.

In the proportion a : b : : b : c,

62 = a c, § 259

(the product of the extremes is equal to the product of the means).

Whence, extracting the square root, b = \fac

Q. E. D.

132 GEOMETRY. BOOK III.

Proposition III.

261. If the product of two quantities be equal to the product of two others, either two may be made the extremes of a proportion in which the other two are made the means.

Let ad = be.

We are to prove a : b : : c : d.

Divide both members of the given equation by b d.

a c b~~d'

Then

or, a : b : : c : d.

Q. E. D.

Proposition IV.

262. If four quantities of the same hind be in propor- tion, they will be in proportion by alternation.

Let a : b : : c : d. We are to prove a : c : : b : d.

kt a c

Now, - = -.

6 d

Multiply each member of the equation by -

c

Then ? = *

c a

or, a : c : : b : d.

Q. E. D.

THEORY OF PROPORTION. 133

Proposition V.

263. If four quantities be in proportion, they will be in proportion by inversion.

Let a : b : : c : d.

We are to prove b : a : : d : c.

Now, £w5.

b d

Divide 1 by each member of the equation.

Then ! = *,

a c

or, b : a : : d : c.

Q. E. o.

Proposition VI.

264. If four quantities be in proportion, they will be in proportion by composition.

Let a : b : : c : d

We are	to prove	a + b	: b : : c + d :	<L
Now		a b	c
Add 1 to each member of the equation.
Then that is.		a a + b	c = 5+1' c + d

b d

a + b : b : : c + d : d.

Q. E D

134 GEOMETRY. BOOK III.

Proposition VII.

265. If font quantities be in proportion, they will be in proportion by division.

Let a : b : : c : d. We are to prove a — b : b : : c — d : d.

Now *=1.

b d Subtract 1 from each member of the equation.

Then that is,

or,

a c
a — b c — d
b d '
a — b : b : : c — d :	d.

Q. E. D.

Proposition VIII. 266. In a series of equal ratios, the sum of the ante- cedents is to the sum of the consequents as any antecedent is to its consequent.

Let a : b = c : d — e :f = g : h. We are to prove a + c + e + g : b + d + f + h : : a : b. Denote each ratio by r.

a c e g Then »— j-3 =/ = T

Whence, a = br, c = dr, e —fr, g = hr. Add these equations.

Then a + c + e + g = (b + d + f + h) r. Divide by (b + d + / + h).

Then a + c+g + -*gara-a>

b+d+f+k r b

or, a + c + e + g : b + d + / + h : : a : b.

Q. E. D.

THEORY OF PROPORTION. 135

Proposition IX. 267. The products of the corresponding terms of two or more proportions are in proportion.

	Let a : b : : c :	*
	e if : : g :	K
	k : I : : m :	n}
We art	• to 'prove aek : bfl i :	cgm	: dhn.
Now	a c e g	k = J	m n
Whence by multiplication,
	aek cgm bfl dhn	t
or,	aek : bfl : : cgm	: dh	n.	Q.	E. D.
	Proposition X.
268.	Like powers, or like roots	j of the terms	•	a j&ra-
portion are	in proportion. Let a : b : : c	: d.
We are	to prove an : bn : : cn	: d»,
and	a* : 6n : : C* :	dn.
Now	a c b = d'
By raising to the n01 power,
	— = : or an : bn bn dn	: : c*	: d»
By extracting the wth root,
	l i a» c» 1,1 = — : or, an : 6* i i	i : :c»	i . d*.
	6*» c?n

Q. E. D.

269. Def. Equimultiples of two quantities are the products obtained by multiplying each of them by the same number. Thus m a and m b are equimultiples of a and b.

136 GEOMETRY. BOOK III.

Proposition XI.

£70. Equimultiples of two quantities are in the same

ratio as the quantities themselves.

Let a and b be any two quantities.

We are to prove ma : mb : : a : b.

at a a

Now _ = _ .

b b Multiply both terms of first fraction by m.

mi m a a

Then =

mb b

or, ma : mb : : a : b.

Q. E. D.

Proposition XII.

271. If two quantities be increased or diminished by like parts of each, the results will be in the same ratio as the quantities themselves.

Let a and b be any two quantities.

We are to prove a ± — a : b ± I b : : a : b. q q

In the proportion,

ma : mb : : a : b}

substitute for m, 1 ± - . 9.

Then (l ± i\ a : (l ± E\ b : : a : b,

a ±P.a : b ±?b

Q. E. D.

272. Dep. Euclid's test of a proportion is as follows : — " The first of four magnitudes is said to have the same ratio to the second which the third has to the fourth, when any equi- multiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ;

THEORY OF PROPORTION. 137

" If the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth ; or,

" If the multiple of the first be equal to that of the second, the multiple of the third is also equal to that of the fourth ; or,

" If the multiple of the first be greater than that of the second, the multiple of the third is also greater than that of the fourth."

Proposition XIII.

273. If four quantities be proportional according to the algebraical definition, they will also be proportional according to Euclid's definition.

Let a, b, c, d be proportional according to the alge-

a c braical definition ; that is -r= -j-

We are to prove a, 6, c, d, proportional according to Euclid '$ definition.

Multiply each member of the equality by — .

n

r™ ma mc

Then — =

nb n d

Now from the nature of fractions,

if m a be less than nb, mc will also be less than n d ;

if m a be equal to nb, mc will also be equal to n d ;

if m a be greater than nb, mc will also be greater than n d.

.'. a, b, c, d are proportionals according to Euclid's def- inition.

Q. E. D.

138 GEOMETRY. BOOK III.

Exercises.

1. Show, that the straight line which bisects the external vertical angle of an isosceles triangle is parallel to the base.

2. A straight line is drawn terminated by two parallel straight lines ; through its middle point any straight line is drawn and terminated by the parallel straight lines. Show that the second straight line is bisected at the middle point of the first.

3. Show that the angle between the bisector of the angle A of the triangle ABC and the perpendicular let fall from A on BG is equal to one-half the difference between the angles B and C.

4. In any right triangle show that the straight line drawn from the vertex of the right angle to the middle of the hypote- nuse is equal to one-half the hypotenuse.

5. Two tangents are drawn to a circle at opposite extremities of a diameter, and cut off from a third tangent a portion A B. If C be the centre of the circle, show that A C B is a right angle.

6. Show that the sum of the three perpendiculars from any point within an equilateral triangle to the sides is equal to the altitude of the triangle.

7. Show that the least chord which can be drawn through a given point within a circle is perpendicular to the diameter drawn through the point.

8. Show that the angle contained by two tangents at the extremities of a chord is twice the angle contained by the chord and the diameter drawn from either extremity of the chord.

9. If a circle can be inscribed in a quadrilateral ; show that the sum of two opposite sides of the quadrilateral is equal to the sum of the other two sides.

10. If the sum of two opposite sides of a quadrilateral be equal to the sum of the other two sides; show that a circle can be inscribed in the quadrilateral,

PROPORTIONAL LINES. 139

On Proportional Lines.

Proposition I. Theorem.

274. If a series of parallels intersecting any two straight lines intercept equal parts on one of these lines, they will intercept equal parts on the other also.

H W

K K>

Let the series of parallels A A', B B', CC, D D', E E', intercept on H' K' equal parts A'B', B'C, CD', etc.

We are to prove

they intercept on H K equal parts A B, B C, C D, etc.

At points A and B draw A m and B n II to H' K'.

Am = A'B', §135

(parallels comprehended between parallels are equal).

Bn = B'C, § 135

.'. A m = Bn. In the A B Am and C B v,

ZA=ZB, § 77

(having their sides respectively II and lying in the same direction from the vertices).

Z m = Z n, § 77

and Am = Bn,

.'. ABAm = A CBn, § 107

(having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

.*. AB = BC,

(being homologous sides of equal A). In like manner we may prove BC = CD, etc.

Q. E. D.

140

GEOMETRY. — BOOK III.

Proposition II. Theorem.

275. If a line be drawn through two sides of a triangle parallel to the third side, it divides those sides propor- tionally.

Fig. 1. Fig. 2.

In the triangle ABC let E F be drawn parallel to B C.

xrr v EB FC

We are to prove = .

AE AF

Case I. — When A E and EB (Fig. 1) are commensurable.

Find a common measure of A E and E B, namely B m.

Suppose B m to be contained in B E three times,

and in A E five times.

Then ££«?:

AE 5

At the several points of division on B E and A E draw straight lines II to B C.

These lines will divide A C into eight equal parts,

of which FC will contain three, and A F will contain five, § 274

(if parallels intersecting any two straight lines intercept equal parts on one

of these lines, they will intercept equal parts on the other also).

. FC __ 3

" AF~ 5'

EB 3

AE 5'

. EB = FC

' ' AE TF'

But

Ax. 1

PROPORTIONAL LINES. 141

Case. II. — When A E and E B (Fig. 2) are incommensurable.

Divide A E into any number of equal parts,

and apply one of these parts to EB as often as it will be contained in E B.

Since A E and E B are incommensurable, a certain number of these parts will extend from E to a point K, leaving a re- mainder KB, less than one of the parts.

Draw KB II to BC. Since A E and E K are commensurable,

EK FH in T.

AE = AF (CaS6L)

Suppose the number of parts into which A E is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to B.

The limit of E K will be E B, and the limit of FH will be FC

.*. the limit of will be ,

AE AE

and the limit of will be

AF AF

•pi jr ~p it

Now the variables and — — are always equal, how-

A E AF

ever near they approach their limits ;

A their limits EJL and L£- are equal, § 199

A& A r

Q. E. D.

276. Corollary. One side of a triangle is to either part cut off by a straight line parallel to the base, as the other side is to the corresponding part.

Now EB : AE : : FC : AF. § 275

By composition,

EB + A E : A E : : FC + A F : A F, § 263

or, A B : A E : : A C : A F.

142 GEOMETRY. BOOK III.

Proposition III. Theorem.

277. If a straight line divide two sides of a triangle proportionally, it is parallel to the third side.

A

In the triangle ABC letEF be drawn so that — = — .

AF AF

We are to prove F F II to B C.

From F draw E H \\ to B C.

(one side of a A is to either part cut off by a line II to the base, as the other side is to the corresponding part).

But 44 - M- HyP-

Ax. 1

.*. F F and F II coincide, (their extremities being the same points).

But FH is II to BC; Cons.

.*. F F, which coincides with F II, is II to BC.

Q. E. D.

278. Def. Similar Polygons are polygons which have their homologous angles equal and their homologous sides proportional.

Homologous points, lines, and angles, in similar polygons, are points, lines, and angles similarly situated.

AF	AF
. AC ' ' AF	AC AH'
.'. AF =	AH.

SIMILAR POLYGONS. 143

On Similar Polygons. Proposition IV. Theorem. 279. Two triangles which are mutually equiangular are

similar.

A A' A

In the A ABC and A1 B1 C let A A, B, C be equal to

A A', B', C respectively.

We are to prove A B : A' B' = AC : A' C = BC : B' C. Apply the A A' B1 C to the A ABC, so that Z A' shall coincide with Z A.

Then the A A' B' C will take the position of A A E H.

Now Z A EH (same as Z B') = Z B.

.'. EH is II to BC, § 69

{when two straight lines, lying in the same plane, are cut by a third straight line, if the ext. int. A be equal the lines are parallel),

.'.AB:AE = AC:AH, §276

(one side of a A is to either part cut off by a line II to tlie base, as tlie other side is to the corresponding part).

Substitute for A E and A H their equals A' B' and A' C.

Then AB : A< B> = AC : A'C.

In like manner we may prove

A B : A' B' = B C : B' C.

.*. the two A are similar. § 278

Q. E. D.

280. Cor. 1. Two triangles are similar when two angles of the one are equal respectively to two angles of the other.

281. Cor. 2. Two right triangles are similar when an acute angle of the one is equal to an acute angle of the other.

144

GEOMETRY.

BOOK III.

Proposition V. Theorem. 282. Two triangles which have their sides respectively proportional are similar.

In the triangles ABG and A' B' C let AB AG BG J^B' ~~ A' G' ~~ B'O' We are to prove A A, B, and G equal respectively to A A', B'y and C'.

Take on A B, A E equal to A' B1, and on AG, AH equal to A' &, Draw EH. AB AG A'B> A'Gr Substitute in this equality, for A' B' and A' G' their equals

Hyp.

A E and A H. Then

AB AG AE AH'

.\EH is II to.SC, • {if a line divide two sides of a A proportionally, it is

Now in the A A BG and A EH

Z ABG = ZAEH,

(being ext. int. angles).

Z ACB = Z A HE,

Z A= Z A.

.-. A AB G and A EH are similar, (two mutually equiangular A are similar).

; AB AB

' " BG ~ EH1

(homologous sides nf simi'ar A are proportional).

§277

to the third side).

§ 70

§70

Iden. § 279

§278

SIMILAR POLYGONS.

145

But

AB BC AE EH

A'B'

Hyp- Ax. 1

Cons.

§108

A'B' B'G1' Since A E — A' B',

EH = B'C. Now in the AAEH and A' B' C,

EH=B'C, AE = A'B', and A H = A' Cf, .-.A AEH=AA'B'C, (having three sides of the one equal respectively to three sides of the other).

But A A E H is similar to A ABC.

.-. A A' B' C is similar to A ABC.

Q. E. D.

283. Scholium. The primary idea of similarity is likeness of form ; and the two conditions necessary to similarity are :

I. For every angle in one of the figures there must be an equal angle in the other, and

II. the homologous sides must be in proportion.

In the case of triangles either condition involves the other, but in the case of other polygons, it does not follow that if one condition exist the other does also.

W

R

Thus in the quadrilaterals Q and Q', the homologous sides are proportional, but the homologous angles are not equal and the figures are not similar.

In the quadrilaterals R and R', the homologous angles are equal, but the sides are not proportional, and the figures are not similar.

146 GEOMETRY. BOOK III.

Proposition VI. Theorem.

284. Two triangles having an angle of the one equal to

an angle of the other, and the including sides proportional,

are similar.

A Af /\

In the triangles ABC and A' B' C let £A= Z.A', and

A'B1 A'C

We are to prove A A B C and A' B' C similar.

Apply the A A' B' C to the A ABC so that Z A' shall coincide with Z. A.

Then the point B' will fall somewhere upon A B, as at E,

the point C will fall somewhere upon A 0, as at H, and

B' C upon E H.

at AB AC jj

Now = Hyp.

A'B' A'C JF

Substitute for A' B' and A1 C their equals A E and A H.

Then i*^^.

AE AH

.'.the line EH divides the sides AB and AC propor- tionally ;

.'.EH is II to BC, § 277

(if a line divide two sides of a A proportionally, it is !l to the third side).

.'. the A A BC and A E H are mutually equiangular and similar.

,'. A A'B' C is similar to A ABC.

Q. E. D.

SIMILAR POLYGONS. 147

Proposition VII. Theorem.

285. Two triangles which have their sides respectively parallel are similar.

In the triangles ABC and A' B' C let AB.AC, and BC be 'parallel respectively to A' B'y A'C, and B'C. We are to prove A A B C and A' B' C similar.

The corresponding A are either equal, § 77

(two A ichose sides arc II, two and two, and lie in the same direction, or opposite directions, from their vertices are equal).

or supplements of each other, § 78

(if two A have two sides II and lying in the same direction from their vertices,

while the other two sides are II and lie in opposite directions, the A are

supplements of each other).

Hence we may make three suppositions :

1st. A + A' = 2rt.A, B + B' = 2vt.A, (7+C" = 2rt. A. 2d. A = Af, B + B' = 2vt.A, C + C = 2 rt. A.

3d. A=A', B = B> .'. C=C.

Since the sum of the A of the two A cannot exceed four right angles, the 3d supposition only is admissible. § 98

.'. the two A A B C and A' B' C are similar, § 279

(two mutually equiangular A are similar).

Q. E. D.

148

GEOMETRY. BOOK III.

Proposition VIII. Theorem.

286. Two triangles which have their sides respectivt perpendicular to each other are similar.

B

In the triangles EFD and B A C, let E F, FD and ED, be perpendicular respectively to AC,'BG and A B.

We are to prove A E ''F D and B AC similar.

Place the A E FD so that its vertex E will fall on A B, and the side E F, JL to A C, will cut A C at F'.

Draw F' D' II to F D, and prolong it to meet B C at H. In the quadrilateral B E D'H, JL E and // are rt. A .

.-.ZB + ZED' H=2 rt. A.

But ZED' F' + Z ED 11=2 rt. A.

.'.ZED' F' = ZB.

Now ZC+ZHF'C=ri.Z,

(in a rt. A the sum of the two acute A = a rt. Z) and ZEF'D' + Z IIF'C = rt. Z.

.'.ZEF'D' = ZC. .'.AEF'D' and B AC are similar. But A EFD is similar to A E F D'.

.'. A E F D and B A C are similar.

§158

§34

Ax. 3.

§103

Ax. 9. Ax. 3.

§280 §279

Q. E. D.

287. Scholium. When two triangles have their sides re- spectively parallel or perpendicular, the parallel, sides, or the perpendicular sides, are homologous.

SIMILAR POLYGONS." 149

Proposition IX. Theorem.

288. Lines drawn through the vertex of a triangle divide proportionally the base and its parallel.

In the triangle ABC let II L be parallel to AC, and let BS and BT be lines drawn through its ver- tex to the base.

We	are	to prove
			AS HO	=	ST on	=	TC RL

A B HO and B AS are similar, § 279 (two & which are mutually equiangular are similar).

A B 0 R and B S T are similar, § 279

A B R L and B T C are similar, § 279

• AJL (^\ $? _ (BT\ TC §278

" HO \0 B/ ~~~ 0 R ~\B RJ ~~ RL' S -

(homologous sides of similar & are proportional).

Q. E. D.

Ex. Show that, if three or more non-parallel straight lines divide two parallels proportionally, they pass through a common point.

150 GEOMETRY. BOOK III.

Proposition X. Theorem.

289. If in a right triangle a perpendicular be drawn from the vertex of the right angle to the hypotenuse :

I. It divides the triangle into two right triangles which are similar to the whole triangle, and also to each other:

II. The perpendicular is a mean proportional between the segments of the hypotenuse.

III. Each side of the right triangle is a mean pro- portional between the hypotenuse and its adjacent segment.

IV. The squares on the two sides of the right triangle have the same ratio as the adjacent segments of the hypote- mise.

V. The square on the hypotenuse has the same ratio to the square on either side as the hypotenuse has to the segment adjacent to that side.

B

- F

In the right triangle ABC, let B F be drawn from the vertex of the right angle B, perpendicular to the hypotenuse A C.

I. We are to prove

the AABF, ABC, and FBC similar. In the rt. A BA F and BA C, .

the acute Z. A is common.

.*. the A are similar, § 281

{two rt. A are similar when an acute Z of the one is equal to an acute Z of the other).

In the rt. ABCFqm&BCA,

the acute Z C is common.

.*. the A are similar. § 281

Now as the rt. AABF and C B F are both similar to A B C, by reason of the equality of their A,

they are similar to each other.

SIMILAR POLYGONS. 151

II. We are to prove A F : BF : : BF : FO. In the similar A A B F and G B F,

A F, the shortest side of the one, B F, the shortest side cf the other, B F, the medium side of the one, F G, the medium side of the other.

III. We are to prove A G : A B : : A B : A F. In the similar A A B G and A B F,

A G, the longest side of the one, A B, the longest side of the other, A B, the shortest side of the one, A F, the shortest side of the other. Also in the similar A A B C aiid F B G,

A C, the longest side of the one, B C, the longest side of the other, B C, the medium side of the one, F G, the medium side of the other.

™- nr , AT? AF

IV. We are to prove = .

EC1 F<J In the proportion A G : A B : : A B : A J?,

A~B2 = AG X AF, § 259

(the product of the extremes is equal to the product of the means).

and in the proportion AG : BG : : BG : FG,

F7?=-ACX FG. §259

Dividing the one by the other,

JTB2 AGX AF

KJ? ~ AGX FG' Cancel the common factor A G, and we have

re _ af

BG2 FG'

V. We are to prove HO. = ££.

AW ^F jfG2=ACX AG.

X22 = AGX AF, (Case III.)

Divide one equation by the other :

then £?* = AGX AG _ AC

J3f AGXAF AF q. e. o.

152 GEOMETRY. BOOK III.

Proposition XI. Theorem.

290. If two chords intersect each other in a circle, their segments are reciprocally proportional.

Let the two chords AB and EF intersect at the point 0.

We are to prove AO : EO : : OF : OB.

Draw ,4^ and E B.

In the AAOFtmdEOB,

ZF=ZB, §203

(each being measured by \ arc A E).

A A=* A E, § 203

(each being measured by £ arc F B).

.*. the A are similar. § 280

(two A are similar when two A of the one are equal to two A of the other).

Whence A 0, the medium side of the one, § 278

E 0, the medium side of the other, 0 F, the shortest side of the one, 0 B, the shortest side of the other.

Q. E. D.

SIMILAR POLYGONS. 153

Proposition XII. Theorem.

291. If from a point without a circle two secants be drawn, the whole secants and the parts without the circle are reciprocally proportional.

Let OB and OC be two secants drawn from point 0. We are to prove OB : OC : : OM : OH.

Draw H C and M B. In the A OIIC and OMB

/. 0 is common,

Z B = Z. C, § 203

{each being measured by $ arc H M).

.'. the two A are similar, § 280

(two & are similar wJicn two A of the one are equal to two A of the other).

Whence 0 B, the longest side of the one, §278

: 0 C, the longest side of the other,

: : 0 M, the shortest side of the one,

: 0 If, the shortest side of the other.

Q. E. D.

154 GEOMETRY. BOOK m.

Proposition XIII. Theorem.

292. If from a point without a circle a secant and a tangent he drawn, the tangent is a mean proportional between the whole secant and the part without the circle.

0

c

Let OB be a tangent and 0 C a secant drawn from the point 0 to the circle MBC.

We are to prove 0 C : 0 B : : 0 B : 0 M.

Draw BM and £ C.

In the A tf£i/and OBG

Z. 0 is common.

A OBMis measured by J arc MB, § 209

(being an Z formed by a tangent and a chord).

A G is measured by \ arc B M, § 203

(being an inscribed Z. ).

.\Z OBM=Z C.

.'.AOBCsmdOBMsive similar, § 280

(having tvjo A of the one equal to two A of the other).

Whence 0 C, the longest side of the one, § 278

0 B, the longest side of the other,

0 B, the shortest side of the one,

0 My the shortest side of the other.

Q. E. D.

SIMILAR POLYGONS.

155

Proposition XIV. Theorem.

293. If two polygons be composed of the same number of triangles which are similar, each to each, and similarly placed, then the polygons are similar. E

C B' C

In the two polygons ABODE and A'B'O'D'E', let the triangles B A E, BEO, and CED be similar respectively to the triangles B' A' E'f B' E' 0\ and C'E' D'.

We are to prove the polygon ABODE similar to the polygon A' B' 0' D' E'.

Z A=Z A', § 278

(being lunnologous A of similar A).

Z ABE = Z A'B'E', §278

Z EBO = Z E'B'O', §278

Add the last two equalities.

Then Z ABE+ Z E B 0 =? Z A1 B' E> + Z E' B' C ; or, ZABO^ZA'B'C.

In like manner we may prove Z BO D = Z B' C D'y etc.

.*. the two polygons are mutually equiangular. AE A

Now

IB /EB\_ BO _(EO\_OD ^ED A'E' ArB,~~\ErB')~BlOl~\E'Cl) CD1 E' D1' (the homologous sides of similar A are proportional). • .'. the homologous sides of the two polygons are proportional.

.*. the two polygons are similar, § 278

(having their homologous A equal, and tlieir homologous sides proportional).

Q. E. D.

156

GEOMETRY.

BOOK III.

Proposition XV. Theorem.

294. If two polygons be similar, they are co (he same number of triangles, which are similar and pla

of ilarly

B C B' a

Let the polygons ABODE and A'B'C D'E' be similar.

From two homologous vertices, as E and E',

draw diagonals EB, EC, and E< B', E C.

We are to prove A A E B, EBG, EC D

similar respectively to A A' E B', E' B' C, E C D'.

In the AAEB and A1 E' B1,

Z A=Z A1, § 278

(being homologous A of similar polygons).

AE^= AB_ § 278

A'E' A'B1'

(being homologous sides of similar polygons).

.'. A A E B and A' E> B1 are similar, § 284

(having an A of the one equal to an A of the oilier, and tJie including sides proportional).

Also, Z ABC=Z A'B'C,

(being homologous A of similar polygons).

Z ABE = Z A'B' E',

(being homologous A of similar A ).

.'.Z ABC- ZABE^Z A'B'C ~Z A'B' E'. That is Z EBC = ZE'B'C.

SIMILAR POLYGONS.

157

Now

also

EB AB

WB' ~ A'B'' (being homologous sides of similar & ) ;

BC = AB

WC' A'B1' {being homologous sides of similar polygons).

EB BC

E' B'

B' C" '

Ax. 1

§284

. ' . A E B C and E' B' C are similar,

(having an Z of the one equal to an A of the other, and the including sides

'proportional).

In like manner we may prove AECD similar to A E' C D'.

Q. E. D.

Proposition XVI. Theorem.

295. The perimeters of two similar polygons have the same ratio as any two homologous sides.

B C

Let the two similar polygons be ABODE and A'B' CD' E', and let P and P' represent their perimeters.

We are to prove P : P' : : A B : A'B'.

AB : A'B' : : BC : B' C : : CD : C^etc. § 278 (the homologous sides of similar polygons are proportional).

.'. AB + BC, etc. : A'B' + B'C, etc. : : AB : A'B', § 26G (in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent).

That is

P : P'

AB : A'B'

Q. E. D.

158

GEOMETBY. BOOK III.

Pboposition XVII. Theobem.

296. The homologous altitudes of two similar triangles have the same ratio as any two homologous sides.

In the two similar triangles ABC and A'B'C, let the altitudes be BO and B'O'.

We are to prove

BO

AB

A< B'

B'O1

In the rt. A B 0 A and B' 0' A',

Z. A= Z A' §278

(being homologous A of the similar A A B C and A' B1 C).

.-. A B 0 A and A B' 0' A' are similar, § 281

(two rt. A having an acute Z of the one equal to an acute Z of the other are

similar).

.'. their homologous sides give the proportion

BO

AB

B'O' A'B'

Q. E. D

297. Cob. 1. The homologous altitudes of similar triangles have the same ratio as their homologous bases.

SIMILAR POLYGONS. 159

§278

In the similar A A B C and A' B' C,

AC AB

A7^' ~ A'B''

(the homologous sides of similar A are proportional).

And in the similar A B 0 A and B' 0' A',

B0 - AB S296

• B° AG Ax.l

B' 0' A> C '

298. Cor. 2. The homologous altitudes of similar triangles have the same ratio as their perimeters.

Denote the perimeter of the first by P, and that of the second by P'.

Then L = AA , S 295

(the perimeters of two similar polygons luivc the same ratio as any two homologous sides).

But 1°. = AJL } § 296

Ax. 1

Ex. 1. If any two straight lines be cut by parallel lines, show that the corresponding segments are proportional.

2. If the four sides of any quadrilateral be bisected, show that the lines joining the points of bisection will form a parallelo- gram.

3. Two circles intersect; the line A H KB joining their centres A, B, meets them in //, K. On A B is described an equilateral triangle ABC, whose sides B C, A C, intersect the circles in F, E. F E produced meets B A produced in P. Show that as PA is to P K so is C F to CE. and so also is PH to PB.

B'O'	A'B''
BO	P
Wo1	" P'

160

GEOMETRY.

BOOK III.

Proposition XVIII. Theorem.

299. In any triangle the product of two sides is equal to the product of the segments of the third side formed by the bisector of the opposite angle together with the square of the bisector.

Let Z BA C of the A A B C be bisected oy cue straight line AD. We are to prove BAXAC = BDXDC+AD2.

Describe the O A B C about the A A B C ; produce A D to meet the circumference in E, and draw E C. Then in the A A B D and AEG,

ZBAD = ZCAE, Hyp.

Z B = Z E, §203

{each being measured by I the arc AC).

.\AABDa,if&AEC are similar, § 280

(two A are similar when two A of the one are equal respectively to two A of tha other).

B A, the longest side of the one, : E A, the longest side of the other, : A D, the shortest side of the one, : A C, the shortest side of the other ; BA AD

EA ~ AC'

(homologous sides of similar A are proportional).

.\BAXAC = EAXAD. But EAX AD = (ED + A D) A A

.'. BA X A C = ED X A D + A D\

But EDXAD = BDXDC,

(the segments of two chords in a Q which intersect each other are reciprocally proportional).

Substitute in the above equality B D X D C for E D X A D, then BAX AC = BDX DC + AD*.

Q. E. D.

Whence

or,

§278

§290

SIMILAR POLYGONS.

161

Proposition XIX. Theorem.

300. In any triangle the product of two rides is equal to the product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle.

Let ABC be a triangle, and AD the perpendiculai from A to BC.

Describe the circumference ABC about the A A BC.

Draw the diameter A E, and draw E C.

We are to prove BAXAC = EAXAD.

In the A ABD and AEC .

Z BDA is art. Z,

Z EC A is art, Z, (Jbeing inscribed in a semicircle).

.'./. BDA =Z EC A.

£B = /.E,

(each being measured by \ the arc A (T). .'. A AB D and A E C are similar,

Cons. §204

§ 203 § 281

(two rt. A having an acute Z of the one equal to an acute Z. of the other are

similar).

Whence

or,

BA, the longest side of the one, E A, the longest side of the other, A D, the shortest side of the one, A C, the shortest side of the other ;

BA _ AD

EA~ AC'

.BAX AC = EAX AD.

§ 278

Q. E. D.

162

GEOMETRY. — BOOK III.

Proposition XX. Theorem.

301. The product of the two diagonals of a quadrilateral inscribed in a circle is equal to the sum of the products of its opposite sides.

Let ABC B be any quadrilateral inscribed in a circle, AC and BB its diagonals.

We are to prove BDXAC = ABXCD + ADXBC.

Construct Z ABE = Z BBC,

and add to each Z E BD.

Then in the A ABB and B C E,

ZABB = ZCBEf Ax. 2

and ZBBA=ZBCE, §203

(each being measured by | the arc A B).

.'. A A B D and B C E, are similar, § 280

(two A are similar when two A of the one are equal respectively to two A of the other).

Whence A B, the medium side of the one, C E, the medium side of the other, B B, the longest side of the one, B C, the longest side of the other,

SIMILAR POLYGONS. 163

§278

AD _ BD CE ~ BC' (the homologoics sides of similar A are proportional).

.'.BD X CE = AD X BC.

Again, in the A A B E and B C D,

Z ABE = Z DBC, Cons.

and ZBAE = ZBDC, §203

(each being measured by $ of the arc B C).

.'. A A B E and B CD are similar, § 280

(two A are similar when two A of the one are equal respectively to two A of the other).

"Whence A B, the longest side of the one, B D, the longest side of the other, A E, the shortest side of the one, CD, the shortest side of the other.

or, ^ = 11, §278

BD CD'

(the homologous sides of similar A are proportional). .'.BD X AE = ABX CD. But BDXCE = ADXBC.

Adding these two equalities,

BD (AE+ CE) = ABX CD + ADX BC, or BDXAC = ABXCD + ADXBC.

Q. E. D.

Ex. If two circles are tangent internally, show that chords of the greater, drawn from the point of tangency, are divided proportionally by the circumference of the less.

164 GEOMETRY. — BOOK III.

On Constructions. Proposition XXI. Problem. 302. To divide a given straight line into equal parts. A^— . 7b

Let A B be the given straight line.

It is required to divide A B into equal parts.

From A draw the indefinite line A 0.

Take any convenient length, and apply it to A 0 as many times as the line A B is to be divided into parts.

From the last point thus found on A 0, as C, draw C B.

Through the several points of division on A 0 draw lines II to CB.

These lines divide A B into equal parts, § 274

(if a series of lis intersecting any two straight lines, intercept equal parts on one of these lines, they intercept equal parts on the other also).

Q. E. F.

Ex. To draw a common tangent to two given circles.

I. When the common tangent is exterior.

II. When the common tangent is interior.

CONSTRUCTIONS. 165

Proposition XXII. Problem.

303. To divide a given straight line into parts pro- portional to any number of given lines.

H K B

■A-*z^ v — " r ,

\ \	n
	0
-X

Let A B, m, n, and o be given straight lines.

It is required to divide A B into parts proportional to the given lines m, n, and o.

Draw the indefinite line A X.

On A X take A C = m,

CE = n,

and EF=o.

Draw FB. From E and C draw E K and C H II to F B. K and // are the division points required.

For f4£V.^.«^» §275

\AE) AC CE EF

(i line drawn through two sides of a A II to tlte third side divides tliosr sides proportional! >t).

.'.AH : HK : KB : : AC : CE : E F. Substitute nt, n, and o for their equals AC, C E, and E F. Then A H : HK : KB : : m : » : o.

Q. E. F

166 GEOMETRY. BOOK III.

Proposition XXIII. Problem.

304. To find a fourth proportional to three given straight lines.

B F m

S '-B

Let the three given lines be m, n, and o.

It is required to find a fourth proportional to m, n, and o.

Take A B equal to n.

Draw the indefinite line A R, making any convenient Z with A B.

On A R take A C = m, and 0 S = o.

Draw CB.

From S draw JSFW to C B, to meet A B produced at F.

B F is the fourth proportional required.

For, AG : AB : : OS : B F, § 275

(a line drawn through two sides of a A II to the third side divides those sides proportionally).

Substitute on, n, and o for their equals AC, AB, and G S.

Then m : n : : o : B F.

Q. E. F.

CONSTRUCTIONS. 167

Proposition XXIV. Problem.

305. To find a third proportional to two given straight

lines.

A

A B

A C

Let A B and A G be the two given straight lines.

It is required to find a third proportional to A B and A G.

Place A B and A G so as to contain any convenient A.

Produce A B to D, making BD = AG.

Join BO.

Through D draw D E II to B G to meet A 0 produced at E.

C E is a third proportional to A B and AG. § 251

£5- £S« §275

(a line drawn throiigh two sides of a A II to the third side divides those sides proportionally).

Substitute, in the above equality, A C for its equal B D ;

Then d^ = ^,

AG GE'

or, A B : A G : : A G : CE.

Q. E. F.

168 GEOMETRY. — BOOK III.

Proposition XXV. Problem.

306. To find a mean proportional between two given lines.

B *

Let the two given lines be m and n. It is required to find a mean proportional between m and n. On the straight line A E

take AG = m, and G B = n.

On A B as a diameter describe a semi-circumference.

At G erect the _L G H.

G H is a mean proportional between m and n.

Draw II B and HA.

The Z A HB is a rt. Z, § 204

(being inscribed in a semicircle),

and HG is a J_ let fall from the vertex of a rt. Z to the

hypotenuse.

.'.AG : GH :: GH : G B, §289

<7/*e _L let fall from the vertex of the rt. Z. to the hypotenuse is a mean pro- portional between the segments of the hypotenuse).

Substitute for A G and G B their equals m and n.

Then m : G H : : GH : n. Q E F

307. Corollary. If from a point in the circumference a perpendicular be drawn to the diameter, and chords from the point to the extremities of the diameter, the perpendicidar is a mean pro- portional between the segments of the diameter, and each chord is a mean proportional between its adjacent segment and the diameter.

CONSTRUCTIONS. 169

Proposition XXVI. Problem.

308. To divide one side of a triangle into two parts proportional to the other two sides.

B E

Let ABC be the triangle.

It is required to divide the side B C into two such parts that the ratio of these two parts shall equal the ratio of the other two sides, A C and A B.

Produce C A to F, making A F = A B. Draw FB. From A draw A E II to FB.

E is the division point required.

For 9A. = 9JL. § 275

AF EB S

(a line drawn through two sides of a AW to the third side divides those sides proportionally).

Substitute for A F its equal A B.

Then £A = C*.

AB EB

Q. E. F.

309. Corollary. The line A E bisects the angle CAB. For /1F=ZABF, §112

(being opposite equal sides).

ZF=ZCAF, §70

(being ext.-int. A ).

ZAJ3F=ZBAFJ, §68

(being alt.-int. A ).

.'.ZCAE=ZBAE. Ax. 1

310. Def. A straight line is said to be divided in extreme and mean ratio, when the whole line is to the greater segment as the greater segment is to the less.

170

GEOMETRY. BOOK III.

Proposition XXVII. Problem. 311. To divide a given line in extreme and mean ratio.

S

/

H B

Let AB be the given line.

It is required to divide A B in extreme and mean ratio.

At B erect a J_ B G, equal to one-half of A B.

From G as a centre, with a radius equal to G B, describe a O.

Since A B is J_ to the radius GB at its extremity, it is tangent to the circle.

Through G draw A J), meeting the circumference in E and D. OnAB take AH = AE. H is the division point of A B required.

For AD : AB :: AB : AE, § 292

{if from a point without the circumference a secant and a tangent be drawn, the tangent is a mean proportional between the whole secant and the part without the circumference).

Then AD- AB - AB : : A B - A E : A E.

265

CONSTRUCTIONS. 171

Since A B — 2 G B, Cons.

and ED = 2 GB,

(the diameter of aO being 'twice the radius),

AB = ED. Ax. 1

.'.AD-AB = AD-ED = AE.

But AE = AH, Cons.

.'. A D - A B = A H. Ax. 1

Also AB-AE = AB-AH = HB.

Substitute these equivalents in the last proportion.

Then AH : AB : : HB : AH.

Whence, by inversion, AB : AH : : AH : HB. § 263

.'. A B is divided at H in extreme and mean ratio.

Q. E. F.

Eemark. A B is said to be divided at H, internally, in extreme and mean ratio. If BA be produced to H', making A H' equal to A D, A B is said to be divided at H', externally, in extreme and mean ratio.

Prove AB : AH' : : AH : W B.

When a line is divided internally and externally in th3 same ratio, it is said to be divided harmonically.

Thus^5 ± £__£ £? is divided harmoni- cally at G and D, if C A :GB::DA:DB; that is, if the ratio of the distances of G from A and B is equal to the ratio of the distances of D from A and B.

This proportion taken by alternation gives :

AG :AD::BG:BD; that is, G D is divided harmoni- cally at the points B and A. The four points A, B, C, D, are called harmonic points ; and the two pairs A, B, and G, D, are called conjugate points.

Ex. 1. To divide a given line harmonically in a given ratio. 2. To find the locus of all the points whose distances from two given points are in a given ratio.

/ /	\ \
	\
	\ \
\	\
\ /	\
\ /
	\ / \/
B'	a

172 GEOMETRY. BOOK III.

Proposition XXVIII. Problem.

312. Upon a given line homologous to a given side of a given polygon, to construct a polygon similar to the given polygon.

E

I I

Let A' E be the given line, homologous to A E of the given polygon ABC D E.

It is required to construct on A1 E' a polygon similar to the given polygon.

From E draw the diagonals E B and EG.

From E' draw E> B', making Z A' E' B' = Z A E B.

Also from A1 draw A' B', making Z B' A' E' = Z B A Ey

and meeting E' B' at B'.

The two A A B E and A1 B' E' are similar, § 280

(two A are similar if they have two A of the one equal respectively to two A of the other).

Also from E' draw E' C", making Z B' E' C = Z B E C.

From B' draw B' C, making Z E' B> ' C ' = Z E B C,

and meeting E' C at C.

Then the two A EB G and E' B' G' are similar, § 280 (two & are similar if they have two A of the one equal respectively to two A of the other).

In like manner construct A E' G' B' similar to A E G D.

Then the two polygons are similar, § 293

(two polygons composed of the same member of A similar to each other and similarly placed, are similar).

.'. A' B' G' D' E' is the required polygon.

Q. E. F.

EXEfcClSES. 173

Exercises.

1. A B C is a triangle inscribed in a circle, and B D is drawn to meet the tangent to the circle at A in D, at an angle ABB equal to the angle ABC; show that A C is a fourth propor- tional to the lines B D, A D, A B.

2. Show that either of the sides of an isosceles triangle is a mean proportional between the base and the half of the segment of the base, produced if necessary, which is cut off by a straight line drawn from the vertex at right angles to the equal side.

3. A B is the diameter of a circle, D any point in the circum- ference, and G the middle point of the arc AD. If A C, A D, B C be joined and A D cut B C in E, show that the circle cir- cumscribed about the triangle A E B will touch A C and its diameter will be a third proportional to B C and A B.

4. From the obtuse angle of a triangle draw a line to the base, which shall be a mean proportional between the segments into which it divides the base.

5. Find the point in the base produced of a right triangle, from which the line drawn to the angle opposite to the base shall have the same ratio to the base produced which the per- pendicular has to the base itself.

6. A line touching two circles cuts another line joining their centres ; show that the segments of the latter will be to each other as the diameters of the circles.

7. Required the locus of the middle points of all the chords of a circle which pass through a fixed point.

8. 0 is a fixed point from which any straight line is drawn meeting a fixed straight line at P ; in 0 P a point Q is taken such that 0 Q is to 0 P in a fixed ratio. Determine the locus of Q.

9. 0 is a fixed point from which any straight line is drawn meeting the circumference of a fixed circle at P ; in 0 P a point Q is taken such that 0 Q is to 0 P in a fixed ratio. Determine the locus of Q.

BOOK IV.

COMPARISON AND MEASUREMENT OF THE SUR- FACES OF POLYGONS.

Proposition I. Theorem.

313. Two rectangles having equal altitudes are to each other as their bases.

D

D

0 " ~ O

Let the two rectangles be AC and A F, having the the same altitude A D.

rect. A 0 _ AB

iecLAF~~ AE'

We are to prove

Then

Case I. — When A B and A E are commensttrable. Find a common divisor of the bases A B and A E, as A 0. Suppose A 0 to be contained in A B seven times and in A E four times.

AB = 7 AE ~ 4'

At the several points of division on A B and A E erect Js . The rect. A C will be divided into seven rectangles, and rect. A F will be divided into four rectangles.

These rectangles are all equal, for they may be applied to each other and will coincide throughout.

But

rect A G	7
rect A F	4
AB	7
AE	4
rect A G rect A F	AB AE

COMPARISON AND MEASUREMENT OF POLYGONS.

175

CASE II. — When A B and A E are incommensurable.

D

D

II

B

K

E

Divide A B into any number of equal parts, and apply one of these parts to A E as often as it will be contained in A E.

Since A B and A E are incommensurable, a certain number of these parts will extend from i to a point K, leaving a re- mainder K E less than one of these parts.

Draw JSTJEMI to E F.

Since A B and A K are commensurable,

rect.AH = AK Case j

rect. AC ~ AB

Suppose the number of parts into which A B is divided to be continually increased, the length of each part will become less and less, and the point K will approach nearer and nearer to E.

The limit of A it will be A E, and the limit of rect. A H will be rect. A F.

.'.the limit of — will be UH, AB AB

j i.u v ■*. c reck AH .11 • rect. A F and the limit ot will be

rect. A C

rect. A C

Now the variables and . are always equal

A B rect. AC J H

however near they approach their limits ;

rect. A F

.'. their limits are equal, namely,

AE

TGct.AC AB

§199

Q. E. D.

314. Corollary. Two rectangles having equal bases are to each other as their altitudes. By considering the bases of these two rectangles A D and A D, the altitudes will be A B and A E. But we have just shown that these two rectangles are to each other as A B is to A E. Hence two rectangles, with the same base, or equal bases, are to each other as their altitudes.

176

GEOMETRY. — BOOK IV.

Another Demonstration. Let A C and A1 C be two rectangles of equal altitudes. P C O Pi

F E D A A' D< E> F> G'

rect. AC AD

We are to prove

rect. A'C A' J)'

Let b and &', S and S' stand for the bases and areas of these rectangles respectively.

Prolong A D and A' D\

Take AD, D E, E F . . . . m in number and all equal,

and A' D', D' E', E' F', F' G' . . . . n in number and all equal.

Complete the rectangles as in the figure.

Then base AF = mb,

and base A' ' G' ' == nb' ;

rect. A P = mS, and rect.^/P/=^^/.

Now we can prove by superposition, that if A F be > A' G', rect. A P will be > rect. A' P' ; and if equal, equal ; and if less, less.

That is, if mb be > nb', m S is > n S' ; and if equal, equal ; and if less, less.

Hence, b : b' : : S : &, Euclid's Def., § 272

Q. E. D.

COMPARISON AND MEASUREMENT OF POLYGONS.

177

Proposition II. Theorem.

315. Two rectangles are to each other as the products of their bases by their altitudes.

a'

Rf

J

b b' b

Let R and R' be two rectangles, having for their bases b and b', and for their altitudes a and a'. R_ aXb R' ~

We are to prove

§314

§ 313

a' X y

Construct the rectangle S, with its base the same as that of R and its altitude the same as that of R'.

(rectangles having the same base are to each other as their altitudes) ;

and 3**'

(rectangles having the same altitude are to each other as their bases). By multiplying these two equalities together R aX b

R' ~~ a' Xbr

Q. E. D.

316. Def. The Area of a surface is the ratio of that surface to another surface assumed as the unit of measure.

317. Def. The Unit of measure (except the acre) is a square a side of which is some linear unit ; as a square inch, etc.

318. Def. Equivalent figures are figures which have equal areas.

Rem. In comparing the areas of equivalent figures the symbol ( = ) is to be read " equal in area."

178

GEOMETRY. BOOK IV.

Proposition III. Theorem.

319. The area of a rectangle is equal to the product of its base and altitude.

b 1

Let R be the rectangle, b the base, and a the alti- tude ; and let U be a square whose side is the linear unit.

We are to prove the area of R = a X b.

R _ aXb U , 1X1*

{two rectangles are to each other as the product of their bases and altitudes).

R

§315

But

U

is the area of R,

§316

the area of R = a X b.

Q. E. D.

320. Scholium. When the base and altitude are exactly- divisible by the linear unit, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of

measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure; and the area of the figure equals 7X4,

COMPARISON AND MEASUREMENT OF POLYGONS. 179

Proposition IV. Theorem.

321. The area of a parallelogram is equal to the product of its base and altitude.

BE C F B C E F

A D

Let A E FD be a parallelogram, A D its base, and G D its altitude.

We are to prove the area of the EJ A E F D = A D X G D.

From A draw A B II to D G to meet F E produced.

Then the figure A BG D will be a rectangle, with the same base and altitude as the O A E F D.

In the rt. A A B E and CD F,

AB = GD, §126

(being opposite sides of a rectangle).

and AE = DF, §134

(being opposite sides of a CD) ;

.'.AABE = AGDF, §109

(two rt. A are equal, when the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other).

Take away the A G D F and we have left the rect. ABG D.

Take away the A A B E and we have left the O A E F D.

.'. rect. ABG D = O A EFD. Ax. 3

But the area of the rect. ABC D = AD X CD, § 319

(the area of a rectangle equals the product of its base and altitude).

.'. the area of the O A EFD = A D X C D. Ax. 1

Q. E. D.

322. Corollary 1. Parallelograms having equal bases and equal altitudes are equivalent.

323. Cor. 2. Parallelograms having equal bases are to each other as their altitudes ; parallelograms having equal alti- tudes are to each other as their bases ; and any two parallelo- grams are to each other as the products of their bases by their altitudes.

180 GEOMETRY. BOOK IV.

Proposition V. Theorem.

324. The area of a triangle is equal to one-half of the product of its base by its altitude.

A B D

Let ABC be a triangle, AB its base, and CD its altitude.

We are to prove the area of the A A B C = J A B X CD.

From Cdraw C ff \\ to A B.

From A draw A H \\ to B C.

The figure A B C H is a parallelogram, § 136

{having its opposite sides parallel),

and A C is its diagonal.

.-.A ABC = A AHC, § 133

(the diagonal ofaO divides it into two equal A ).

The area of the ED ABC H is equal to the product of its base by its altitude. § 321

.'.the area of one-half the O, or the A A B C, is equal to one-half the product of its base by its altitude,

or, IABXCD.

2 Q. E. D.

325. Corollary 1. Triangles having equal bases and equal altitudes are equivalent.

326. Cor. 2. Triangles having equal bases are to each other as their altitudes ; triangles having equal altitudes are to each other as their bases ; any two triangles are to each other as the product of their bases by their altitudes.

COMPARISON AND MEASUREMENT OF POLYGONS. 181

Proposition VI. Theorem.

327. The area of a trapezoid is equal to one-half the sum of the parallel sides multiplied by the altitude. H E C

A F B

Let A B G II be a trapezoid, and EF the altitude. We are to prove area of A B G H ' = \ {EG + A B) E F. Draw the diagonal A C. Then the area of the A A HG = \ HG X EF, § 324

(the area of a A is equal to one-half of the 'product of its base by its altitude),

and the area of the A A B C = J A B X EF, § 324 .-.AAHC+ AABG, or, area ofABGH=i(HG+AB) EF.

Q. E D.

328. Corollary. The area of a trapezoid is equal to the product of the line joining the middle points of the non-parallel sides multiplied by the altitude ; for the line 0 P, joining the middle points of the non-parallel sides, is equal to \ (HG + AB). §142

.'.by substituting 0 P for %(H G + A B), we have, the area of A B G H = OPX E F.

329. Scholium. The area of an irregular polygon may be found by dividing the polygon into triangles, and by finding the area of each of these tri- angles separately. But the method generally employed in practice is to draw the longest

diagonal, and to let fall perpendiculars upon this diagonal from the other angular points of the polygon.

The polygon is thus divided into figures which are right triangles, rectangles, or trapezoids ; and the areas of each of these figures may be readily found.

182 GEOMETRY. — BOOK IV.

Proposition VII. Theorem.

330. The area of a circumscribed poly goyi is equal to one- half the product of the perimeter by the radius of the in- scribed circle.

B

Let ABSQ, etc., be a circumscribed polygon, and G the centre of the inscribed circle.

Denote the perimeter of the polygon by P, and the radius of the inscribed circle by R.

We are to prove

ihe area of the circumscribed polygon = \ P X R.

Draw G A, G B, OS, etc.;

also draw 0 0, G D, etc., _L to A B, B S, etc.

The area of the A CA B = \A B X C 0, § 324

{the area of a A is equal to one-half the product of its base and altitude).

The area of the A CBS = \ B S X CD, § 324

.*. the area of the sum of all the A C A B, CBS, etc., = i(AB + BS, etc.) GO, § 187

(for 0 0, CD, etc., are equal, being radii of the same O).

Substitute for A B + BS + SQ, etc., P, and for G 0, R ;

then the area of the circumscribed polygon = |PX R.

Q. E. D.

COMPARISON AND MEASUREMENT OP POLYGONS.

183

Proposition VIII. Theorem.

331. The sum of the squares described on the two sides of a right triangle is equivalent to the square described on the hypotenuse.

Let ABC be a right triangle with its right angle atC. We are to prove AC2 + CB2 = AB2 Draw 0 0 JL to A B.

Then AC2 = AOXAB, § 289

{the square on a side of a rt. A is equal to the product of the hypotenuse by the adjacent segment made by Oic _L let fall from the vertex of the rt. Z) ;

and fit'2 = BOX AB,

By adding, AT? + F7?= (A 0 + B 0) A B, = ABX AB,

332. Corollary. The side and diagonal A of a square are incommensurable.

Let ABGD be a square, and AC the diagonal.

Then AB2 + FV2 = A~C\

or, 2 AB2 = ATC2. B

Divide both sides of the equation by AB2,

AB2

§ 289

Q. E. D.

Extract the square root of both sides the equation, then

AC ,_

AB

= s/Y.

Since the square root of 2 is a number which cannot be exactly found, it follows that the diagonal and side of a square nro two inoouimpnsurable linos.

184

GEOMETRY. BOOK IV.

Another Demonstration.

833. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two

^y^ / n	^ \
/	\
/
/ \L	1

D L E

Let ABC be a light A, having the right angle BAG.

We are to prove BG2 = BA + AG .

OnB G, GA, A B construct the squares B E, CH, A F.

Through A draw A L II to C E.

Draw A D and FG.

Z BACissirt. Z,

Z BAGi&nrt. Z,

. •. G A G is a straight line.

Z CAHis&Tt. Z, .'. B A H is a straight line.

and

Also

Hyp.

Cons.

Cons.

Now

Z DBC = Z FBA,

(each being art. £).

Cons.

COMPARISON AND MEASUREMENT OF POLYGONS. 185

Add to each the A A B G ; then ZABD = ZFBG,

.\AABB = AFBC. § 106

Now O B L is double A A B D,

(being on the same base B D, and between the same lis, A L and BD),

and square A F is double A F B C, (being on the same base FB, and between the same lis, FB and GO);

.-. O BL = square A F.

In like manner, by joining A E and BK, it may be proved that

O CL = square G H.

Now the square onBG = 0 BL + O G L,

= square A F + square G Ht

.-. BG* = FT + AG\

Q. E. D.

On Projection. 334. Def. The Projection of a Point upon a straight line of indefinite length is the foot of the perpendicular let fall from the point upon the line. Thus, the projection of the point G upon the line A B is the point P.

C C

P R * n^P DD

Fig. 1. Fig. 2.

The Projection of a Finite Straight Line* as G D (Fig. 1), upon a straight line of indefinite length, as A B, is the part of the line A B intercepted between the perpendiculars G P and D B, let fall from the extremities of the line G D.

Thus the projection of the line G D upon the line A B is the line P R.

If one extremity of the line G D (Fig. 2) be in the line A B, the projection of the line G D upon the line A B is the part of the line A B between the point D and the foot of the perpendicular G P ; that is, D P.

186

GEOMETRY. BOOK IV.

Proposition IX. Theorem.

335. In any triangle, the square on the side opposite an acute angle is equivalent to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side.

Let C be an acute angle of the triangle ABC, and D C the projection of AC upon B C.

We are to prove U? = WD2 + J~C* — 2 B C X D C. If D fall upon the base (Fig. 1),

DB = BC-Z>C; If D fall upon the base produced (Fig. 2),

DB = DC~BC.

In either case B~B2 = BC* + IfC* - 2 B C X D C. Add A D to both sides of the equality ;

then, JHf + fiB2 = FV2 + ID2 + UC2 -2BCXDC.

But

AD2 + WB2= £B2,

331

(the sum of the squares on two sides of a rt. A is equivalent to the square on the hypotenuse) ;

and

ad2 + irc2 = jrc*

331

Substitute JTB and A C for their equivalents in the above equality ;

then, AB2 = FV2 + J~C2-2BCXDC.

Q. E. D.

COMPARISON AND MEASUREMENT OF POLYGONS. 187

Proposition X. Theorem.

336. In any obtuse triangle, the square on the side opposite the obtuse angle is equivalent to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other on that side,

A

Let G be the obtuse angle of the triangle ABC, and G D be the projection of A C upon BC produced.

We are to prove IB* = Blf + £Jf + 2 B G X D G.

DB=BC+ DG

Squaring, 1TB2 = Blf + Blf + 2 B G X D G

Add A~lf to both sides of the equality ;

then, AD2 + D~B2 = E~02 + AD2 + DC2 + 2BGXDG.

But ID2 + DB2 = IB2 , § 331

{the sum of the squares on two sides of a rt. A is equivalent to the square on the hypotenuse) ;

and JTD2 + Blf = JTG2. § 331

Substitute A^B and J~D for their equivalents in the above equality;

then, A~B2 = Blf + A~G2 + 2 BC X DG.

Q. E. D.

337. Definition. A Medial line of a triangle is a straight line drawn from any vertex of the triangle to the middle point of the opposite side.

188 GEOMETRY. BOOK IY.

Proposition XI. Theorem.

338. In any triangle, if a medial line be drawn from the vertex to the base :

I. The sum of the squares on the two sides is equivalent to twice the square on half the base, increased by twice the square on the medial line ;

IT. The difference of the squares on the two sides is equivalent to twice the product of the base by the projection of the medial line upon the base.

A

In the triangle ABC let A M be the medial line and M D the projection of A M upon the base B C. Also let AB be greater than A C.

We are to prove

i. rtf ¥ nf = 2 em2 + 2 in*.

II. IB* - AG1 =2BCX MD. Since A B > A C, the Z A MB will be obtuse and the Z. A M C will be acute. § HQ

Then JTB2 = BM1 + AM1 + 2 BM X M D, §336

{in any obtuse A the square on the side opposite the obtuse Z. is equivalent to the sum of the squares on the other two sides increased by twice the product of one of those sides and the projection of the, other on that side) ;

and JT7? = m)2 + AM2 - 2 MCX MD, §335

in any A the square on the side opposite an acute Z is equivalent to the sum of the squares on the other two sides, diminished by twice the product of one of those sides and the projection of the other upon that side).

Add these two equalities, and observe that B M — MC. Then AB2 + A~C2 = 2 BM2 + 2 A~M2. Subtract the second equality from the first. Then AJ?-A~C2=2BCXMD.

Q. E. D.

COMPARISON AND MEASUREMENT OF POLYGONS. 189

Proposition XII. Theorem. 339. The sum of the squares on the four sides of any quadrilateral is equivalent to the sum of the squares on the diagonals together with four times the square of the line joining the middle points of the diagonals. A

In the quadrilateral A BCD, let the diagonals be A C and B D, and F E the line joining the middle points of the diagonals. We are to prove

JTff + BJf + (TT)2 + DA2 = AC2 + Bl? + 4 El*' Draw BE and D E.

Now iO2 + BC1 = 2 (— Y + 2 rf, § 338

(the sum. of the squares on the two sides of a A is equivalent to twice the square on half t/ie base increased by twice the square on the medial line to the base),

and CI? + m2 = 2 (A-£Y + 2DE2. § 338

Adding these two equalities,

.O2 + BO2 + 01? + DA2 = 4 (4^V + 2 (^ + E~E\

But BE2 + JTE1 = 2 (^V + 2 EF2, § 338

(the sum of the squares on the two sides of a A is equivalent to twice the square on half the base increased by twice the square on t/te medial line to the base).

Substitute in the above equality for (BE2 + DE2) its equivalent ;

theni^4-^2 + Z7Z52+ro2 = 4(^2 + 4(^)2+4^a

= IC2 + BD2 + 4 ET

Q. E. D.

340. Corollary. The sum of the squares on the four sides of a parallelogram is equivalent to the sum of the squares on the diagonals.

190

GEOMETRY.

BOOK IV.

Proposition XIII. Theorem.

341. Two triangles having an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles.

Let the triangles ABC and ADE have the common angle A.

We are to prove

Draw BE.

Now

AABC ABXAG

AADE ADXAE

AABC AC

AABE AE

(A having the same altitude are to each other as their bases).

Also

AABE

AB AD

AADE

(A leaving the same altitude are to each other as their bases).

Multiply these equalities ;

326

§ 326

then

AABC = ABX AG AADE ADX AE

Q. E. D.

COMPARISON AND MEASUREMENT OF POLYGONS. 191

Proposition XIV. Theorem. 342. Similar triangles are to each other as the squares on their homologous sides.

0 ~ 0' B'

Let the two triangles be AG B and A'C'B'.

w . AACB Aff

We are to prove =

A A'C'B' AT^

Draw the perpendiculars C 0 and C O1.

Then AACB = *BXC0 Z ** X «£-, § 326

A A'C'B' A'B'XC'O' A' B' CO' *

(two A are to each other as the products of their bases by their altitudes).

But ±2=™, §297

A'B' CO'' S

(the homologous altitudes of similar & have the same ratio as their homolo- gous bases).

Substitute, in the above equality, for its equal ;

1 J CO' l A'B'

,, AACB AB AB _ A^

then = v

a a i ni m a i »/ ^

Q. E. D.

192

GEOMETRY. BOOK IV.

Proposition XV. Theorem.

343. Two similar polygons are to each other as the squares on any two homologous sides. B C

F E

Let the two similar polygons be ABC, etc.. and

A' BO, etc. .

m • ABC, etc. ATE2

We are to prove : = .

A' B' C, etc. jrrfi

From the homologous vertices A and A' draw diagonals.

AB BC

A'B' ~

Now

CD ,

, etc.,

B'C CD'

(similar polygons have their homologous sides proportional) ;

.'.by squaring,

CD1

ATB2 Ftf _

jFB'2 Ftf CU'2

, etc.

The AABC,ACD, etc., are respectively similar to A' B'C,

A' C D\ etc.,

294

(two similar polygons are composed of the same number of & similar to each other and similarly placed).

A ABC

nf

A A' B'C AHS'2 (similar A are to each other as the squares on their homologous sides)

AACD 0~D2

§ 342

and

A A' CD'

CD'2

§ 34?

COMPARISON AND MEASUREMENT OF POLYGONS. 193

WW1 JJlr

AABC AACD

AA'B'C A A' CD1

In like manner we may prove that the ratio of any two of the similar A is the same as that of any other two.

AABC AACD A APE AAEF

' ' AA'B'C "" A A' C" D' ~ A A' D' E' ™ AA'E'F'

AABC + ACD + ADE-V AEF AABC

AA'B'C' + A'C'D' + A'D'E' + A' E' F'~ A A1 B' C'

(in a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent).

But AABC -I*-, §342

AA'B'C jrgi' S

(similar & are to each other as the squares on their homologous sides) ;

. the polygon ABC, etc. _ £1? the polygon A' B' C, etc. . A7H'2 '

Q. E. D.

344. Corollary 1. Similar polygons are to each other as the squares on any two homologous lines.

345. Cor. 2. The homologous sides of two similar poly- gons have the same ratio as the square roots of their areas.

Let S and S' represent the areas of the two similar polygons A B G, etc., and A' B' C, etc., respectively.

Then S : S' : : A& : A7!?,

(similar polygons are to each other as the squares of their homologous sides).

\[S : v^ : : . iB : A' B'f 268

or, AB : A' B' : : yfi : fl?.

194

GEOMETRY. BOOK IV.

On Constructions.

Proposition XVI. Problem.

346. To construct a square equivalent to the sum of two given squares.

		B'h 1 N Al X.._	s
		Rf
R

Let R and R' be two given squares. It is required to construct a square = R+ R'. Construct the rt. Z A.

Take A B equal to a side of Rf

and A C equal to a side of R'.

Draw 5(7.

Then B C will be a side of the square required.

For 130* = AB2 + ~AC2, § 331

(the square on the hypotenuse of a rt. A is equivalent to the sum of the squares on the two sides).

Construct the square S, having each of its sides equal to BC.

Substitute for BC2, J~B* and jR?, S, R, and Rl re- spectively ;

then 8 = R + R'.

.*. # is the square required.

Q. E. F.

CONSTRUCTIONS.

195

Proposition XVII Problem.

347. To construct a square equivalent to the difference of two given squares.

\ * I

y<rx

Let R be the smaller square and R' the larger. It is required to construct a square = R! — R. Construct the rt. Z. A.

Take A B equal to a side of R.

From B as a centre, with a radius equal to a side of Rf,

describe an arc cutting the line A X at C.

Then A C will be a side of the square required.

For draw.SC.

A-B* + A-C2 = B7?, §331

{the sum, of the squares on the two sides of a rt. A is equivalent to the square on the hypotenuse).

By transposing, AC1 = BC2 — AT?.

Construct the square S, having each of its sides equal to A C.

Substitute for AC*, BC2, and IT?, S, R', and R re- spectively ;

then S = R' — R.

.*. S is the square required.

Q. E. F

196 GEOMETRY. — BOOK IV.

Proposition XVIII. Problem.

348. To construct a square equivalent to the sum of any number of given squares.

	y\
	s \
F/	\ \
/ \	\
/ \	\
	\ \ \ \ \ \
€t<^	\ \ \
	-C:^
A^	j:-^«

Let m, n, o, p, r be sides of the given squares. It is required to construct a square = m? + n2 + o2 + p2 + r3. Take A B = m.

Draw A C = n and _L to A B at A. Draw B C. Draw C E = o and i. to B C at C, and draw B E. Draw EE = ptmd JL to BE at E, and draw BE. Draw EH = r and J_ to B E at F, and draw B H. The square constructed on B H is the square required.

For BE2 = FTP + £~F*,

- FTP + ^T2 + ^g2,

= EH2 + JTF2 + E~G2 + tf^2,

= #7Z2 + EE2 + .tftf2 + cT + l^2, § 331

(the sum of the squares on two sides of a rt. A is equivalent to the square on the hypotenuse).

Substitute for AB, C A, EC, E E, and F H, m, n} o, p, and r respectively;

then BH2 = m2 + n2 + o2 + p2 + r*.

Q. E. F.

CONSTRUCTIONS.

197

Proposition XIX. Problem.

849. To construct a polygon similar to two given similaf polygons and equivalent to their sum.

Let R and R' be two similar polygons, and AB and A' B' two homologous sides.

It is required to construct a similar polygon equivalent to R+ R'.

Construct the rt. Z P.

Take PH= A' £', and P 0 = A B.

Draw 0 H.

Take A" B" = 0 H.

Upon A" B", homologous to A B, construct the polygon R" similar to R.

Then R" is the polygon required.

For R' : R : : A1"^ : ATff, § 343

{similar polygons ore to each other as the squares on their homologous sides).

Also R" : R' : : A" B"6 \ A' B'\

In the first proportion, by composition,

R' + R : R' :: AHB* + Ftf : A1^'2, FH2 + FO2 : FE2, IH)2 : PH\ : AfrW1'1

But

R" : R'

§343 § 264

A'B1

: ED2 : PH2.

\R" : R'

.'. R"

R'+R R' + R.

R'\

q. e. r.

198

GEOMETRY.

BOOK IV.

Proposition XX. Problem.

350. To construct a polygon similar to two given polygons and equivalent to their difference.

/ • I X....

A' B' A B A'f Bii P 0

Let R and R' be two similar polygons, and AB and

A' B' two homologous sides.

It is required to construct a similar polygon which shall be equivalent to R' — R.

Construct the rt. Z P,

and take PO = AB.

From 0 as a centre, with a radius equal to A' B'>

describe an arc cutting P X at H.

Draw 0 H.

Take A" B" = P H.

On A" B", homologous to A B, construct the polygon R" similar to R.

Then R" is the polygon required.

For R' : R : : A7^ : AB*, § 343

(similar polygons are to each other as the squares on their homologous sides).

§343 § 265

Also R" : R : : A" B"' : A B\

In the first proportion, by division,

R : R

But

R" : R

A1^ - AB2 : ITS',

on2 - op* : oy*,

PH2 : UP*.

2 yJT&t

: A" B

: PH2 : U~P. .R".R.:R'-R:R', .'. R" = R1 — R.

Q. E. F.

CONSTRUCTIONS.

199

polygon.

Proposition XXI. Problem. 351. To construct a triangle equivalent to a given

I A E F

Let ABC DH E be the given polygon.

It is required to construct a triangle equivalent to the given polygon.

From D draw D E, and from H draw HF II to D E.

Produce A E to meet HF at F, and draw D F.

The polygon A BG D F has one side less than the polygon ABC D H E, but the two are equivalent.

For the part A B CD E is common,

and the A D EF= A D EH, fox the base D E is common, and their vertices F and H are in the line FH II to the base, § 325 (/& having the same base and equal altitudes are equivalent).

Again, draw C F, and draw D K II to 0 F to meet A F produced at K.

Draw GK.

The polygon ABG K has one side less than the polygon ABG D F, but the two are equivalent.

For the part A B GF is common,

and the A G FK = A GFD, for the base GF is common, and their vertices K and D are in the line KD II to the base. § 325

In like manner we may continue to reduce the number of sides of the polygon until we obtain the A G I K.

Q. E. F.

200 GEOMETRY. — BOOK IV.

Proposition XXII. Problem.

352. To construct a square which shall have a given ratio to a given square.

S

.--"""TV-.. \

m fa.

\ \

^o

n Let R be the given square, and - the given ratio.

m

It is required to construct a square which shall be to R as n is to m.

On a straight line take AB = m, and BC = n.

On A G as a diameter, describe a semicircle.

At B erect the J_ B S, and draw SA and SG.

Then the A A S G is a rt. A with the rt. Z at S, § 204 (being inscribed in a semicircle. )

On S A, oi S A produced, take SE equal to a side of R.

Draw EF li io AG.

Then S F is a side of the square required.

For S-'^'i §289

(£Ae squares on the sides of a rt. A have the same ratio as the segments of the hypotenuse made by the JL let fall from the vertex of the rt. Z).

Also M » :**i § 275

SG SF' *

(a straight line dravm through two sides of a A, parallel to the third side, divides those sides proportionally).

Square the last equality ;

then 12.5*

5T2 ST*

ft J2 <V W2

Substitute, in the first equality, for its equal j

then ^ = ^=-,

S~F2 BG n

that is, the square having a side equal to SF will have the same ratio to the square R, as n has to m.

Q. E. F.

CONSTRUCTIONS. 201

Proposition XXIII. Problem.

853. To construct a polygon similar to a given polygon and having a given ratio to it.

m v /

n

At &

n Let R be the given polygon and - the given ratio.

m

It is required to construct a polygon similar to R, which shall be to R as n is to m.

Find a line, A' B', such that the square constructed upon it shall be to the square constructed upon A B as n is to m. § 352

Upon A' B' as a side homologous to A B, construct the polygon S similar to R.

Then S is the polygon required. S! A7!?'2

(similar polygons are to each other as the squares on their homologous sides).

But ^ _ - J Cons.

ATtf m

.*.— = _, or, S : R '. : n : m. R m

Q. E. F

202 GEOMETRY. BOOK IV.

Proposition XXIV. Problem.

354. To construct a square equivalent to a given paral- lelogram.

P

B C r ] ^"T N

/ \

" i / \

A b D J Ml N g-3C

Let ABCD be a parallelogram, b >.bs base, and a its altitude.

It is required to construct a square = EJ A B C D.

Upon the line MX take M N = a, and N 0 = b.

Upon if 0 as a diameter, describe a semicircle.

At N eiect NP± to MO.

Then the square R, constructed upon a line equal to N P, is equivalent to the O A B G D.

For MN : NP : : NP : NO, § 307

(a A. let fall from any point of a circumference to the diameter is a mean proportional between the segments of the diameter).

.'. NP* = MN X NO = aXb, §259

(the product of the means is equal to the product of the extremes).

Q. E. F.

355. Corollary 1. A square may be constructed equiva- lent to a triangle, by taking for its side a mean proportional between the base and one-half the altitude of the triangle.

356. Cor. 2. A square may be constructed equivalent to any polygon, by first reducing the polygon to an equivalent tri- angle, and then constructing a square equivalent to the triangle.

CONSTRUCTIONS. 203

Proposition XXY. Problem.

357. To construct a parallelogram equivalent to a given square, and having the sum of its base and altitude equal to a given line.

Ss"

It

A ^

if LX J n

Let R be the given square, and let the sum of the base and altitude of the required parallelogram be equal to the given line M N.

It is required to construct a O = R, and having the sum of its base and altitude = M N.

Upon M N as a diameter, describe a semicircle. At M erect a J_ MP, equal to a side of the given square JR. Draw P Q II to M N, cutting the circumference at S. Draw SO± to M N. Any O having G M for its altitude and G N for its base, is equivalent to R.

For tftfisll toPJf, §65

{two straight lines _L to the same straight line are II ).

.\SC = PM, §135

(lis comprehended between lis arc equal).

.'. Slf = PM2 = R.

But MG : SC : : SC : C N, §307

(a _L let fall from any point in a circumference to the diameter is a mean proportional between the segments of the diameter).

Then SGl = MCXGN, §259

(the product of the means is equal to the product of the extremes).

204

GEOMETRY. BOOK IV,

Proposition XXVI. Problem.

859. To construct a parallelogram equivalent to a given square, and having the difference of its base and altitude equal to a given line. S

\C

M

r

r

-/

R>

\

>N

/

B

Let R be the given square, and let the difference of the base and altitude of the required parallelo- gram be equal to the given line M N.

It is required to construct a E3 = R, with the difference of the base and altitude = M N.

Upon the given line M N as a diameter, describe a circle.

From M draw MS, tangent to the O, and equal to a side of the given square R.

Through the centre of the O, draw #2? intersecting the circumference at G and B.

Then any O, as R', having SB for its base and SC for its altitude, is equivalent to R.

For SB : SM : : SM : S C, § 292

(if from a point without a O, a secant and a tangent be drawn, the tangent is a mean proportional between the whole secant and the part without the O).

Then 8~M2 = SBXSC; §259

and the difference between SB and SC is the diameter of the O, that is, M N.

Q. E. F.

CONSTRUCTIONS. 205

Proposition XXVII. Problem.

860. Given x = fa, to construct x.

E

x,

D B

Let m represent the unit of length.

It is required to find a line which shall represent the square root of 2.

On the indefinite line A B, take A C = m, and CD = 2 m.

On A D as a diameter describe a semi-circumference.

At C erect a JL to A B, intersecting the circumference at E.

Then C E is the line required.

For AC : CE : : CE : CD, § 307

{the ± let fall from any point in the circumference to the diameter, is a mean proportional between the segments of the diameter) ;

.-.C~E2 = ACX CD, §259

.\CE=slACX CD, = vfTx~2 = fa .

Q. E. F.

Ex. 1. Given x = y/5, y = y/7, z — 2 fa> ; to construct x, y, and z.

2. Given 2 : x : : x : 3 ; to construct #.

3. Construct a square equivalent to a given hexagon.

206 GEOMETRY. BOOK IV.

Proposition XXVIII. Problem.

361. To construct a polygon similar to a given polygon P, and equivalent to a given polygon Q.

A

\ P' I

A' B'

m A " "b

Let P and Q be two given polygons, and A B a side of polygon P.

It is required to construct a polygon similar to P and equiva- lent to Q.

Find a square equivalent to Pf § 356

and let m be equal to one of its sides.

Find a square equivalent to Q, § 356

and let n be equal to one of its sides.

Find a fourth proportional to m, n, and A B. § 304

Let this fourth proportional be A' B'.

Upon A1 B', homologous to A B, construct the polygon F similar to the given polygon P.

Then P' is the polygon required.

CONSTRUCTIONS. 207

For ™ = — • Cons.

n A'B'

Squaring,		m> AB'
But		P = m?,
and		Q = n*;
	. P " Q	m2 r& P A&

Cons. Cons.

But T, = ^> §343

P> JTff*

(similar polygons are to each other as the squares on their homologous sides) ;

.-.- = — ; Ax. 1

Q P'

.'. P' is equivalent to Q, and is similar to P by construction.

Q. E. F.

Ex. 1. Construct a square equivalent to the sum of three given squares whose sides are respectively 2, 3, and 5.

2. Construct a square equivalent to the difference of two given squares whose sides are respectively 7 and 3.

3. Construct a square equivalent to the sum of a given tri- angle and a given parallelogram.

4. Construct a rectangle having the difference of its base and altitude equal to a given line, and its area equivalent to the sum of a given triangle and a given pentagon.

5. Given a hexagon ; to construct a similar hexagon whose area shall be to that of the given hexagon as 3 to 2.

6. Construct a pentagon similar to a given pentagon and

equivalent to a given trapezoid.

208 GEOMETEY. BOOK IV.

Proposition XXIX. Problem.

862. To construct a polygon similar to a given polygon, and having two and a half times its area.

Y

A/I

B M C 0 N

Let P be the given polygon.

It is required to construct a polygon similar to P, and equivalent to 2 J P.

Let A B be a side of the given polygon P. Then ^T : s/T% : : A B : x,

or \J2 : \Jd : : A B : x, § 345

{the homologous sides of similar polygons are to each other as the square roots of their areas).

Take any convenient unit of length, as MC, and apply it six times to the indefinite line M N.

On M 0 (= 3 M C) describe a semi-circumference ;

and on M N (= 6 M C) describe a semi-circumference.

At C erect a _L to M N, intersecting the semi-circumfer- ences at D and H.

Then CD is the sj% and 0 H is the \/E. § 360

Draw C 7, making any convenient Z with C H. On CY take C E = A B. From D draw D E, and from H draw H Y \\ to D E.

CONSTRUCTIONS.

209

Then C Y will equal x, and be a side of the polygon re- quired, homologous to A B.

For CD : CH : : CE : C Y, §275

(a line drawn through two sides of a A, II to the third side, divides the two sides proportionally).

Substitute their equivalents for CD, C H, and C E ;

then \J2 : V^ : : A B : C Y.

On C Y, homologous to A B, construct a polygon similar to the given polygon P ;

and this is the polygon required.

Q. E. F.

Ex. 1. The perpendicular distance between two parallels is 30, and a line is drawn across them at an angle of 45° ; what is its length between the parallels 1

2. Given an equilateral triangle each of whose sides is 20 ; find the altitude of the triangle, and its area.

3. Given the angle A of a triangle equal to | of a right angle, the angle B equal to J of a right angle, and the side a, opposite the angle A, equal to 10 ; construct the triangle.

4. The two segments of a chord intersected by another chord are 6 and 5, and one segment of the other chord is 3 ; what is the other segment of the latter chord %

5. If a circle be inscribed in a right triangle : show that the difference between the sum of the two sides containing the right angle and the hypotenuse is equal to the diameter of the circle.

6. Construct a parallelogram the area and perimeter of which shall be respectively equal to the area and perimeter of a given triangle.

7. Given the difference between the diagonal and side of a square; construct the square.

BOOK V.

REGULAR POLYGONS AND CIRCLES.

363. Def. A Regular Polygon is a polygon which is equilateral and equiangular.

Proposition I. Theorem.

364. Every equilateral polygon inscribed in a circle is a regular polygon.

Let ABC, etc., be an equilateral polygon inscribed in a circle.

We are to prove the polygon ABC, etc., regular.

The arcs A B, B C, C D, etc., are equal, § 182

(in the same O, equal chords subtend equal arcs).

.'. arcs ABC, BCD, etc., are equal, Ax. 6

.*. the A A, B, C, etc., are equal, (being inscribed in equal segments).

.'. the polygon ABC, etc., is a regular polygon, being equilateral and equiangular.

Q. E. D.

REGULAR POLYGONS AND CIRCLES. 211

Proposition II. Theorem.

365. I. A circle may be circumscribed about a regular polygon.

II. A circle may be inscribed in a regular polygon.

Let A BCD, etc., be a regular polygon. We are to prove that a O mag be circumscribed about this regular polygon, and also a O mag be inscribed in this regular polygon.

Case I. — Describe a circumference passing through A, B, and C.

From the centre 0, draw 0 A, 0 D,

and draw 0 s _L to chord B C.

On 0 s as an axis revolve the quadrilateral 0 ABs,

until it comes into the plane of 0 sC D.

The line s B will fall upon s C, (for ZOsB = ZOsC, both being rt. A ).

The point B will fall upon C, § 183

(since s B — s C).

The line BA will fall upon CD, § 363

(since /. B = Z C, being A of a regular polygon).

The point A will fall upon D, § 363

(since B A = C D, being sides of a regular polygon).

.'. the line 0 A will coincide with line 0 D, (their extremities being the same points).

.'. the circumference will pass through D. In like manner we may prove that the circumference, pass- ing through vertices B, C, and D will also pass through the vertex E, and thus through all the vertices of the polygon in succession.

Case II. — The sides of the regular polygon, being equal chords of

the circumscribed O, are equally distant from the centre, § 1 85

.'.a circle described with the centre 0 and a radius Os

will touch all the sides, and be inscribed in the polygon. § 174

212 GEOMETRY. BOOK V.

366. Def. The Centre of a regular polygon is the common centre 0 of the circumscribed and inscribed circles.

367. Def. The Radius of a regular polygon is the radius 0 A of the circumscribed circle.

368. Def. The Apothem of a regular polygon is the radius 0 s of the inscribed circle.

369. Def. The Angle at the centre is the angle included by the radii drawn to the extremities of any side.

Proposition III. Theorem.

370. Each angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon.

B Let 'ABC, etc., be a regular polygon of n sides.

_ 4 rt. A We are to prove Z. A 0 B — — — •

Circumscribe a O about the polygon.

The AAOB,BOC, etc., are equal, § 180

(in the same O equal arcs subtend equal A at tJie centre).

.'. the Z A 0 B = 4 rt. A divided by the number of A about 0.

But the number of A about 0 = n, the number of sides of the polygon.

4 rt. A

******* nr»

Q. E. D.

371. Corollary. The radius drawn to any vertex of a regular polygon bisects the angle at that vertex.

REGULAR POLYGONS AND CIRCLES. 213

Proposition IV. Theorem.

372. Two regular polygons of the same number of sides are similar.

Let Q and Q' be two regular polygons, each having n sides.

We are to prove Q and Q1 similar polygons.

The sum of the interior A of each polygon is equal to 2 rt. A (n - 2), § 157

(tlie sum of tlie interior A of a polygon is equal to 2 rt. A taken as many times less 2 as the polygon has sides).

Each A of the polygon Q = — ' > § 158

(for the A of a regular polygon are all equal, and hence each Z is equal to the sum of the A divided by their number).

Also, each A of Q' = 2 rt A(n~2) . § 153

n

.'. the two polygons Q and Q' are mutually equiangular.

Moreover, = ^ 5 363

(the sides of a regular polygon are all equal) ;

and ^JL = 1, § 3G3

B'C

...£*«£*, Ax.l

B G B'C

.'. the two polygons have their homologous sides proportional ;

.'. the two polygons are similar. § 278

Q. E. D.

214

GEOMETRY.

BOOK V.

Proposition V. Theorem.

373. The 7wmologous sides of similar regular polygons have the same ratio as the radii of their circumscribed cir- cles, and, also as the radii of their inscribed circles.

Let 0 and 0' be the centres of the two similar regu- lar polygons ABC, etc., and A'B'C, etc.

From 0 and 0' draw 0 E, 0 D, O'E', 0' D', also the Js 0 m and 0' m'.

0 E and 0' E' are radii of the circumscribed (D, § 367

and Om and O'm' are radii of the inscribed (D. § 368

ED OE Om

= O'E "

We are to prove

ED' O'E' O'm'

In the AOED and 0' E> D'

the A 0 E D, 0 D E, 0' E D' and 0' D' E' are equal, § 371 {being halves of the equal A F E D, E D C, F' E> Df and E> D' O) ;

.'. the A 0 ED and 0' E D' are similar, § 280

(if two A have two A of the one equal respectively to two A of the other, they are similar).

ED

OE

Also,

E'D' O'E' (the homologous sides of similar A are proportional).

ED Om

E'D'

O'm'

§278

§297

(the homologous altitudes of similar A have the same ratio as their homolo-

qous bases).

Q. E. D.

REGULAR POLYGONS AND CIRCLES.

215

Proposition VI. Theorem.

374. The perimeters of similar regular polygons have the same ratio as the radii of their circumscribed circles, and. also as the radii of their inscribed circles.

A>^- ~-\B'

Let P and P' represent the perimeters of the two similar regular polygons ABC, etc., and A'B'C, etc. From centres 0, 0' draw 0 E, 0' E', and J§ 0 m and 0' m'.

Om

P OP

We are to prove — =

1 P> 0> E'

ED

O'm'

§ 295 F E' D' *

(the perimeters of similar polygons have the same ratio as any two homolo-

gous

Moreover,

OE ED 0> E' ~~ EH)'9

(the homologous sides of similar regular polygons have the radii of their circumscribed (D).

Also

0

O'm'

ED ~ErD}i

§373 ratio as the

§373

(the homologous sides of similar regulmr polygons have the same ratio as the radii of their inscribed (D).

OE O'E'

Om

Wm'

Q. E. D.

216

GEOMETRY. BOOK V.

Proposition VII. Theorem.

375. The circumferences of circles have the same ratio as their radii.

Let C and C be the circumferences, R and R' the radii of the two circles Q and Q'.

We are to prove G : C : : R : R'.

Inscribe in the (D two regular polygons of the same number of sides.

Conceive the number of the sides of these similar regular polygons to be indefinitely increased, the polygons continuing to be inscribed, and to have the same number of sides.

Then the perimeters will continue to have the same ratio as the radii of their circumscribed circles, § 374

(the perimeters of similar regular polygons have the same ratio as the radii of their circumscribed (D),

and will approach indefinitely to the circumferences as their limits.

.'. the circumferences will have the same ratio as the radii of their circles, § 1 99

.'.C : C0:: R : R'.

Q. E. O

REGULAR POLYGONS AND CIRCLES. 217

376. Corollary. By multiplying by 2, both terms of the ratio R : R', we have

0 ; C : : 2 R i 2 # j

that is, the circumferences of circles are to each other as their diameters.

Since C : C : : 2 R : 2 R',

G : 2 7? : : C : 2 R', § 262

or, = — — .

2 R 2R'

That is, the ratio of the circumference of a circle to its diameter is a constant quantity.

This constant quantity is denoted by the Greek letter ir.

377. Scholium. The ratio tt is incommensurable, and there- fore can be expressed only approximately in figures. The let- ter 77, however, is used to represent its exact value.

Ex. 1. Show that two triangles which have an angle of the one equal to the supplement of the angle of the other are to each other as the products of the sides including the supplementary angles.

2. Show, geometrically, that the square described upon the sum of two straight lines is equivalent to the sum of the squares described upon the two lines plus twice their rectangle.

3. Show, geometrically, that the square described upon the difference of two straight lines is equivalent to the sum of the squares described upon the two lines minus twice their rectangle.

4. Show, geometrically, that the rectangle of the sum and difference of two straight lines is equivalent to the difference of the squares on those lines.

218 GEOMETRY. BOOK V.

Proposition VIII. Theorem.

378. If the number of sides of a regular inscribed poly- gon be increased indefinitely, the apothem, will be an increas- ing variable whose limit is the radius of the circle.

In the right triangle OCA, let 0 A he denoted by R, OC byr, and A G by b.

We are to prove Urn. (r) = R.

r<R, §52

(a _L is the shortest distance from a point to a straight line).

And R-r<b, §97

{one side of a A is greater than the difference of the other two sides).

By increasing the number of sides of the polygon indefi- nitely, A B, that is, 2 b, can be made less than any assigned quantity.

.'. b, the half of 2 b, can be made less than any assigned quantity.

.'. R — r, which is less than b, can be made less than any assigned quantity.

.'.Urn. (R — r) = 0.

.'.R-lim. (r) = 0. § 199

.*. Urn. (r) = R.

Ql. e. d.

REGULAR POLYGONS AND CIRCLES.

219

Proposition IX. Theorem.

379. The area of a regular polygon is equal to one-half the product of its apothem by its perimeter.

B

Let P represent the perimeter and R the apothem of the regular polygon ABC, etc.

We are to prove the area of A BG, etc., = J R X P.

Draw OA, OB, OC, etc.

The polygon is divided into as many A as it has sides.

The apothem is the common altitude of these A,

and the area of each A is equal to J R multiplied by the base. § 324

.*. the area of all the A is equal to J R multiplied by the sum of all the bases.

But the sum of the areas of all the A is equal to the area of the polygon,

and the sum of all the bases of the A is equal to the perimeter of the polygon.

.". the area of the polygon = J R X P.

Q. E. D.

220 GEOMETRY. BOOK V.

Proposition X. Theorem.

380. The area of a circle is equal to one-half the product of its radius by its circumference.

Let R represent the radius, and 0 the circumference of a circle.

We are to prove the area of the circle = J R X C.

Inscribe any regular polygon, and denote its perimeter by P, and its apothem by r.

Then the area of this polygon = J r X P, §379

(the area of a regular polygon is equal to one-hulf the product of its apothem by the perimeter).

Conceive the number of sides of this polygon to be indefi- nitely increased, the polygon still continuing to be regular and inscribed.

Then the perimeter of the polygon approaches the circum- ference of the circle as its limit,

the apothem, the radius as its limit, § 378

and the area of the polygon approaches the O as its limit.

But the area of the polygon continues to be equal to one- half the product of the apothem by the perimeter, however great the number of sides of the polygon.

.'. the area of the O = J R X 0. § 199

Q. E. D.

I

REGULAR POLYGONS AND CIRCLES. 221

c

381. Corollary 1. Since — as »« $ 376

2 R *

.-. C=2ttE. In the equality, the area of the O = \ R X C, substitute 2 ir R for C ; then the area of the O = \ R X 2 tt i?,

That is, ?Ae area o/ a O = tt tames ?Ae sgware ora tfc radius.

382. Cor. 2. 7%e area o/ a sector eawafe | <Ae product of its radius by its arc ; for the sector is such part of the circle as its arc is of the circumference.

383. Def. In different circles similar arcs, similar sectors, and similar segments, are such as correspond to equal angles at the centre.

Proposition XI. Theorem.

384. Two circles are to each other as the squares on their radii.

Let R and R' be the radii of the two circles Q and Q'.

We are to prove — = — . * Q' R'2

Now Q = ttR2, §381

(the area of aO = ir times the square on its radius),

and Q, = 7rR/2. §381

Q x /.'■-' _ m^

Q. E. D.

Then « *

385. Corollary. Similar arcs, being like parts of their re- spective circumferences, are to each other as their radii ; similar sectors, being like parts of their respective circles, are to each other as the squares on their radii.

222

GEOMETRY. BOOK V.

Proposition XII. Theorem.

386. Similar segments are to each other as the squares

on their radii. C

O

P' p

Let A C and A' C be the radii of the two similar seg- ments ABP and A' B' P'.

w * AB? AC1

We are to prove

A'B'P' j[rQfl

The sectors A CB and A' C B> are similar, § 383

{having the A at the centre, C and O, equal).

In the AACB and A' C B'

£C = /.C, § 383

{being corresponding A of similar sectors).

AC=CB, § 163

A'C' = C'B'; §163

.-. the A A C B and A' C B' are similar, § 284

{having an 4- of the one equal to an Z. of the other, and the including sides 'proportional).

Now sector ACB _ AC2 § 385

sector A'C'B' jrrjt {similar sectors are to each other as the squares on their radii) ;

and AACB =A°L, §342

A A'C'B'

A'C

{similar A are to each other as the squares on their homologous sides).

-rr sector ACB- A ACB AC2

Hence = - ,

sector A' C B' - A A' C B' j7!?2

or, segment A BP = IV2 . j 271

segment A' B' P' AJU'2 {if two quantities be increased or diminished by like parts of each, the results will be in the same ratio as the quantities themselves).

Q. E. D.

EXERCISES. 223

Exercises.

1. Show that an equilateral polygon circumscribed about a circle is regular if the number of its sides be odd.

2. Show that an equiangular polygon inscribed in a circle is regular if the number of its sides be odd.

3. Show that any equiangular polygon circumscribed about a circle is regular.

4. Show that the side of a circumscribed equilateral triangle is double the side of an inscribed equilateral triangle.

5. Show that the area of a regular inscribed hexagon is three-fourths of that of the regular circumscribed hexagon.

6. Show that the area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and cir- cumscribed equilateral triangles.

7. Show that the area of a regular inscribed octagon is equal to that of a rectangle whose adjacent sides are equal to the sides of the inscribed and circumscribed squares.

8. Show that the area of a regular inscribed dodecagon is equal to three times the square on the radius.

9. Given the diameter of a circle 50 ; find the area of the circle. Also, find the area of a sector of 80° of this circle.

10. Three equal circles touch each other externally and thus inclose one acre of ground ; find the radius in rods of each of these circles.

11. Show that in two circles of different radii, angles at the centres subtended by arcs of equal length are to each other in- versely as the radii.

12. Show that the square on the side of a regular inscribed pentagon, minus the square on the side of a regular inscribed decagon, is equal to the square on the radius.

224 GEOMETRY. BOOK V.

On Constructions.

Proposition XIII. Problem.

387. To inscribe a regular polygon of any number of sides in a given circle.

Let Q be the given circle, and n the number of sides of the polygon.

It is required to inscribe in Q, a regular polygon having n sides.

Divide the circumference of the O into n equal arcs.

Join the extremities of these arcs.

Then we have the polygon required.

For the polygon is equilateral, § 181

{in the same O equal aros are subtended by equal chords) ;

and the polygon is also regular, § 364

{an equilateral polygon inscribed in a O is regular).

Q. E. F

CONSTRUCTIONS.

225

Proposition XIV. Problem.

388. To inscribe in a given circle a regular polygon which has double the number of sides of a given inscribed regular polygon.

Let ABCD be the given inscribed polygon.

It is required to inscribe a regular polygon having double the number of sides of ABC D.

Bisect the arcs A B, BC, etc.

Draw AH, E B, B F, etc.,

The polygon AE B FC, etc., is the polygon required.

For the chords AB, BC, etc., are equal, § 363

(being sides of a regular polygon).

.*. the arcs AB, BC, etc., are equal, § 182

(in the same O equal chords subtend equal arcs).

Hence the halves of these arcs are equal,

or, AE, EB, B F, FC, etc., are equal j

.'. the polygon A EB F, etc., is equilateral.

The polygon is also regular, § 364

(an equilateral polygon inscribed in a O is regular) ;

and has double the number of sides of the given regular polygon.

Q. E. F.

226

GEOMETRY. BOOK V.

Proposition XV. Problem. 389. To inscribe a square in a given circle.

Let 0 be the centre of the given circle.

It is required to inscribe a square in the circle.

Draw the two diameters A C and B D _L to each other.

Join AB, BC, CD, and DA.

Then A B C D is the square required.

For, the A ABC, BCD, etc., are rt. A, § 204

(being inscribed in a semicircle) ,

and the sides A B, B C, etc., are equal, § 181

(in the same O equal arcs are subtended by equal chords) ;

.*. the figure A B CD is a square, (having its sides equal and its A rt. A ).

§ 127

Q. E. F.

390. Corollary. By bisecting the arcs AB, BC, etc., a regular polygon of 8 sides may he inscribed ; and, by continuing the process, regular polygons of 16, 32, 64, etc., sides may be inscribed.

CONSTRUCTIONS. 227

Proposition XVI. Problem. 391. To inscribe in a given circle a regular hexagon.

/ / V

/ / \

/ / \

/j

Let 0 be the centre of the given circle.

It is required to inscribe in the given O a regular hexagon.

From 0 draw any radius, as 0 0.

From G as a centre, with a radius equal to 0 G,

describe an arc intersecting the circumference at F.

Draw Oi^and C F.

Then G F is a side of the regular hexagon required.

For the A 0 F C is equilateral, Cons.

and equiangular, § 112

.'. the Z FO 0 is J of 2 rt. A, or, J of 4 rt. A. § 98

.*. the arc F C is \ of the circumference ABC F,

.''. the chord FC, which subtends the arc FG, is a side of a regular hexagon ;

and the figure G FD, etc., formed by applying the radius

six times as a chord, is the hexagon required.

Q. E. F.

392. Corollary 1. By joining the alternate vertices A, C, B, an equilateral A is inscribed in a circle.

393. Cor. 2. By bisecting the arcs AB, B C, etc., a regu- lar polygon of 12 sides may be inscribed in a circle; and, by continuing the process, regular polygons of 24, 48, etc., sides may be inscribed.

228 GEOMETRY. — BOOK V.

Proposition XYII. Problem. 394. To inscribe in a given circle a regular decagon.

Let 0 be the centre of the given circle.

It is required to inscribe in the given O a regular decagon.

Draw the radius 0 G, and divide it in extreme and mean ratio, so that 0 0 shall be to 0 S as 0 S is to SG. §311

From G as a centre, with a radius equal to 0 S, * describe an arc intersecting the circumference at B. Drawee, £#, and BO. Then B G is a side of the regular decagon required. For OG : OS : : OS : SC, Cons.

and • BG=OS. . Cons.

Substitute for 0 S its equal B G, then OG : BG :: BG : S G.

Moreover the Z 0 G B = Z S G B, Iden.

.*. the A OGB and B GS are similar, § 284

(having an Z of the, one equal to an Z. of the other, and the including sides proportional).

But the A 0 GB is isosceles, § 160

(its sides 0 C and 0 B being radii of the same circle).

.•.the A B G S, which is similar to the A 0 GB, is isosceles,

CONSTRUCTIONS. 229

and BS = BC. §114

But OJS = BC, Cons.

.\OS = BS, Ax. 1 .*. the A S 0 B is isosceles,

and theZ 0 = ZSBOy §112

(being opposite equal sides).

But the Z C S B = Z 0 + Z SB 0, § 105

(the exterior A of a A is equal to the sum of the two opposite interior A ).

.\theZ CSB = 2Z 0. ZSCB(=Z CSB) = 2Z 0, §112

and Z OBC (== Z SCB) = 2 Z 0. §112

.'. the sum of the A of the A 0 0 B = 5 Z 0.

.\5 Z 0 = 2rt. A, §98

and Z 0 = £ of 2 rt. A, or ^ of 4 rt. A

.'. the arc B C is ^ of the circumference, and .'. the chord B G is a side of a regular inscribed decagon.

Hence, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater segment ten times as a chord.

Q. E. F.

395. Corollary 1. By joining the alternate vertices of a regular inscribed decagon, a regular pentagon may be inscribed.

396. Cor 2. By bisecting the arcs BC, OF, etc., a regular polygon of 20 sides may be inscribed, and, by continuing the process, regular polygons of 40, 80, etc., sides may be inscribed.

230 GEOMETRY. BOOK V.

Proposition XVIII. Problem.

397. To inscribe in a given circle a regular pentedecagon, or polygon of fifteen sides.

F Let Q be the given circle. It is required to inscribe in Q a regular pentedecagon. Draw EH equal to a side of a regular inscribed hexagon, § 391 and E F equal to a side of a regular inscribed decagon. § 394

Join FH. Then FH will be a side of a regular inscribed pentedecagon. For the arc E H is £ of the circumference,

and the arc E F is ^ of the circumference ; .'. the arc FH is £ — ^ or -fa, of the circumference.

.'. the chord FH is a side of a regular inscribed pente- decagon,

and by applying FH fifteen times as a chord, we have the polygon required.

Q. E. F.

398. Corollary. By bisecting the arcs FH, HA, etc., a regular polygon of 30 sides may be inscribed ; and by con- tinuing the process, regular polygons of 60, 120, etc. sides may be inscribed.

CONSTRUCTIONS. 231

Proposition XIX. Problem.

399. To inscribe in a given circle a regular polygon similar to a given regular polygon.

cLr \#' C D

Let A BCD, etc., be the given regular polygon, and C D' E the given circle.

It is required to inscribe in C D' E1 a regular polygon similar to A B G D, etc.

From 0, the centre of the polygon ABC D, etc.

draw 02) and 0 C.

From 0' the centre of the O C D' ®,

draw O1 C and O1 D',

making the Z 0' = Z 0.

Draw CD'.

Then C D' will be a side of the regular polygon required.

For each polygon will have as many sides as the Z. 0 (=Z 0') is contained times in 4 rt. A.

.'. the potygon C D' E', etc. is similar to the polygon CDE, etc., §372

{two regular polygons of the same number of sides are similar).

Q. E. F.

232

GEOMETRY. BOOK V.

Proposition XX. Problem.

400. To circumscribe about a circle a regular polygon similar to a given inscribed regular polygon.

BMC

Let HMRSy etc., be a given inscribed regular polygon.

It is reqtrired to circumscribe a regular polygon similar to HMRS, etc.

At the vertices H, M, R, etc., draw tangents to the O, intersecting each other at A, B, C, etc.

Then the polygon ABC D, etc. will be the regular poly- gon required.

Since the polygon A BC D, etc.

has the same number of sides as the polygon If MRS, etc.,

•it is only necessary to prove that ABC D, etc. is a regular polygon. § 372

In the A BHMfmd C M R,

HM=MR,

(being sides of a regular polygon).

§ 363

CONSTBUCTIONS. 233

the A BHM, BMH, C M R, and C It Mane equal, § 209 (being measured by halves of equal arcs) ;

.-. the A BHM and CM R are equal, § 107

(having a side and two adjacent A of the one equal respectively to a side and two adjacent A of the other).

.'.ZB = ZC,

(being homologous A of equal A ).

In like manner we may prove Z C = Z D, etc

.'. the polygon ABC D, etc., is equiangular.

Since the A BHM, C MR, etc. are isosceles, § 241 (two tangents drawn from the same point to aO are equal),

the sides B H, B M, CM, C R, etc. are equal, (being homologous sides of equal isosceles & ).

.'.the sides AB, BC, C D, etc. are equal, Ax. 6

and the polygon ABC D, etc. is equilateral.

Therefore the circumscribed polygon is regular and similar to the given inscribed polygon. § 372

Q.E F.

Ex. Let R denote the radius of a regular inscribed polygon, r the apothem, a one side, A one angle, and C the angle at the centre ; show that

1. In a regular inscribed triangle a = R ^3, r = \ R, A =00°, C= 120°.

2. In an inscribed square a = R \f2, r = £ R V2, A = 90°, C = 90°.

3. In a regular inscribed hexagon a = R, r = J R tfS, ^ = 120°, (7 = 60°.

R (V^ - 1)

4. In a regular inscribed decagon a = ^ >

r = J R VlO + 2 y/5, ^ = 144°, (7 = 36°.

234

GEOMETRY.

BOOK V.

Proposition XXI. Problem.

401. To find the value of the chord of one-half an arc,

in terms of the chord of the whole arc and the radius of the

circle.

D

Let AB be the chord of arc A B and A D the chord of one- half the arc A B.

It is required to find the value of A D in terms of A B and R {radius).

From D draw D H through the centre 0,

and draw 0 A.

HI) is A. to the chord A B at its middle point C, § 60

(two points, 0 and D, equally distant from the extremities, A and B, de- termine the position of a A. to the middle point of A B).

The Z HAD is a rt. Z, § 204

(being inscribed in a semicircle),

.\A~ff = DHX DC, §289

(the square on one side of a rt. A is equal to the product of the hypotenuse by the adjacent segment made by the ± let fall from the vertex of the rt. Z ).

Now DH=2R,

and J)C = DG-CO = B-CO;

.\A~D2 = 2R(R-CO).

CONSTRUCTIONS. 235

Since A C 0 is a rt. A,

AO* = At? + CD2 ; §331

.-. CO2 = AO* - AC1.

00 = ^(AO2-AG2),

= SlR>-{\ABf, ^S/lP-lJ-B*,

» 4

= V4 R? - .Q2. 2

In the equation if2? = 2R (R — CO),

substitute for C 0 its value ^4 R2 ~ A °

then jTD2 = 2r(r-^EII),

= 2 7?2 - R N± R2 - AB2\ .

.'.AD = JiR?-rN±R?- AB2\ .

Q. E. F.

402. Corollary. If we take the radius equal to unity,

the equation A D = J2R2 - R N±R?- IB2\ becomes AD = ^2-^i-ABi.

236

GEOMETRY.

BOOK V.

Proposition XXII. Problem.

403. To compute the ratio of the circumference of a circle to its diameter, approximately.

Since

§376

Let C be the circumference and R the radius of a circle.

. G

C when B = 1, ir = «- •

It is required to find the numerical value of ir.

We make the following computations by the use of the formula obtained in the last proposition,

A Z> = i/2 — V4 - A B2,

No. Sides.

12

24

48

96

192

384

768

when A B is a side of a regular hexagon : In a polygon of

Form of Computation.

AD — VI

Length of Side.

V^-P .51763809

AD = ^-\J±- (.51 763809)2 .26105238

AD = \J2-sJl- (.26105238)2 .13080626

A D = sj'2 - y/4 - (. 1 3080626)2 .06543817

A D = \J2 - y/4 - (.06543817)2 .03272346

A D = V^2 — y/4 — (.03272346)2 .01636228

AD = \j2-\l^- (.01636228)2 .00818121

Perimeter.

6.21165708 6.26525722 6.27870041 6.28206396 6.28290510 6.28311544 6.28316941

Hence we may consider 6.28317 as approximately the cir- cumference of a O whose radius is unity.

. , , t C 6.28317 . . 7r, which equals — , = .

2 2

tt= 3.14159 nearly.

Q. E. F

ISOPERIMETRICAL POLYGONS.

237

On Isoperimetrical Polygons. — Supplementary.

404. Def. Isoperimetrical figures are figures which have equal perimeters.

405. Def. Among maguitudes of the same kind, that which is greatest is a Maximum, and that which is smallest is a Minimum.

Thus the diameter of a circle is the maximum among all inscribed straight lines; and a perpendicular is the minimum among all straight lines drawn from a point to a given straight line.

Proposition XXIIL Theorem.

406. Of all triangles having two sides respectively equal, that in which these sides include a right angle is the maxi-

Let the triangles ABC and EBC have the sides AB and BG equal respectively to EB and BC; and let the angle ABC be a right angle.

We are to prove A ABO A EBC.

From E, let faU the ± ED.

The A ABC and EBC, having the same base B C, are to each other as their altitudes A B and ED, § 326

(& having the same base are to each other as their altitudes).

Now E D is <EB, §52

(a _L is the shortest distance from a point to a straight line).

But EB = AB, Hyp.

..EDis<AB.

A ABO A EBC.

Q. E. D.

238 GEOMETRY. BOOK V.

Proposition XXIV. Theorem.

407. Of all polygons formed of sides all given but one, the polygon inscribed in a semicircle, having the undetermined side for its diameter, is the maximum.

C

Let AB, BO, 0 D, and DE be the sides of a polygon inscribed in a semicircle having A E for its di- ameter.

We are to prove the polygon ABODE the maximum of the sides A B, B 0, 0 D, and D E.

From any vertex, as 0, draw 0 A and C E.

Then the A A OE is a rt. Z. , § 204

(being inscribed in a semicircle).

Now the polygon is divided into three parts, ABO, ODE, and A OE.

The parts ABO and ODE will remain the same, if the Z A 0 E be increased or diminished ;

but the part AGE will be diminished, § 406

{of all &. having two sides respectively equal, that in which these sides in- clude art. A is the maximum).

.*. A B 0 D E is the maximum polygon.

Q. E. D.

ISOPERIMETRICAL POLYGONS.

239

Proposition XXV. Theorem.

408. The maximum of all polygons formed of given sides can be inscribed in a circle.

W D

A A'

Let ABODE be a polygon inscribed in a circle, and A' B' C D' E' be a polygon, equilateral with re- spect to ABODE, but which cannot be inscribed in a circle.

We are to prove

the polygon A B 0 D E > the polygon A' B' C D' E'.

Draw the diameter A H.

Join OH and D H.

Upon C D> (= 0 D) construct the A C H' D1 = A C HD,

and draw A' W.

Now the polygon A B 0 H > the polygon A' B' C H', § 407

(of all polygons formed of sides all given but one, the polygon inscribed in a semicircle having the nndetermhied side for its diameter, is the maximum).

And the polygon A E D H > the polygon A' E' D' H'. § 407

Add these two inequalities, then

the polygon A BO HDE> the polygon A' B' C'H'D'E'.

Take away from the two figures the equal A 0 H D and C'H'D'.

Then the polygon A B 0 D E > the polygon A' B' C D' E'.

Q. E. O,

240

GEOMETRY. BOOK V.

Proposition XXVI. Theorem.

409. Of all triangles having the same base and equal perimeters, the isosceles triangle is the maximum.

^H

Let the AACB and ABB have equal perimeters, and let the A AC B be isosceles.

We are to prove AAOB>AADB.

Draw the Js CE and B F.

A A OB

CE BF

A ABB

(A having the same base are to each other as their altitudes).

Produce AC to ff, making C ff= AC. Draw H B.

§ 326

The Z A B H is a rt. Z, for it will be inscribed in the semicircle drawn from C as a centre, with the radius C B<

ISOPERIMETEICAL POLYGONS. 241

From C let fall the X C K ; and from D as a centre, with a radius equal to D Bf describe an arc cutting H B produced, at P. Draw DP and A P, and let fall the ± D M. Since AH = AC+CB = AD + DB, and AP<AD+ DP;

.:AP<AD + DB; .\AH>AP. .\BH>BP. §56

Now BK=%BH, §113

(a _L drawn from the vertex of an isosceles A bisects the base),

and BM=iBP. §113

But CE = BK, §135

(lb comprehended between lis are equal);

and DF=BM, §136

.-. CE> DF. .\AACB>AADB.

Q. E. D.

242

GEOMETRY. — BOOK V.

Proposition XXVII. Theorem.

410. The maximum of isoperimetrical polygons of the same number of sides is equilateral.

Let ABC D, etc., be the maximum of isoperimetrical polygons of any given number of sides.

We are to prove AB, BC, CD, etc., equal. Draw A G.

The A AB C must be the maximum of all the A which are formed upon A G with a perimeter equal to that of A ABG.

Otherwise, a greater A A KG could be substituted for A A B G, without changing the perimeter of the polygon.

But this is inconsistent with the hypothesis that the poly- gon ABC D, etc., is the maximum polygon.

.*. the A A B G, is isosceles, § 409

(of all & having the same base and equal perimeters, the isosceles A is the maximum).

In like manner it may be proved that B C = CD, etc.

Q. E. D.

411. Corollary. The maximum of isoperimetrical poly- gons of the same number of sides is a regular polygon.

For, it is equilateral, § 410

(the maximum of isoperimetrical polygons of the same number of sides is equilateral).

Also it can be inscribed in a O, § 408

(the maximum of all polygons formed of given sides can be inscribed in a O).

Hence it is regular, § 364

(an equilateral polygon inscribed in a O is regular).

ISOPERIMETRICAL POLYGONS.

243

Proposition XXVIII. Theorem.

412. Of isoperimetrica I regular polygons, tha t is greates t which has the greatest number of sides.

Let Q be a regular polygon of three sides, and Q' be a regular polygon of four sides, each having the same perimeter.

We are to prove Q' > Q.

In any side A B of Q, take any point D.

The polygon Q may be considered an irregular polygon of four sides, in which the sides A D and D B make with each other an Z equal to two rt. A .

Then the irregular polygon Q, of four sides is less than the

regular isoperi metrical polygon Q' of four sides, § 411

(the maximum of isoperimetrical polygons of the same number of sides is a regular polygon).

In like manner it may be shown that Qf is less than a regular isoperimetrical polygon of five sides, and so on.

Q. E. D.

413. Corollary. Of all isoperimetrical plane figures the circle is the maximum.

244

GEOMETRY. — BOOK V.

Proposition XXIX. Theorem.

414. If a regular polygon be constructed with a given area, its perimeter will be the less the greater the number of its sides.

Let Q and Q' be regular polygons having the same area, and let Q' have the greater number of sides.

We are to prove the perimeter of Q > the perimeter of Ql.

Let Q" be a regular polygon having the same perimeter as Q', and the same number of sides as Q.

Then Q is > Q", § 412

{of isoperimetrical regular polygons, that is the greatest which has the greatest number of sides).

But Q = Q',

.-.Qis> Q". .\ the perimeter of Q is > the perimeter of Q". But the perimeter of Q' = the perimeter of Q", Cons. .*. the perimeter of Q is > that of Q'.

Q. E. D.

415. Corollary. The circumference of a circle is less than the perimeter of any other plane figure of equal area.

SYMMETRY.

245

On Symmetry. — Supplementary.

416. Two points are Symmetrical when they are situated on opposite sides of, and at equal distances from, a fixed point, line, or plane, taken as an object of reference.

417. When a point is taken as an object of reference, it is called the Centre of Symmetry ; when a line is taken, it is called the Axis of Symmetry ; when a plane is taken, it is called the Plane of Symmetry.

418. Two points are symmetrical with re- spect to a centre, if the centre bisect the straight line terminated by these points. Thus, P, P' are symmetrical with respect to C, if C bisect the straight line PP.

419. The distance of either of the two symmetrical points from the centre of symmetry is called the Radius of Symmetry. Thus either C P or C P' is the radius of symmetry.

420. Two points are symmetrical with P respect to an axis, if the axis bisect at right

angles the straight line terminated by these X X?

points. Thus, P, P' are symmetrical with re- spect to the axis XX', if XX' bisect P P' at p, right angles.

421. Two points are symmetrical with respect to a plane, if the plane bisect at right angles the straight line terminated by these points. Thus P, P' are symmetrical with respect to M N, if M N bisect P P' at right angles.

W

P>

N

246

GEOMETRY. BOOK V.

422. Two plane figures are symmetrical with respect to a centre, an axis, or a plane, if every point of either figure have its corresponding symmetrical point in the other.

A

A'

Fig. 2.

Fig. 3.

Thus, the lines A B and A' B' are symmetrical with respect to the centre G (Fig. 1), to the axis XX (Fig. 2), to the plane M N (Fig. 3), if every point of either have its corresponding symmetrical point in the other.

	\ '		Z)
		\
*'■/	/!	\	1
Ml—		I	1
	V	j /b\

N

A' D'

Fig. 6.

Also, the triangles ABB and A1 B' D' are symmetrical with respect to the centre C (Fig. 4), to the axis XX' (Fig. 5), to the plane MN (Fig. 6), if every point in the perimeter of either have its corresponding symmetrical point in the perimeter of the other.

423. Def. In two symmetrical figures the corresponding symmetrical points and lines are called homologous.

SYMMETRY.

247

Two symmetrical figures with respect to a centre can be brought into coincidence by revolving one of them in its own plane about the centre, every radius of symmetry revolving through two right angles at the same time.

Two symmetrical figures with respect to an axis can be brought into coincidence by the revolution of either about the axis until it comes into the plane of the other.

424. Def. A single figure is a symmetrical figure, either when it can be divided by an axis, or plane, into two figures symmetrical with respect to that axis or plane ; or, when it has a centre such that every straight line drawn through it cuts the perimeter of the figure in two points which are symmetrical with respect to that centre.

Fig. 1.

Fig. 2.

Thus, Fig. 1 is a symmetrical figure with respect to the axis XX't if divided by XX' into figures ABC D and AB'C'D which are symmetrical with respect to XX'.

And, Fig. 2 is a symmetrical figure with respect to the centre 0, if the centre 0 bisect every straight line drawn through it and terminated by the perimeter.

Every such straight line is called a diameter.

The circle is an illustration of a single figure symmetrical with respect to its centre as the centre of symmetry, or to any diameter as the axis of symmetry.

248 GEOMETRY. — BOOK V.

Proposition XXX. Theorem.

425. Two equal and parallel lines are symmetrical with respect to a centre.

A B>

B A'

Let A B and A1 B' be equal and parallel lines. We are to prove A B and A' B' symmetrical. Draw A A' and B &t and through the point of their inter- section G, draw any other line EG H', terminated in A B and A'B'.

In the A GABzxAGA'B'

AB = A'B', Hyp.

also, A A and B = A A' and B' respectively, § 68

(being alt. -int. A ),

.'.AGAB = A GA'B'; § 107

.'. GA and G B = G A' and G B' respectively, (being homologous sides of equal &).

Now in the A A G R and A' C H'

AC = A'G,

A A and AC H — A A' and A' G II1 respectively,

. .'.AAGH^AA'GH', § 107

(having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

.'.GH^CH',

(being homologous sides of equal A ).

.*• EP is the symmetrical point of H.

But H is any point in A B ;

.'. every point in A B has its symmetrical point in A' B1.

.'. A B and A! B' are symmetrical with respect to G as a centre of symmetry.

Q. E. D.

426. Corollary. If the extremities of one line be re- spectively the symmetricals of another line with respect to the same centre, the two lines are symmetrical with respect to that centre.

SYMMETRY.

249

Proposition XXXI. Theorem.

427. If a figure be symmetrical with respect to two axes 'perpendicular to each other, it is symmetrical with respect to their intersection as a centre.

Let the figure ABODE FGH be symmetrical to the

two axes XX', YY' which intersect at 0.

We are to prove 0 the centre of symmetry of the figure.

Let / be any point in the perimeter of the figure.

Draw IKL ± to XX', and I M N J_ to YY.

JoinZO, ON, and KM.

Now KI=KL,

(tJie figure being symmetrical with respect to X XI).

But KI=OM,

(lis comprehended between lis are egual).

,'.KL = OM.

.\ K LO M is a O,

{having two sides equal and parallel).

.'. LO is equal and parallel to KM, {being opposite sides of a O).

In like manner we may prove 0 N equal and parallel to KM. Hence the points L, 0, and N are in the same straight line drawn through the point 0 11 to KM.

Also L0 = 0N,

(since each is equal to KM). *, any straight line LO N, drawn through 0, is bisected at 0. .'. 0 is the centre of symmetry of the figure. § 424

Q. E. D.

§420

§ 135

Ax. 1 § 136

§ 134

250

GEOMETRY. — BOOK V.

Exercises. 1. The area of any triangle may be found as follows : From half the sum of the three sides subtract each side severally, mul- tiply together the half sum and the three remainders, and extract the square root of the product.

Denote the sides of the tri- angle A B G by a, b, c, the alti- a+ 6 + c

tude by p, and

by s.

Show that

,2 =

+ c2-2cXAD,

and show that P2 = b2-

(b2+<?-a2y 4 c2

p =

p =

V/462c2-(62+c2-a2)2

2c

\/ (b + c+ a) (b + c— a) (a + b - c) (a— b + c)

2c

Hence, show that area of A A B C, which is equal to

cXp

= i^ (b + c + a) (b + c-a) (a + b-c) (a-b + c), = \/ s(s — a)(s — b)(s — c).

2. Show that the area of an equilateral triangle, each side of which is denoted by a, is equal to — j- .

3. How many acres are contained in a triangle whose sides are respectively 60, 70, and 80 chains 1

4. How many feet are contained in a triangle each side of which is 75 feet 1

BOOK VI.

PLANES AND SOLID ANGLES.

On Lines and Planes.

428. Def. A Plane has already been denned as a surface such that the straight line joining any two points in it lies wholly in the surface.

The plane is considered to be indefinite in extent, so that however far the straight line be produced, all its points lie in the plane. A plane is usually represented by a quadrilateral supposed to lie in the plane.

429. Def. The Foot of a line is the point in which it meets the plane.

430. Def. A straight line is perpendicular to a plane if it be perpendicular to every straight line of the plane drawn through its foot.

In this case the plane is perpendicular to the line.

431. Def. The Distance from a point to a plane is the perpendicular distance from the point to the plane.

432. Def. A line is parallel to a plane if all its points be equally distant from the plane.

In this case the plane is parallel to the line.

433. Def. A line is oblique to a plane if it be neither per- pendicular nor parallel to the plane.

434. Def. Two planes are parallel if all the points of either be equally distant from the other.

435. Def. The Projection of a point on a plane is the foot of the perpendicular from the point to the plane.

436. Def. The projection of a line on a plane is the locus of the projections of all its points.

437. Def. The plane embracing the perpendiculars which project the points of a straight line upon a plane is called the projecting plane of the line.

J

252 GEOMETRY. — BOOK VI.

438. Def. The angle which a line makes with a plane is the angle which it makes with its projection on the plane.

This angle is called the Inclination of the line to the plane.

439. Def. A plane is determined by lines or points, if no other plane can embrace these lines or points without being coincident with that plane.

440. -Def. The intersection of two planes is the locus of all the points common to the two planes.

441. An infinite number of planes may embrace the same straight line.

Thus, if the plane M N em- M_

brace the line AB it may be made to revolve about A B as an axis,

and to occupy an infinite number ^ ^^ /f

of positions, each of which is the / /

position of a plane embracing the line A B.

442. A plane is determined by a straight line and a point without that line.

Thus, let any plane em- bracing the straight line A B revolve about the line as an axis until it embraces the point C.

Now if the plane revolve either way about the line A B as an axis, it will cease to embrace the point G.

Hence any other plane embracing the line A B and the point C must be coincident with the first plaie. § 439

443. Three points not in a straight line determine a p>lane. For, by joining any two of the points , we have a straight

line and a point which determine a plane. § 442

444. Two intersecting straight lines determine a plane.

For, a plane embracing one of these straight lines and any point of the other line (except the point of intersection) is deter- mined. § 442

445. Two parallel straight lines determine a plane.

For, a plane embracing either of these parallels and any point in the other is determined. § 442

LINES AND PLANES.

253

Proposition I. Theorem.

446. If two planes cut one another their intersection is a straight line.

Let MN and PQ be two planes which cut one another.

We are to prove their intersection a straight line.

Let A and B be two points common to the two planes.

Draw the straight lino A B.

Since the points A and B are common to the two planes, the straight line A B lies in both planes. § 428

Now, no point out of this line can be in both planes ;

for, if it be possible, let C be such a point. But there can be but one plane embracing the point 0 and the line A B. § 442

.'. C does not lie in both planes.

.'. every point in the intersection of the two planes lies in the straight line A B.

Q. E. D

254

GEOMETRY. — BOOK VI.

Proposition II. Theorem.

447. From a point without a plane only one perpendic- ular can be drawn to the plane ; and at a given point in a plane only one perpendicular can be erected to the plane.

IV

7

Fig. l.

N

Fig. 2.

Let G D (Fig.l) be a, perpendicular let fall from the point G to the plane MN.

We are to prove that no other J_ can be drawn from the point C to the plane MN

If it be possible, let G B be another _L to the plane MN,

and let a plane P Q pass through the lines G B and 0 D.

The intersection of P Q with the plane MN is a straight line BD. §446

Now if CD and GB be both J_ to the plane, the A GBD would have two rt. A, GBD and G D B, which is impos- sible. § 102

Let D G {Fig. 2) be a perpendicular to the plane MN at the point D.

If it be possible, let D A be another JL to the plane from the point D}

and let a plane P Q pass through the lines D G and DA.

The intersection of P Q with the plane JOris a straight line.

Now if D G and DA could both be J_ to the plane MN at D, we should have in the plane P Q two straight lines _L to the line D Q at the point D, which is impossible. § 61

Q. E. D.

448. Corollary. A perpendicular is the shortest distance from a point to a plane.

LINES AND PLANES.

255

Proposition III. Theorem.

449. If a straight line be perpendicular to each of two straight lines drawn through its foot in a plane it is perpen. dicular to the plane.

Let DC be perpendicular to each of the two lines AC A' and BCB' drawn through its foot in the plane M N.

We are to prove DC A- to the plane MN.

Take CA = C A' and C B = CB'.

JomABttidiA'B'.

Then A B and A' B' are symmetrical with respect to C, § 426 {their extremities being symmetrical). Through C draw any line HC H' in the plane M N.

Then H and H' are symmetrical, § 422

(being corresponding points in the symmetrical lines A B and A1 2?').

About C, the centre of symmetry, revolve A B, keeping A C and B C _L to CD, until it comes into coincidence with A' B'.

Then the point H will coincide with its symmetrical point H\

and Z DCH will coincide with, and be equal to, Z DCH1.

.*. A DCHandi DCW are rt. A. § 25

.'.DC is ± to HCH'.

Now since DC is _L to any line, HCH\ drawn through its foot in the plane MN, it is _L to every such line.

§430. a e. d.

\ DC is i_ to the plane MN.

256

GEOMETRY. BOOK VI.

Proposition IV. Theorem. 450. Oblique lines drawn from a point to a plane at equal distances from the foot of the perpendicular are equal; and of two oblique lines unequally distant from the foot cfthe perpendicular the more remote is the greater.

Let the oblique lines BC, B D, and BE, be drawn at equal distances, AC, AD, and A E, from the foot of the perpendicular BA ; and let BC be drawn more remote from the foot of the perpendicular than BC.

We are to prove I. BC=BD = BE. II. BOBC.

I. In the rt. ABAC and BA D

BA=BA, AC = AD, and rt. A BA C = rt. Z BA D.

.'.ABAC = ABAZ>,

,\BC = BD,

(being homologous sides of equal &).

II. Since A C is > A C,

BC'is>BC,

Iden. Hyp.

§106

§55

Q. E. 6.

451. Cor. 1. Equal oblique lines from a point to a plane meet the plane at equal distances from the foot of the perpendic- ular; and of two unequal oblique lines, the greatermeets the plane at the greater distance from the foot of the perpendicular.

452. Cor. 2. All equal oblique lines BC, B D, etc., drawn from a point to a plane terminate in the circumference CDE described from A as a centre with a radius equal to A C. Hence, to draw a perpendicular from a point to a plane, draw any ob- lique line from the given point to the plane ; revolve this line about the point, tracing the circumference of a circle in the plane, and draw a line from the point to the centre of the circle.

LINES AND PLANES. 257

Proposition Y. Theorem.

453. If three straight lines meet at one pointy and a straight line be perpendicular to each of them at that point, the three straight lines lie in the same plane.

P:" A

Let the straight line A B be perpendicular to each of the straight lines BC, BD, and B E, at B.

We are to prove B C, B D, and BE in the same plan-e M N.

If not, let B D and BE be in the plane M N, and BC with- out it ; and let P H, passing through A B and B C, cut the plane M N in the straight line B H.

Now A B, BC, and B U are all in the plane P H, and since A B is _L to B D and B E, it is _L to the plane M N, § 449

(if a straight line be _L to each of two straight lines drawn through its foot in a plane, it is ± to tlie plane).

.'. A B is _L to B II, a straight line in the plane M N, § 430

(a _L to a plane is 1. to every straight line in that plane drawn through

Us foot).

That is Z ABU is art. Z.

But Z ABC is a rt. Z. Hyp.

r.Z ABC = Z ABU.

.'. BC and B1I coincide.

.*. B C is not without the plane M N.

Q. E. D

454. Corollary. The locus of all perpendiculars to a given straight line at a given point is a plane perpendicular to this given straight line at the given point.

455. Scholium. In the geometry of space the term locus has the same signification as in plane geometry, only it is not limited to lines, but is extended to include surfaces.

258

GEOMETRY. BOOK VI.

Proposition VI. Theorem.

456. If from the foot of a perpendicular to a plane a straight line be drawn at right angles to any line of the plane, the line drawn from its intersection with the line of the plane to any point of the perpendicular is perpendicular to the line of the plane.

Let P F be a perpendicular to the plane M N, FC a perpendicular from the foot of P F to any line AB, in the plane M ' N, and CP a line drawn from its intersection with A B to any point P in the perpendicular P F.

We are to prove G P -L to A B.

Take GA = CB and draw FA, FB, P A, P B.

Now FA = FB, § 53

(two oblique lines drawn from a point in a X cutting off equal distances from the foot of the ± are equal),

and PA= PB, §450

(oblique lines drawn from a point to a plane at equal distances from the foot of the J- are equal).

.'. PG is ±to A B, §60

(two points equally distant from the extremities of a straight line determine the X at the middle point of the line).

Q. E. D

LINES AND PLANES.

259

Proposition VII. Theorem.

457. If a line be perpendicular to a plane, every line which is parallel to this perpendicular is likewise perpendic- ular to the plane.

A C

/J

^\7F 7

ML n

Let AB be perpendicular to the plane M N, and CD

any line parallel to AB.

We are to prove C D perpendicular to the plane M N.

Draw B D in the plane M N, and through D draw E F in the plane M N _L to B D, and join D with any point in A B, as A.

BDis±toAB, §430

(if a straight line be ± to a plane it is ± to every line of tJie plane drawn through its foot) ;

itisalso_LtoCZ), §67

(if a straight line be X to one of two lis, it is A. to the oilier).

Now EF is X to AD, § 456

(if from the foot of a JL to a plane a straight line be drawn at right angles to

any line of the plane, the line drawn from its intersection with the line

of the plane to any point in the ± is ± to the line of the plane),

and is also _L to B D. Cons.

.-. E F is _L to the plane ABDC, § 449

(a straight line X to two straight lines drawn through its foot in a plane is JL to the plane),

.\EFis±to CD, § 430

(if a straight line be J- to a plane it is _L to every line of the plane drawn through its foot).

.'. CD is _L to BD and EF, and consequently to the plane MN. § 449

Q. E. D.

458. Corollary 1. Two lines which are perpendicular to the same plane are parallel.

459. Cor. 2. Two lines parallel to a third straight line not in their own plane are parallel to each other.

260

GEOMETRY. — BOOK VI.

Proposition VIII. Theorem.

460. If a straight line and a plane be perpendicular to the same straight line, they are parallel.

Let the straight line B G and the plane M N be per- pendicular to the straight line A B.

We are to prove B C II to M N.

From any point G of the line BC let G D be drawn per- pendicular to M N.

Join A D.

B A and C D are parallel, § 458

(two straight lines _L to the same plane are II ).

ADisLtoBA, §430

(if a straight line be ± to a plane it is _L to every line of the plane drawn through its foot).

.*. A D and B C are parallel, § 65

(two straight lines JL to the same straight line are II ).

.-. A BCD is aO. §125

.\CD = AB. §134

Now, since G is any point in the line B G, all the points in B G are equally distant from the plane M N.

.'.BG is II to MN, § 432

(a line is II to a plane if all its points be equally distant from the plane).

Q. E. D.

LINES AND PLANES. 261

Proposition IX. Theorem.

461. If two planes be perpendicular to the same straight line they are parallel. P

7

I

Q

In

Let the two planes M N and PQ be perpendicular to the straight line A B.

We are to prove P Q II to M N.

From any point G in the plane P Q draw G D _L to M N.

Join B C.

BO is J_ to A B, §430

{if a straight line be A. to a jylane it is JL to every line of the plane drawn through its foot).

.'. B C is II to the plane MN, § 460

{if a. straight line and a plane be ± to the same straight line they are II ).

.•.CiHsequalto^, § 432

{if a straight line be II to a plane, all its points are equally distant from tlie

plane).

Since G is any point in the plane P Q, all the points in the plane P Q are at equal distances from M N.

.'.PQ is II to MN, § 434

{two x>lanes arc II if all the points of either be equally distant from the other).

Q. E. D.

262

GEOMETRY. — BOOK VI.

Proposition X. Theorem.

462. If two angles not in the same plane have their sides respectively parallel and lying in the same direction, they are equal.

M

Let A A and A' be respectively in the planes M N and P Q and have A D parallel to A' D' and A C parallel to A' C and lying in the same direction.

We are to prove Z. A = Z A'.

Take AD = A' D1 and A C = A' C. Join A A', D D', C C, CD, C D1. Since A D is equal and II to A' D', the figure ADD' A' is a O, § 136

.\AA' = DD'. §134

In like manner AA' = CC',

.'. CCf = DD'. Ax. 1

Also, since 0 C and D D1 are respectively II to A A', they

are II to each other, § 459

(two straight lines II to a third straight line not in their own plane are II to

each other).

.'.CDD'C'iBa.n. § 136

.'.CD = C'D', §134

.\AADC = AA'D'C', §108

(having three sides of the one equal respectively to three sides of the other). .'./.A = ZA',

(being homologous A of equal A).

Q. E. D.

463. Corollary. If two angles lie in different planes and have their sides parallel and extending in the same direction, the planes are parallel. For the intersecting lines, A C and A D, which determine the plane M N are parallel respectively to the lines A' C and A1 D' which determine the plane P Q, therefore the planes are determined parallel.

LINES AND PLANES.

263

Proposition -XI. Theorem. 464. Two parallel lines comprehended between two par- allel planes are equal.

Let the two parallel lines AB and C D be included between the parallel planes M N and P Q.

We are to prove A B = C D.

IfAB and CD be J_ to the two II planes they are equal, § 434 {if two planes be II, all the points of either are equally distant from tlie other).

If A B and G Z) be not JL to the two II plane3, draw from the points A and C the lines A E and C F _L to the plane M N.

A Bis II to CF, §458

(two lilies ± to tlie same plane are II ).

Join BE and D F.

In AAEB&nd C F D,

AE=CF, §434

ZAEB = ZCFD, §430

(if a straight line be ± to a plane it is ± to any line of the plane drawn through its foot) ;

and ZBAE = ZDCF, §462

(if two A not in the same plane have their sides II and lying in the same direction they are equal).

.'.AAEB = A CFD, § 107

{having a side and two adj. A of the one equal respectively to a side and two adj. A of the other).

Hence A B = 0 B,

(being homologous sides of equal ▲ ).

Q. E. D.

264

GEOMETRY. — BOOK VI.

Proposition XII. Theorem.

465. The intersections of two parallel planes by a third plane are parallel lines.

Let the plane 0 S intersect the parallel planes P Q and M N in the lines AC and B D respectively.

We are to prove AC II to B D.

Through the points A and C draw the II lines A B and C D in the plane 0 S.

Now AB = CD,

(II lines comjjrehended between II planes are equal).

.'. ABC D is aO,

(having two sides equal and II ).

.'.AC is II to BD, (being opposite sides of a O ).

§ 464 § 136 §125

Q. E. D.

LINES AND PLANES.

265

Proposition XIII. Theorem.

466. If a straight line be perpendicular to one of two parallel planes it is perpendicular to the other.

Let MN and PQ be parallel planes and A B be per- pendicular to PQ.

We are to prove A B _L to M N.

Let two planes embracing AB intersect the planes M N and P Q in A C, B E and A D, B F respectively.

Then A C is II to B E and A D to B F, §465

(the intersections of two II planes by a third plane are II lines).

But EB and FB are J_ to A B, § 430

(if a straight line be ± to a plane it is ± to every straight line of the plane drawn through its foot).

.-.AC and A D which are respectively II to BE and B F are ±.toAB, § 67

(if a straight line be ± to one of two II lines, it is A. to the other).

.-. A B is _L to MN, § 449

(if a line be _L to two straight lines in a plane drawn through its foot it is _L to the plane).

Q. E. D.

467. Corollary. If two planes be parallel to a third plane they are parallel to each other. For, every line perpendicular to this third plane is perpendicular to the other planes ; and two planes perpendicular to a straight line are parallel.

266 GEOMETRY. BOOK VI.

Proposition XIV. Theorem.

468. If a straight line be parallel to another straight line drawn in a plane, it is parallel to the plane.

7

ML ! I i

< — t r—^F

N Let AG be parallel to the line B I) in the plane M N.

We are to prove AG II to the plane MN.

From A and G, any two points in A G, draw A B and G D ±to£D, and A E and G F _L to the plane M N.

Join BE and D F.

Now ABkHi to GD, § 65

{two straight lines ± to the same line art II ).

Also AB = CD, §135

(II lines comprehended between II lines are equal),

and A E is \\ to GF, §458

(two straight lines ± to the same plane are II ).

.-. Z BAE = Z £>GF, §462

(if two A not in the same plane have tlieir sides II and lying in the same direction, they are equal).

.-. rt. AAEB = it. A GFD, §110

(two rt. A are equal when an acute Z and the hypotenuse of the one are equal respectively to an acute A and the hypotenuse of the other).

r.AE=GF,

(being homologous sides of equal A).

Now since the points A and C, any two points in the line A G, are equally distant from the plane MN,

all the points in A G are equally distant from the plane MN.

.'. A G is II to the plane MN. § 432

Q. E. D.

LINES AND PLANES.

267

Proposition XV. Theorem. 469. If two straight lines be intersected by three par- allel planes their corresponding segments are proportional.

^^^ A	C	/
P 1		N
jy*^ e\		1
r\		Q
/ fj[	\	\J
/ B "/c

Let A B and C D be intersected by the parallel planes MN, PQ,RS, in the points A, E, B, and C, F, D.

AE ^ OF EB~ FD'

We are to prove

Draw A D cutting the plane P Q in G. Join E G and FG.

Then E G is II to B D, §465

{the intersections of two II planes by a third plane are II lines).

• AE. -AG_ " EB ~ GD'

(a line drawn through two sides of a A II to the third side divides those sides proportionally).

§275

Also,

GFis II to AC,
. CF FD ~	A G GD
. AE = ' ' EB	CF FD

§465 §275

Ax. 1

Q. E. D.

268

GEOMETRY. BOOK VI.

On Dihedral Angles.

470. Def. The amount of rotation which one of two inter- secting planes must make about their intersection in order to coincide with the other plane is called the Dihedral angle of the planes.

The Faces of a dihedral angle are the intersecting planes.

The Edge of a dihedral angle is the intersection of its faces.

The Plane angle of a dihedral angle is the plane angle formed by two straight lines, one in each plane, perpendicular to the edge at the same point.

Thus, in the diagram, C-A B-D is a dihedral an- gle, CB and DA are its faces, A B is its edge, 0 PH is its plane angle if OP and HP in the faces be perpendicular to the edge A B at the same point P.

471. The plane angle of a dihedral angle has the same mag- nitude from whatever point in the edge we draw the perpendicu- lars. For every pair of such angles have their sides respectively parallel (§ 65), and hence are equal (§ 462).

Two equal dihedral angles, DA B-C, and D-A B-E', have corresponding equal plane angles, DAG and DAE. This may be shown by superposi- tion.

Any two dihedral angles, C-A B-E' and E-A B-H', have the same ratio as their corre- sponding plane angles, C A E and E A H. This may be shown by the method employed in ^ §200 and §201.

Hence a dihedral angle is measured by its plane angle.

It must be observed that the sides of the plane angle which measures the dihedral angle must be perpendic- ular to the edge. Thus in the rectangular solid A H, Fig. 1, the

SOLID ANGLES.

269

dihedral angle F-B A-H, is a right dihedral angle, and is meas- ured by the angle CED, if its sides CE and ED, drawn in the planes ^Li^and AG respectively, be perpendicular to AB. But angle C'E'D', drawn as represented in the diagram, is acute, while angle C"EnD", drawn as represented, is obtuse.

F
V	D D"
\	\

II

Fig. 1.

Fig. 2.

Many properties of dihedral angles can be established which are analogous to propositions relating to plane angles. Let the student prove the following :

1. If two planes intersect each other, their vertical dihedral angles are equal.

2. If a plane intersect two parallel planes, the exterior- interior dihedral angles are equal ; the alternate-interior dihedral angles are equal ; the two interior dihedral angles on the same side of the secant plane are supplements of each other.

3. When two planes are cut by a third plane, if the exterior- interior dihedral angles be equal, or the alternate dihedral angles be equal, or the two interior dihedral angles on the same side of the secant plane be supplements of each other, and the edges of the dihedrals thus formed be parallel, the two planes are parallel.

4. Two dihedral angles are equal if their faces be respec- tively parallel and lie in the same direction, or opposite direc- tions, from the edges.

5. Two dihedral angles are supplements of each other if two of their faces be parallel and lie in the same direction, and the other faces be parallel and lie in the opposite direc- tion, from the edges.

270

GEOMETRY. BOOK VI.

Proposition XVI. Theorem.

472. If a straight line be perpendicular to a plane every plane embracing the line is perpendicular to that plane.

M

I

B

I

D Let AB be perpendicular to the plane MN

We are to prove any plane} P Q, embracing A B, perpen- dicular to M N.

At B draw, in the plane MN, B C J_ to the intersection D Q.

Since A B is J_ to MN, it is _L to D Q and B G, § 430

(if a straight line be A. to a plane, it is ± to every straight line in that plane draum through its foot).

Now /.ABC is the measure of the

dihedral Z P-D Q-N. § 470

But Z ABC is sl right angle,

.'. the Z P-D Q-N is a right dihedral,

.'.PQis X to MN.

Q. E. D.

SOLID ANGLES.

271

Proposition XVII. Theorem.

473. If two planes be perpendicular to each other, a straight line drawn in one of them perpendicular to their intersection is perpendicular to the other plane.

A

.1/

vz 7

* c In

D

bet the planes M N and PQ be perpendicular to each other, and at any point B of their intersection DQ let BA be drawn in the plane PQ, perpendicular to DQ.

We are to prove A B _L to the plane M N.

Draw B 0 in the plane MN 1. to DQ.

Then Z. A B G is a right angle,

{being the plane Z. of the rt. dihedral Z formed by the two planes).

,\ABisJ\-to the two straight lines D Q and B C.

.'. A B is _L to the plane M N, § 449

{if a straight line be ± to two straight lines drawn through its foot in a plane, it is _L to the plane).

Q. E. D

272

GEOMETRY.

BOOK VI.

Proposition XVIII. Theorem.

474. If two planes be perpendicular to each other, a straight line drawn through any point of intersection per- pendicular to one of the planes will lie in the other plane.

C

Fig. 1.

Fig. 2.

Let PQ {Fig. l)be perpendicular to the plane M N, C Q their intersection, and B A be drawn through any point B in C Q perpendicular to the plane M N.

We are to prove that B A lies in the plane P Q.

At the point B draw B A' in the plane P Q J_ to the inter- section C Q.

The line B A' will be _L to the plane M N, § 472

(if two planes be ± to each other, a straight line drawn in one of them ± to their intersection is _L to the other).

Now B A is J_ to the plane MN\

.'. B A and B A' coincide,

Hyp. §447

(at a given point in a plane only one ± can be erected to that plane). But B A' lies in the plane P Q ;

.'. B A, which coincides with BA/, lies in the plane P Q.

Q. E. D.

Scholium. Through a line parallel or oblique to a plane, as A C, Fig. 2, only one plane can be passed perpendicular to the given plane.

SOLID ANGLES.

273

Proposition XIX. Theorem.

475. Jf two intersecting planes be each perpendicular to a third plane, their intersection is also perpendicular to that plane.

I

Let the planes BD and BC intersecting in the line A B be perpendicular to the plane PQ.

We are to prove A B J_ to the plane P Q.

A perpendicular erected at B, a point common to the three planes, will lie in the two planes B C and B I), § 473

(if two planes be ± to each other, a straight line drawn through any point of intersection ± to one of the planes will lie in the other plane).

And, since this _L lies in both the planes, B C and B D, it must coincide with their intersection.

.'. A B is J_ to the plane P Q.

Q. E. D.

476. Corollary. If a plane be perpendicular to each of two intersecting planes, it is perpendicular to the intersection of those planes. •

274 GEOMETRY. — BOOK VI.

Proposition XX. Theorem.

477. Every point in the plane which bisects a dihedral angle is equally distant from the faces of that angle.

A

Let plane A M bisect the dihedral angle formed by the planes A D and A C ; and let P E and P F be perpendiculars drawn from any point P in the plane A M to the planes A C and A D.

We are to prove P E = P F.

Through P E and P F pass a plane intersecting the planes A G and A D in 0 E and 0 F.

Join P 0.

Now the plane P E F is J_ to each of the planes A C and A D, § 471

(if a straight line be J. to a plane, any plane embracing the line is A. to that

plane) ;

.*. the plane PE F is JL to their intersection A 0. § 476

(If a plane be _L to each of two intersecting planes, it is ±to the intersection of these planes) .

.-.Z POE = Z POF,

(being measures respectively of the equal dihedral A M-OA-C and M-OA-D). .'.rt. APOE = vt. A POF, § 110

.'.PE = PF,

(being homologous sides of equal A ).

Q. E. D.

SOLID ANGLES.

275

Supplementary Propositions. Proposition XXI. Theorem.

478. The acute angle which a straight line makes with its own projection on a plane is the least angle which it makes with any line of that plane.

A

Let B A meet the plane M N at B, and let B A' be its projection upon the plane M N, and BO any other line drawn through B in the plane.

We are to prove Z ABA' <Z ABC.

Take BC=BA'.

Join A C.

In the A A B A' and A B C,

AB = AB,

BA' = BC,

but AA' < AC,

(a ± is the shortest distance from a point to a plane).

.\ Z ABA' <Z ABC, § 116

{if two sides of a Abe equal respectively to two sides of another, but the third

side of the first A be greater than the third side of the second, ilien the

Z opposite the third side of the first A is greater than tlie Z ojjposite the

third side of the second).

Q. E. D.

Exercise. — The angle included by two perpendiculars drawn from any point within a dihedral angle to its faces, is the supple- ment of the dihedral angle.

Iden. Cons. § 448

276 GEOMETRY. BOOK VI.

Proposition XXII. Theorem.

479. If two straight lines be not in the same plane, one and only one common perpendicular to the lines can be drawn. C E D

M A

b\ -fl— £--<?

7

^;. ?.: ^../at

Let AB and C D be two given straight lines not in the same plane. We are to prove one and only one common perpendicular to the two lines can be drawn.

Since A B and C D are not in the same plane they are not II, § 474

(two lis lie in the same plane).

Through the line A B pass the plane M N II to CD.

Since CD is II to the plane M N, all its points are equally distant from the plane M ' N \ § 432

hence CD', the projection of the line C D on the plane MN, will be It to CD, § 76

and will intersect the line A B at some point as O.

Now since C O is the line which projects the point C upon the plane MN, it is J_ to the plane MN; § 435

hence C C is _L to C D1 and A B, § 430

(if a line be ± to a plane, it is ± to every line drawn through its foot in the

plane).

Also, CO is J_ to CD, §67

.*. C C is the common J_ to the lines CD and A B.

Moreover, line C C is the only common _L

For, if another line E B, drawn between A B and CD, could

be J_ to A B and C D, it would also be J_ to a line B G drawn

II to CD in the plane M N, § 67

and hence _L to the plane M N. § 449

But EH, drawn in the plane CD1 II to CO, is _L to the

plane MN. § 457

Hence we should have two Je. from the point E to the plane

M N, which is impossible, § 44/

.*. CO is the only common J_ to the lines CD and A B.

Q. E. D

POLYHEDRAL ANGLES.

277

On Polyhedral Angles.

480. Def. A Polyhedral angle is the extent of opening of three or more planes meeting in a common point.

Thus the figure S-A EC BE, formed by the planes A SB, BSC, CSD, DSE, ESA, meeting in the common point S, is a polyhedral angle.

The point S is the vertex of the angle.

The intersections of the planes SA, SB, etc., are its edges.

The portions of the planes bounded by the edges are its faces.

The plane angles A SB, BSC, etc., formed by the edges are its face angles.

481. Def. Polyhedral angles are classified as trihedral, quad- rahedral, etc., according to the number of the faces.

482. Def. Trihedral angles are rectangular, bi-rectangular, or tri-rectangular, according as they have one, two, or three right dihedral angles.

483. Def. Trihedral angles are scalene, isosceles, or equilateral, according as the face angles are all unequal, two equal, or three equal.

484. Def. A polyhedral angle is convex, if the polygon formed by the intersections of a plane with all its faces be a convex polygon.

485. Def. Two polyhedral angles are equal when they can be applied to each other so as to coincide in all their parts.

Since two equal polyhedral angles coincide however far their edges and faces be produced, the magnitude of a polyhedral angle does not depend upon the extent of its faces. But, in order to represent the angle in a diagram, it is usual to pass a plane, as

278 GEOMETRY. — BOOK VI.

ABODE, cutting all its faces in the straight lines, A B, B C, etc. ; and by the face A S B is meant the indefinite surface in- cluded between the lines S A and SB indefinitely produced.

486. Def. Two polyhedral angles are symmetrical if they have the same number of faces, and the successive dihedral and face angles respectively equal but arranged in reverse order.

Thus, if the edges AS, B S,

a Ft1

etc., of the polyhedral angle, * :-N

S-A BCD, be produced, there is m.S.'jJF

formed another polyhedral angle, / // y*'*

S-A1 B1 C D>, which is symmetri- l/y'y

cal with the first, the vertex S Sj£'

being the centre of symmetry. //n V^N

If we take S A' = S A, and //// \ V\ through the points A and A' the A<^ y (nj \ d>\^—^.a

parallel planes A B C D and \z___V v. \/

A'B'C D be passed, we shall B C c b

have SB' = SB, SC' = SC, etc. For if we conceive a third parallel plane to pass through S, then A A', B B', etc., are divided proportionally, § 469. And if any one of them be bisected at S, the others are also bisected at S. Hence, the points A', B', etc., are symmetrical with A, B, etc.

Moreover, the two symmetrical polyhedral angles are equal in all their parts. Tor their face angles A SB and A' SB', B SO and B' S C are equal each to each, being vertical plane angles. And the dihedral angles formed at the edges S A and SA', SB and SB', are equal each to each, being vertical dihedral angles.

Now if the polyhedral angle S-A' B' C D1 be revolved about the vertex S until the polygon A' B' C D is brought into the position abed, in the same plane with ABC D, it will be evident that while the parts A SB, B SC, etc., succeed each other in the order from left to right, the corresponding equal parts a Sb, b Sc, etc., succeed each other in the order from rigid to left. Hence the two figures cannot be made to coincide by superposition, but are said to be equal by symmetry.

SOLID ANGLES. 279

Proposition XXIII. Theorem. 487. The sum of any two face angles of a trihedral angle is greater than the third. ^

Let S-ABC be a trihedral angle in which the face angle ASC is greater than either angle A S B or angle BSC.

We are to prove Z ASB + Z BSC > Z ASC. In the face A SG draw S D, making Z A SD = Z A SB. Through any point D of S D draw any straight line ADC cutting A S and S C.

TakeSB = SD. Pass a plane through A C and the point B. In the A A S D and A S B

AS=AS, Iden.

SD = SB, Cons.

ZASD = ZASB. Cons.

.'.AASD = A A SB, § 10G

.'.AD = AB,

{being homologous sides of equal A).

lathe A ABC, AB + BOAC.

Subtract the equals A B and A D. Then BO DC.

Now in the A BSC and DSC

SB=SD, Cons.

SC = SC, Iden.

but BO DC,

.\Z BSOZ DSC. §116

.'.ZASB + Z BSC > ZASD + Z DSC, that is ZASB+ZBSOZASC.

280

GEOMETRY.

BOOK VI.

Proposition XXIV. Theorem.

488. The sum of the face angles of any convex polyhe- dral angle is less than four right angles.

S

Let the polyhedral angle S be cut by a plane, mak- ing the section ABC DE a convex polygon.

We are to prove Z A SB + Z BSC etc. < 4 rt. A.

From any point 0 within the polygon draw 0 A,0 B, 0 C, OD, OE.

The number of the A having their common vertex at 0 will be the same as the number having their common vertex at S.

.*. the sum of all the A of the A having the common vertex at S is equal to the sum of all the A of the A having the com' mon vertex at 0.

But in the trihedral A formed at A, B, C, etc.

ZSAE+ ZSAB>Z OAE+ Z 0 A B, § 487 (the sum of any two face A of a triliedral Z is greater than the third).

and Z SBA + Z SBOZ OBA + Z OBC. §487

.'. the sum of the A at the bases of the A whose common vertex is S is greater than the sum of the A at the bases of the A whose common vertex is 0.

.'. the sum of the A at S is less than the sum of the A at 0.

But the sum of the A at 0 = 4 rt. A . § 34

.*. the sum of the A at 8 is less than 4 rt. A .

Q. E. D,

SOLID ANGLES.

281

Proposition XXV. Theorem. 489. An isosceles trihedral angle and its symmetrical trihedral angle are equal.

Let S-A B C be an isosceles trihedral angle, having; ZASB = ZBSC. And let S-A' B' C be its sym- metrical trihedral angle.

We are to prove trihedral ZS-ABC = trihedral Z S-A' B' C.

Revolve Z S-A' B' C about S until SB' falls on SB and the plane SB' A' falls on the plane SBC.

Now the dihedral Z C-SB-A = dihedral Z A'-SB'-C, {being vertical dihedral A ).

.*. the plane SB' C will fall on the plane SB A.

Now ZBSC = ZBSA, Hyp.

and Z B'SA' = ZBSAf

(being vertical A ).

.\Z BSO = Z B'SA'; Ax. 1

.'.SA' will fall on SO.

In like manner S C will fall on S A,

.'. the two trihedral A will coincide and be equal.

q. e. o.

282

GEOMETRY. BOOK VI.

Proposition XXVI. Theorem. 490. Two symmetrical trihedral angles are equivalent.

Let the trihedral Z S-ABC and Z S-A' B' C be sym- metrical.

We are to prove trihedrat Z S-ABC =c=* trihedral Z S-A'B'C.

Draw D' D making the A DS A, DSC, and DSB equal.

Then ZD'SA' = ZD'SC' = A D'SB',

{being vertical A of (lie equal A D S A, DSC, and D SB).

Then the trihedral Z S-D CB = trihedral Z S-D' C'B'} § 489 (tvjo isosceles symmetrical trihedral A are equal).

And trihedral Z S-D C A = trihedral Z S-D' C A',

and trihedral ZS-ADB = trihedral Z S-A1 D' B'.

Adding the first two equalities, the polyhedral Z S-A BCD

ro= polyhedral Z S-A' B' CD1.

Take away from each of these equals the equal trihedral A S-ADB and S-A' D' B'.

Then trihedral Z S-ABC '^ trihedral Z S-A' B' C.

Q. E. D.

491. Scholium. If D D' fall within the given trihedral angles these trihedral angles would be composed of three isosceles trihedral angles which would be respectively equal, and hence the given trihedral angles would be equivalent.

* The symbol (o) is to be read " equivalent to."

SOLID ANGLES. 283

Exercises.

1. If a plane be passed through one of the diagonals of a parallelogram, the perpendiculars to the plane from the extremi- ties of the other diagonal are equal.

2. If each of the projections of a line A B upon two inter- secting planes be a straight line, the line A B is a straight line.

3. The height of a room is eight feet, how can a point in the floor directly under a certain point in the ceiling be deter- mined with a ten-foot pole ]

4. If a line be drawn at an inclination of 45° to a plane, what is the greatest angle which any line of the plane, drawn through the point in which the inclined line pierces the plane, makes with the line.

5. Through a given point pass a plane parallel to a given plane.

6. Find the locus of points in space which are equally distant from two given points.

7. Show that the three planes embracing the edges of a tri- hedral angle and the bisectors of the opposite face angles re- spectively intersect in the same straight line.

8. Find the locus of the points which are equally distant from the three edges of a trihedral angle.

9. Cut a given quadrahedral angle by a plane so that the section shall be a parallelogram.

10. Determine a point in a given plane such that the sum of its distances from two given points on the same side of the plane shall be a minimum.

11. Determine a point in a given plane such that the differ- ence of its distances from two given points on opposite sides of a plane shall be a maximum.

284

GEOMETRY.

BOOK VI.

Proposition XXVII. Theorem.

492. Two trihedral angles are equal or symmetrical when the three face angles of the one are respectively equal to the three face angles of the other.

S' S S>

In the trihedral A S and S', let Z A S B = Z A' S' B>, Z ASC = ZA'S'C') and Z BSC = ZB'S'C.

We are to prove that the homologous dihedral angles are equal, and hence the trihedral angles S and S' are either equal or symmetrical.

On the edges of these angles take the six equal distances SA,SB, SC,S'A',S'B',S'C'.

Draw A B, B C, A C, A'B', B'C, A'C.

The homologous isosceles A SAB, S> A' B', SAC, S1 A' C, SBC, S'B' C are equal, respectively. § 106

.'.AB,AC,BC equal respectively A'B', A' C, B' C,

{being homologous sides of equal A).

.-. A ABC = A A'B' O. § 108

At any point D in SA draw D E and D F _L to SA in the faces AS B and ASC respectively.

These lines meet A B and A C respectively, (since the A SAB and SAC are acute, each being one of the equal A of an isosceles A).

Join EF.

QnS'A'takQA'D' = AD.

SOLID ANGLES. 285

Draw D' E' and D' F in the faces A1 S' B' and A' S' C re- spectively ± to S' A', and join E' F'. In the rt. A A DE and A' D' E'

AD = A>D', Cons.

ZDAE=Z D'A'E', (being homologojts A of the equal & SAB and & A* B1).

.-.rt. A ADE = vt. A A' D' E\ § 111

.-. AE = A'E' and D E = D' E't

{being homologous sides of equal &).

In like manner we may prove AF ' = A' F' and DF~D'F.

Hence in the A A E F and A1 E' F' we have

A E and A F = respectively A' E' and A1 F,

and ZEAF=ZE,A/F,

(being homologous A of the equal A ABC and Af B1 C).

.:AAEF = AA'E'F', §106

.-. EF= E'F (being homologous sides of the equal A AEF and A1 E' F).

Hence, in the A E D F and E' D* F we have

ED,DF, and EF= respectively E' D', b' P,'slti& E1 F'.

.\AEDF=A E'jyP, § 108

.-. Z EDF = Z E'D'P,

(being homologous A of equal &).

.-. the dihedral Z B-A S-C = dihedral Z B'-A' S'-C, (since A E D F and E' D' F, the measures of these dihedral A, are equal).

In like manner it may be proved that the dihedral A A-B S-C and A-C SB are equal respectively to the dihedral A A'-B'S'-C and A'-C S'-B'.

Q. E. D.

This demonstration applies to either of the two figures de- noted by S'-A' B' C, whicli are symmetrical with respect to each other. If the first of these figures be given, S and S' are equal, for they can be applied to each other so as to coincide in all their parts. If the second be given, S and S' are symmetrical. § 486

BOOK VII.

POLYHEDRONS, CYLINDERS, AND CONES.

General Definitions.

493. Def. A Polyhedron is a solid bounded by four or more polygons.

A polyhedron bounded by four polygons is called a tetra- hedron; by six, a hexahedron; by eight, an octahedron; by twelve, a dodecahedron; by twenty, an icosahedron.

494. Def. The Faces of a polyhedron are the bounding polygons.

495. Def. The Edges of a polyhedron are the intersec- tions of its faces.

496. Def. The Vertices of a polyhedron are the intersec- tions of its edges.

497. Def. A Diagonal of a polyhedron is a straight line joining any two vertices not in the same face.

498. Def. A Section of a polyhedron is a polygon formed by the intersection of a plane with three or more faces.

499. Def. A Convex polyhedron is a polyhedron every section of wnich is a convex polygon.

500. Def. The Volume of a polyhedron is the numerical measure of its magnitude referred to some other polyhedron as a unit of measure.

501. Def. The polyhedron adopted as the unit of measure is called the Unit of Volume.

502. Def. Similar polyhedrons are polyhedrons which have the same form.

503. Def. Equivalent polyhedrons are polyhedrons which have the same volume.

504. Def. Equal polyhedrons are polyhedrons which have the same/orra and volume.

On Prisms.

505. Def. A Prism is a polyhedron two of whose faces are equal and parallel polygons, and the other faces are parallelo- grams.

PRISMS.

287

506. Def. The Bases of a prism are the equal and parallel polygons.

507. Def. The Lateral faces of a prism are all the faces except the

508. Def. The Lateral or Con- vex Surface of a prism is the sum of its lateral faces.

509. Def. The Lateral edges of a prism are the intersections of its lateral faces ; the Basal edges of a prism are the intersections of the bases with the lateral faces.

510. Def. Prisms are triangular, quadrangular, pentag- onal, etc., according as their bases are triangles, quadrangles, pentagons, etc.

511. Def. A Right prism is a prism whose lateral edges are perpendicular to its bases.

512. Def. An Oblique prism is a prism whose lateral edges are oblique to its bases.

513. Def. A Regular prism is a right prism whose bases are regular polygons, and hence its lateral faces are equal rectangles.

514. Def. The Altitude of a prism is the perpendicular distance between the planes of its bases. The altitude of a right prism is equal to any one of its lateral edges.

515. Def. A Truncated prism is a por- tion of a prism included between either base and a section inclined to the base and cutting R|QHT pr|sm. all the lateral edges.

516. Def. A Right section of a prism is a section perpen- dicular to its lateral edges.

517. Def. A Parallelopiped is a prism whose bases are parallelograms.

518. Def. A Right parallelopiped is a parallelopiped whose lateral edges are perpendicular to its bases ; hence its lateral faces are rectangles.

519. Def. An Oblique parallelopiped is a parallelopiped whose lateral edges are oblique to its bases.

520. Def. A Rectangular parallelopiped is a right paral- lelopiped whose bases are rectangles.

521. Def. A Cube is a rectangular parallelopiped all of whose faces are squares.

288 GEOMETRY. — BOOK VII.

Proposition I. Theorem.

522. The sections of a prism made by parallel plat are equal polygons.

Let the prism A D be intersected by the parallel planes G K, G' K'.

We are to prove section G H I K L = section G' H' I' K' L'.

G H, III, IK, etc., are parallel respectively to G' H', H' I', I'K',etc, §465

{the intersections of two II planes by a third plane are II lines).

.'. AG HI, H IK, etc., are equal respectively to A G1 H' I', H,rKr9 etc., §462

{two A not in the same plane, having their sides respectively parallel and lying in the same direction, are equal).

Also, sides GH, HI, IK, etc., are equal respectively to G'W, 11'1',1'K1, etc., § 135

( II lines comprehended between II lines are equal).

.'. section GHIKL = section G' W I> K' L', § 155

(being mutually equiangular and equilateral).

Q. E. D.

523. Corollary. Any section of a prism parallel to the base is equal to the base ; and all right sections of a prism are equal.

prisms. 289

Proposition II. Theorem. 524. The lateral area of a prism is equal to the product of a lateral edge by the perimeter of the right section.

E>

D>

Let GH I KL be a right section of the prism AD'.

We are to prove lateral area of prism A D1 = A A' X perim- eter G H I K L.

Consider the lateral edges A A', B B', etc., to be the bases of the U] AB', B C'} etc., which form the convex surface of the prism.

Then the altitudes of these HI will be the J» GH, HI, IK, etc.,

and the area of each O is the product of its base and alti- tude. § 321 Now the bases of these Z17 are all equal, § 464 (II lines comprehended between II planes are equal) ; and the sum of the altitudes GH, HI, IK, etc., is the perimeter of the right section.

Hence, the sum of the areas of these ZI7 is the product of a lateral edge A A' by the perimeter of the right section.

That is, the lateral area of the prism is equal to the product of a lateral edge by the perimeter of a right section.

Q. E. D.

525. Corollary. The lateral area of a right prism is equal to the altitude multiplied by the perimeter of the base.

290 GEOMETRY. — BOOK VII.

Proposition III. Theorem. 526. Two prisms are equal if the three faces including a trihedral angle of the one be respectively equal to the three corresponding faces including a trihedral angle of the other , and similarly placed.

J J*

F

Let AD, AG, A J, be respectively equal to A' D',

A ' J1, an d similarly pla ced.

We are to prove prism A 1 = prism A' I'.

Now trihedral Z A = trihedral Z A1, § 492

(two trihedrals arc equal, when the three face A of the one are equal respec- tively to tlic three face A of the other and are similarly placed).

Apply trihedral Z A to trihedral Z A'.

Then the base A D will coincide with the base A' D',

face A G with A' G',

and face A J with A' J' ;

.'. FG will coincide with F'G', and F J with FJ'.

.'. the upper bases, F I and F' I1, will coincide, (being equal polygons, since they arc equal to the equal lower bases).

.*. the remaining edges will coincide,

(their extremities being the same points).

.'. the prisms will coincide and be equal.

Q. E. D.

527. Corollary 1. Two truncated prisms are equal, if the three faces including a trihedral of the one be respectively equal to the three faces including a trihedral of the other, and be similarly placed.

528. Cor. 2. Two right prisms having equal bases and altitudes are equal. If the faces be not similarly placed, if one be inverted, the faces will be similarly placed and the prisms can be made to coincide.

PRISMS. 291

Proposition IV. Theorem.

529. An oblique prism is equivalent to a right prism whose bases are equal to right sections of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism.

B/'

B - C Let A D' be an oblique prism, and F I a right section.

Complete the right prism F I', making its edges equal to those of the oblique prism.

We are to prove oblique prism A D' ^= right prism F I'.

In the solids A I and A' I'

trihedral Z A = trihedral Z A', § 492

{two trihedrals are equal when three face A of the one are respectively equal to three face A of the other, and are similarly placed).

Xow face A D = face A1 D', § 505

(being the two bases of the oblique prism A D') ;

face A J = face A' J', Cons.

and face A G = face A' G'. Cons.

.'. solid AI= solid AT, § 527

(two truncated prisms are equal wlicn the three faces including a trihedral of the one are respectively equal to tJie three faces including a trihedral of the other, and are similarly placed).

To each of these equal solids add the solid F D'. Then oblique prism A D' *> right prism F I'.

Q. E. D.

292

GEOMETRY. BOOK VII.

Proposition V. Theorem.

530. Any two opposite faces of a parallelopiped are equal and parallel.

Let AG be a parallelopiped. We are to prove faces A F and D G equal and parallel.

Since A GYisaO, § 517

A B and D C are equal and II line9. § 125

Also, since A H is a O, § 505

A E and D H are equal and II lines. § 125

.'.Z EAB = Z HDC, §462

(two A not in the same plane having their sides II and lying in the same direction are equal).

.'. face AF= face D G.

§ 140 § 463

allel.

Moreover, face A F is II to D G

the same plane have their sides II an direction their planes are parallel).

In like manner we may prove A H and B G equal and par-

(if two A not in the same plane have their sides II and lying in the same direction their planes are parallel,).

Q. E. D.

531. Scholium. Any two opposite faces of a parallelo- piped may be taken for bases, since they are equal and parallel parallelograms.

prisms. 293

Proposition VI. Theorem.

532. The plane passed through two diagonally opposite edges of a parallelopiped divides the parallelopiped into two equivalent triangular prisms.

II

Let the plane A E GC pass through the opposite edges A E and C G of the parallelopiped A G.

We are to prove that the parallelopiped A G is divided into two equivalent triangular prisms, A B C-F, and A D C-H.

Let I J KL be a right section of the parallelopiped made by a plane ± to the edge A E.

The intersection IK of this plane with the plane A EGC is the diagonal of the O I J KL.

.-.A IKJ=A IKL. § 133

But prism A B C-F is equivalent to a right prism whose base is UK and whose altitude is A E, § 529

(any oblique prism is ^ to a right prism whose, bases arc equal to right sec- tions of the oblique prism, and whose altitude is equal to a lateral edge oftlic oblique prism).

The prism A D C-H is equivalent to a right prism whose base is ILK, and whose altitude is A E. § 529

Now the two right prisms are equal, § 528

{two right prisms having equal bases and altitudes arc equal).

.'.ABC-Fo ADC-H.

Q. E. D

294

GEOMETRY. BOOK VII.

Proposition VII. Theorem.

533. Two rectangular parallelepipeds having equal bases are to each other as their altitudes.

/	B / P J
i	< /
&
/ pi J	m ( '1 ,	/ /
/	' /
/	/ /
)*	/A )

1

Let AB and A'B' be the altitudes of the two rectangu- lar parallelopipeds, P, and Pf, having equal bases.

w * p AB

We are to prove — = .

P' A'B' CASE I. — When A B and A1 B' are commensurable.

Find a common measure m, of A B and A' B'. Suppose m to be contained in A B 5 times, and in A' B' 3 times.

AB 5 A'B'~3'

At the several points of division on AB and A' B' pass planes _L to these lines.

The parallelopiped P will be divided into 5,

and P' into 3, parallelopipeds equal, each to each, § 528 (two right prisms having equal bases and altitudes are equal).

Then LLt

P' 3

. P _AB

"P'~A'Br

Then we have

PRISMS.

295

IV

Case II. — When A B and A' B' are incommensurable.

ȣ

t

\

\

Let A B be divided into any number of equal parts,

and let one of these parts be applied to A' B' as many times as A' B' will contain it.

Since A B and A' B' are incommensurable, a certain number of these parts will extend from A' to a point Df leaving a ra mainder D B' less than one of these parts.

Through D pass a plane _L to A' B', and denote the parallel- opiped whose base is the same as that of P'y and whose altitude is 4' 2) by Q.

Now, since A B and A' D are commensurable,

Q :P = A'D :AB. (Case I.)

Suppose the number of parts into which A B is divided to be continually increased, the length of each part will become less and less, and the point D will approach nearer and nearer to B'.

The limit of Q will be P',

and the limit of A' D will be A' B'y

.'. the limit of Q : P will be P' : P,

and the limit of A' D : A B will be A' B' : A B,

Moreover the corresponding values of the two variables Q : P and A1 D : A B are always equal, however near these variables approach their limits.

.*. their limits P' : P = A' B' : A B. § 199

Q. E. D.

534. Scholium. The three edges of a rectangular parallelo- piped which meet at a common vertex are its dimensions. Hence two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions.

296

GEOMETRY. BOOK VII.

Proposition VIII. Theorem.

535. Two rectangular parallelopipeds having equal alti- tudes are to each other as their bases.

( Q (	* I	1 p> 1	1	/p( c / I
(.	<•
)	1 /„ (

7

Let a, b, and c, and a', b', c, be the three dimensions re- spectively of the two rectangular parallelopipeds P and P.

We are to prove — = .

* P' a'X b'

Let Q be a third rectangular parallelopiped whose dimen- sions are a', b and c.

Now Q has the two dimensions b and c in common with P, and the two dimensions a' and c in common with P.

Then -=-, §534

Q a'' *

(two rectangular parallelopipeds which have two dimensions in common are to each other as their third dimensions) ;

and |ll.

P' ¥

Multiply these two equalities together ;

P _ aXb

P' ~ a' XV '

then

§534

Q. E. D.

536. Scholium. This proposition may be stated thus : two rectangular parallelopipeds which have one dimension in common are to each other as the products of the other two dimensions.

PRISMS.

297

Proposition IX. Theorem.

537. Any two rectangular parallelopipeds are to each other as the products of their three dimensions.

X

	\ p. \
c \ \	C
\ \

III

Let a, b,c, and a,' b', cf, be the three dimensions respec- tively of the two rectangular parallelopipeds P and P.

We are to prove = ___

Let Q be a third rectangular parallelopiped whose dimen- sions are a, b, and d.

Then £ = £., §534

Q cf

{two rectangular parallelopipeds which have two dimensions in common are

to each other as their third dimensions) ;

and

§536

Q_ _ aXb P' ~~ a' XV'

{two rectangular parallelopipeds which have one dimension in common are to each other as tlic products of tlicir oilier two dimensions).

Multiply these equalities together ; aXbXc

then

P

a' XUXd

Q. E. D.

298

GEOMETRY. — BOOK VII.

Proposition X. Theorem. 538. The volume of a rectangular parallelopiped is equal to the product of its three dimensions, the unit of volume being a cube whose edge is the linear unit.

/	' / / / /
/ / / / 2
/ /
				/ /
				/ /
				/ t A
				y

1

Let a, b, and c be the three dimensions of the rectan- gular parallelopiped P, and let the cube U be the unit of volume.

We are to prove volume of P = aXbX c. P aXbXc U~

1X1X1 p

But _ is the volume of P ;

.'. the volume of P = a X b X c.

§537 §500

Q. E. D.

539. Corollary I. Since a cube is a rectangular parallelo- piped having its three dimensions equal, the volume of a cube is equal to the third power of its edge.

540. Cor. II. The product a X b represents the base when c is the altitude ; hence : The volume of a rectangular parallelo- piped is equal to the product of its base by its altitude.

541. Scholium. When the three dimensions of the rec- tangular parallelopiped are each exactly divisible by the linear unit, this proposition is rendered evident by dividing the solid into cubes, each equal to the unit of volume. Thus, if the three edges which meet at a common vertex contain the linear unit 3, 4 and 5 times respectively, planes passed through the several points of division of the edges, and perpendicular to them, will divide the solid into cubes, each equal to the unit of volume ; and there will evidently be 3 X 4 X 5 of these cubes.

PRISMS.

299

Proposition XL Theorem. 542. The volume of any parallelopiped is equal to the product of its base by its altitude.

KB J

Let A BC D-F be a parallelopiped having all its faces oblique, and HE its altitude.

We are to prove A B G D-F =ABGDX HE.

By making the right section HUN and completing the

parallelopiped HIJ N-GLKM we have a right parallelopiped

equivalent to, A B G D-F. § 529

(an oblique prism is equivalent to a right prism whose base is a right section

of the oblique prism and whose altitude is equal to a lateral edge of

the oblique jJrism).

Through the edge I L make the right section ILPO, and complete the right parallelopiped ILPO-HGQE, and we have a rectangular parallelopiped equivalent to H I J N-G LKM>\ 529

and hence equivalent to A B G D-F.

Kow O ILGH^O FFGH,

O 0PQE = (OILGH) = OJKMN;

and O ABGD = 0 EFGH.

.'.n OPQE-O ABCD.

Moreover, the three parallelopipeds have the common alti- tude HE.

But OPQE-ILGH=GPQEX HE; §540

.\ABGD-F=ABGD X HE.

Q. E. D.

§322 §530 §530

300

GEOMETRY. BOOK VII.

Proposition XII. Theorem. 543. The volume of any prism is equal to the product

of its base by its altitude. Ef

Case I. — When the base is a triangle. Let V denote the volume, B the base, and H the altitude of the triangular prism A EC-E'. We are to prove V = B X H.

Upon the edges A E, EC, E E', construct parallelopiped AEG D-E'.

Then A E C-E> =s= \ A E G D-E', § 532

(the plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms),

and AEC=\AEGD. §133

But A E CD-E' = 2BX H, § 542

(the volume of any parallelopiped is equal to the product of its base by its

altitude).

.'. V= J (2 BX H) = BX H.

Case II. — - When the base is a polygon of more than three sides.

Planes passed through the lateral edge A A', and the several diagonals of the base will divide the given prism into triangular prisms,

which have for a common altitude the altitude of the prism.

Hence, the volume of the entire prism is the product of the sum of their bases by the common altitude ,

that is the entire base by the altitude of the prism.

Q. E. D.

544. Corollary. Prisms having equivalent bases are to each other as their altitudes ; prisms having equal altitudes are to each other as their bases ; and any two prisms are to each other as the product of their bases and altitudes. Any two prisms having equivalent bases and equal altitudes are equivalent.

PRISMS. 301

Proposition XIII. Theorem. 545. The four diagonals of a parallelopiped bisect each

n

Let AG, EC, B H, and FD% be the four diagonals of the parallelopiped A G. We are to prove these four diagonals bisect each other.

Through the opposite and II edges A E and C G pass a plane intersecting the II bases in the II lines A C and E G.

The section A C G E is a O,

(having Us op})ositc sides II ) ;

.'. its diagonals AG and EC bisect each other in the point 0. § 138

In like manner a plane passed through the opposite and II edges FG and A D will form a O AFGD,

whose diagonals A G and F D will bisect each other in the point 0. § 138

Also, a plane passed through the opposite and II edges E H and B C will form a O E B G H,

whose diagonals E C and B H will bisect each other in the point 0.

,\ the four diagonals bisect each other at the point 0.

Q. E. D.

546. Corollary. The diagonals of a rectangular parallelo- piped are equal.

547. Scholium. The point 0, in which the four diagonals intersect, is called the centre of the parallelopiped ; and it is evi- dent that any straight line drawn through the point 0 and terminated by two opposite faces of the parallelopiped is bisected at that point. Hence 0 is the centre of symmetry.

802

GEOMETRY. BOOK VII.

On Pyramids.

548. Def. A Pyramid is a polyhedron one of whose faces is a polygon, and whose other faces are triangles having a com- mon vertex and the sides of the polygon for bases.

549. Def. The Base of a pyramid is the face whose sides are the bases of the triangles having a common vertex.

550. Def. The Lateral faces of a pyramid are all the faces except the base.

551. Def. The Lateral surface of a pyramid is the sum of its lateral faces.

552. Def. The Lateral edges of a pyramid are the intersec- tions of its lateral faces.

553. Def. The Basal edges of a pyramid are the intersec- tions of its base with its lateral faces.

554. Def. The Vertex of a pyramid is the common vertex of its lateral faces.

555. Def. The Altitude of a pyramid is the perpendicular distance from its vertex to the plane of its

Thus, V-ABCDE is a pyramid ; ABCDEis its base; AVB,BVC, etc. are its lateral faces, and their sum ^ is its lateral surface; V A, V B, etc. are its lateral edges ; A B, B C, etc. @ B

its basal edges ; V is its vertex ; V 0 is its altitude.

PYRAMIDS. 303

556. Def. A Regular pyramid is a pyramid whose base is a regular polygon, and whose vertex is in the perpendicular to the base at its centre.

557. Def. The Axis of a regular pyramid is the straight line joining its vertex with the centre of the base.

558. Def. The Slant height of a regular pyramid is the altitude of any lateral face.

559. Def. A pyramid is triangular, quadrangular, pentag- onal, etc., according as its base is a triangle, quadrilateral, pentagon, etc. A triangular pyramid formed by four faces (all of which are triangles) is a tetrahedron.

560. Def. A Truncated pyramid is the portion of a pyramid included be- tween its base and a section cutting all its lateral edges.

561. Def. A Frustum of a pyramid is a truncated pyramid in which the cut- ting section is parallel to the base.

562. Def. The base of the pyramid

is called the Lower base of the frustum, and the parallel sec- tion, its Upper base.

563. Def. The Altitude of a frustum is the perpendicular distance between the planes of its bases.

564. Def. The lateral faces of a frustum of a regular pyra- mid are the trapezoids included between its bases ; the lateral surface is the sum of the lateral faces; the Slant height of a frustum of a regular pyramid is the altitude of any lateral face.

304

GEOMETRY. — BOOK VII.

Proposition XIV. Theorem.

565. If a pyramid be cnt by a plane parallel to its base, I. The edges and altitude are divided proportionally j II. The section is a polygon similar to the base.

vi y

Let the pyramid V-A B 0 D E, whose altitude is V 0, be cut by a plane abode parallel to its base, in- tersecting the lateral edges in the points a, b, c, d, e, and the altitude in o.

We are to prove I Va__ Vb Vo .

VA~ VB VO*

II. The section abcde similar to the base ABODE.

I. Suppose a plane passed through the vertex V II to the base.

Then the edges and the altitude will be intersected by three II planes.

. Va_ Vb_ Vo_

" VA~ VB VO'

{if straight lines be intersected by three II planes, their corresponding segment* are -proportional).

II. The sides ab, be etc. are parallel respectively to A B, B C,

etc., § 465

(the intersections of\\ planes by a third plane are II lines) ; .'. A a be, bed etc. are equal respectively to A ABC,

BOD etc., §462

(if two A not in the same plane have their sides respectively II and lying in the same direction, they are equal).

.*. the two polygons are mutually equiangular.

§469

PYRAMIDS. 305

Also, since the sides of the section are II to the correspond- ing sides of the base,

A Vab, Vbc etc. are similar respectively to A VA B, VB C etc.

... Jtl -(™Xi±L=(I±) = l± etc.

AB \VB/~ BC~ \VCJ~ CD

.'.the polygons have their homologous sides proportional; .*. section a b c d e is similar to the base ABC D E. § 278

Q. E. D.

566. Corollary 1. Any section of a pyramid, parallel to its base is to the base as the square of its distance from the ver- tex is to the square of the altitude of the pyramid.

Since £» (™)_£i.

VO \VB/ AB

Squaring TW=TW-

a b c d e a b

Eut ABCDE= TJ?' §344

(similar polygons are to each other as the squares of their homologous sides).

a b c d e V o

"ABCDE Yd2'

567. Cor. 2. If two pyramids having equal altitudes be cut by planes parallel to their bases, and at equal distances from their vertices, the sections will have the same ratio as their bases.

For

and

Now, since Vo = V o1 , and VO = V 0',

abode : ABCDE : -.a'b'd : A1 B' C. Whenceabcde \a'b'd : : AB C D E : A' B' C. § 262 568. Cor. 3. If two pyramids have equal altitudes and equivalent bases, sections made by planes parallel to their bases and at equal distances from their vertices are equivalent.

abode	To2
ABCDE	VO2'
a'V d	Vo*
A'B'C1	VO*'

306

GEOMETRY. — BOOK VII.

Proposition XV. Theorem. 569. The lateral area of a regular ^ one-half the product of the perimeter of its height. v

is equal to by its slant

Let V-ABCDE be a regular pyramid, and VH its slant height.

We are to prove the sum of the faces V AB, V BG, etc. = J (AB + BO, etc.) X VH.

Now AB=BC= CD, etc., §363

(being sides of a regular polygon).

VA= VB = VC, etc., § 450

(oblique lines dravm from any point in a ± to a plane at equal distances from the foot of the A. are equal).

.'. A VAB, VBG, etc. are equal isosceles A, § 108

whose bases are the sides of the regular polygon and whose common altitude is the slant height VH.

Now the area of one of these A, as VAB,= ^ base AB X altitude VH, § 324

.*. the sum of the areas of these A, that is, the lateral area of the pyramid, is equal to J the sum of their bases (AB + BC + CD, etc.) X V H.

Q. E. D.

570. Corollary 1. The lateral area of the frustum of a regular pyramid, being composed of trapezoids which have for their common altitude the slant height of the frustum, is equal to one-half the sum of the perimeters of the bases multiplied by the slant height of the frustum.

571. Cor. 2. The dihedral angles formed by the intersec- tions of the lateral faces of a regular pyramid are all equal. § 492

PYRAMIDS.

307

Proposition XYI. Theorem. 572. Two triangular pyramids having equivalent bases and equal altitudes are equivalent. X

Let S-ABC and S'-A' B' C be two triangular pyramids having equivalent bases ABC and A' B'C situated in the same plane, and a common altitude A X. We are to prove S-ABC ^ S'-A' B' C. Divide the altitude A X into a number of equal parts, and through the points of division pass planes II to the

planes of their bases, intersecting the two pyramids.

In the pyramids S-ABC and S'-A' B' C inscribe prisms

whose upper bases are the sections D E F, G U I, etc., D' E' F',

G'HT, etc.

The corresponding sections are equivalent, § 568

(if two pyramids have equal altitudes and equivalent bases, sections made by planes II to their bases and at equal distances from their vertices are equivalent).

.'. the corresponding prisms are equivalent, § 544

(prisms having equivalent bases and equal altitudes are equivalent).

Denote the sum of the prisms inscribed in the pyramid S-A B C, and the sum of the corresponding prisms inscribed in the pyramid S'-A' B' C by V and V respectively.

Then F= V.

Now let the number of equal parts into which the altitude A X is divided be indefinitely increased ;

The volumes V and V are always equal, and approach to

the pyramids S-A B C and S'-A' B' C respectively as their limits.

Hence S-A B C o S'-A' B' C. § 1 99

Q. E. D.

308 GEOMETEY. BOOK VII.

Proposition XVII. Theorem. 573. The volume of a triangular pyramid is equal to one- third of the product of its base and altitude.

Let S-ABC be a triangular pyramid, and H its altitude.

We are to prove S-A B G «= J A B G X H.

On the base ABC construct a prism ABG-SED, having its lateral edges II to SB and its altitude equal to that of the pyramid.

The prism will be composed of the triangular pyramid S-A B G and the quadrangular pyramid S-A G D E.

Through S A and S D pass a plane SAD.

This plane divides the quadrangular pyramid into the two triangular pyramids, S-A G D and S-A E D , which have the same altitude and equal bases. § 133

r.S-AG D=o= S-A ED, §572

(two triangular pyramids having equivalent bases and equal altitudes are equivalent).

Now the pyramid S-A E D may be regarded as having ESD for its base and A for its vertex.

.'. pyramid S-A E D =©= pyramid S-A BG, § 572

(having equal bases SED and ABC and the same altitude).

.'. the three pyramids into which the prism A B G-SE D is divided are equivalent.

.*. pyramid S-A B G is equivalent to J of the prism.

But the volume of the prism is equal to the product of its base and altitude ; § 543

.*. S-ABG = iABCX H.

Q. E. D.

PYRAMIDS. 309

Proposition XVIII. Theorem. 574. The volume of any pyramid is equal to one-third the product of its base and altitude.

Let S-A BC D E be any pyramid.

We are to prove S-A B C D E = \ABCDEXSO.

Through the edge SB, and the diagonals of the base DA, D B, pass planes.

These divide the pyramid into triangular pyramids, whose bases are the triangles which compose the base of the pyramid,

and whose common altitude is the altitude SO of the pyramid.

The volume of the given pyramid is equal to the sum of the volumes of the triangular pyramids.

But the sum of the volumes of the triangular pyramids is

equal to \ the sum of their bases multiplied by their common

altitude, § 573

{the volume of a triangular pyramid is equal to one-third the product of its

base and altitude),

that is, the volume of the pyramid S-A BC D E = J ABCDE X SO.

Q. E. D.

575. Corollary. Pyramids having equivalent bases are to each other as their altitudes ; pyramids having equal altitudes are to each other as their bases. Any two pyramids are to each other as the products of their bases and altitudes.

576. Scholium. The volume of any polyhedron may be found by dividing it into pyramids, and computing the volumes of these pyramids separately.

310

GEOMETRY. — BOOK VII.

Proposition XIX. Theorem.

577. Two tetrahedrons having a trihedral angle of the one equal to a trihedral angle of the other are to each other as the products of the three edges of these trihedral angles. V.

Let V and V denote the volumes of the two tetra- hedrons D-ABC, D'-AB'C, having the trihedral A of the one equal to the trihedral A of the other.

^ t V AB X ACX AD

We are to prove — =

r V AB'XAC'X AD'

Place the tetrahedrons so that their equal trihedral A shall be in coincidence.

Consider ABC and A B' C the bases of the two tetrahe- drons,

and from D and D1 draw D 0 and D' 0' J_ to the base ABO.

Now

ABC X DO ABO DO AB'C'X D' 0' ~ AB'C'X D< O1

§ 575

{any two pyramids are to each other as the products of their bases and

altitudes).

But

and

ABC ABX AG

AB' C AB'XAC DO AD

D' 0' AD'

{being homologous sides of the similar &.ADO and A D1 0').

§341 § 278

V

ABX ACX AD A B' X ACX AD'

Q. E. D.

exercises. 811

Exercises.

1. Given a cubical tank holding one ton of water ; find its length in feet, if a cubic foot of water weigh 1000 ounces.

2. At 17 cents a square foot, what is the cost of lining with zinc a rectangular cistern 5 ft. 7 in. long, 3 ft. 11 in. broad, 2 ft.

8 J in. deep 1

3. Find the side of a cubical block of cast iron weighing a ton, if iron weigh 7.2 as much as water, and a cubic foot of water weigh 1000 ounces.

4. How many cubic yards of gravel will be required for a walk surrounding a rectangular lawn 200 yards long, and 100 yards wide ; the walk to be 3 feet wide and the gravel 3 inches deep]

5. The volume of a rectangular solid is the sum of two cubes whose edges are 10 inches and 2 inches respectively, and the area of its base is the difference between 2 squares whose sides are 1£ feet and 1£ feet respectively ; find its altitude in feet.

6. A rectangular cistern whose length is equal to its breadth is 22 decimetres deep, and contains 10 tonneaux of water; find its length.

7. Given a regular prism whose base is a regular hexagon in- scribed in a circle 6 metres in diameter, and whose altitude is 8.7 metres ; find the number of kilolitres it will contain, if the thickness of the walls be 1 decimetre.

8. A pond whose area is 11 hectares, 21 ares, 22.2 centares, is covered with ice 21 centimetres thick. What is the weight of this body of ice in kilogrammes, the weight of ice being 92 % that of water.

9. Given two hollow oblique prisms, whose interior dimen- sions are as follows : the area of a right section of the first is 18 sq. ft., of the second 2.1 sq. metres ; a lateral edge of the first is

9 ft., of the second 2.1 metres ; find the volume of each in cubic yards, cubic metres, cubic decimetres, and cubic centimetres; find the capacity of each in gallons and litres, in bushels and hectolitres ; and find the weight of water in pounds and in kilo- grammes, required to fill each prism.

312

GEOMETRY. BOOK VII.

Proposition XX. Theorem.

578. The frustum of a triangular pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum and, whose bases are the lower base, the tipper base, and a mean proportional between the two bases of the frustum.

J

Let B and b denote the lower and upper bases of the frustum ABC-DEF, and H its altitude.

Through the vertices A, E, G and E, D, G pass planes dividing the frustum into three pyramids.

Now the pyramid E-A B G has for its altitude H, the alti- tude of the frustum, and for its base B, the lower base of the frustum.

And the pyramid C-E D F has for its altitude H, the alti- tude of the frustum, and for its base b, the upper base of the frustum. Hence, it only remains

To prove E-A D G equivalent to a pyramid, having for its altitude H, and for its base \B X b.

E-A B C and E-A D G, regarded as having the common ver- tex G, and their bases in the same plane B D, have a common altitude.

.'. E-A B G : E-A D G : : A A E B : A A E D. § 575

(pyramids having equal altitudes are to each other as their bases).

Now since the AAEB and A E D have a common altitude, (that is, the altitude of the trapezoid A BED),

we have AAEB : A AED : :AB : D E,

326

PYRAMIDS. 313

.-.E-ABC : E-A DC : : A B : D E.

In like manner E-A D C and E-D F C, regarded as having the common vertex E and their bases in the same plane D C, have a common altitude.

.'.E-A DC : E-DFC ::AADC :A DEC. § 575

But since the A A D C and DEC have a common altitude, (the altitude of the trapezoid A CF D),

we have A A D C : A D EC : : A C : D F. §326

Now A D E F is similar to A A B C, § 565

(the section of a pyramid made by a plane II to the base is a polygon similar to the base) ;

.'.AB :DE : : A C : D F. § 278

.'.E-ABC : E-A DC : : E-A DC : E-DFC. Now E-ABC = i HX B, § 573

and E-DFC = C-E D F = \ H X b. §573

.'. E-A DC = \J%HXBX%HXb = J # ^ X 6.

Q. E. D.

579. Corollary 1. Since the volume of the frustum is de- noted by V, the lower base by B, the upper base by b, and the altitude by Hy

we have V=\HXB+\HXb + \HX ^Wx~b = $HX(B+b+ \jTx~b).

580. Cor. 2. The frustum of any pyramid is equivalent to the sum of three pyramids whose common altitude is the altitude of the frustum, and whose bases are the loiver base, the upper base, and a mean proportional between the bases of the frustum.

For the frustum of any pyramid is equivalent to the corre- sponding frustum of a triangular pyramid having the same alti- tude and an equivalent base (§ 578) ; and the bases of the frustum of a triangular pyramid being both equivalent to the correspond- ing bases of the given frustum, a mean proportional between the triangular bases is equivalent to a mean proportional between their equivalents.

314

GEOMETRY. BOOK VII.

Proposition XXL Theorem.

581. A truncated triangular prism is equivalent to the sum of three pyramids whose common base is the base of the prism, and whose vertices are the three vertices of the inclined section.

Let AB C-D E F be a truncated triangular prism whose base is ABC, and inclined section D E F.

We are to prove A B C-D E F =0= three pyramids, E-A B G, D-A B C and F-A B G.

Pass the planes AEG and DEC, dividing the truncated prism into the three pyramids E-A B G, E-A G D, and EG D F.

Now the pyramid E-A B G has the base ABC and the vertex E.

E-AGDoB-ACD, §574

(for they have the same base AC D and the same altitude, since their vertices E and B are in the line EB II to the base A CD),

But pyramid B-A CD, which is equivalent to pyramid E-A C I), may be regarded as having the base ABC and the vertex D.

Again, E-CDF*> B-A C F,

for their bases CDF and AG F, in the same plane, are

equivalent, § 325

{for the A CDF and A CFhave the common base C F and equal altitudes, their vertices lying in the line A D\\ to C F).

PYRAMIDS.

315

Moreover, E-C D F and B-A C F have the same altitude,

(since their vertices E and B are in the line E B II to the plane of their bases A CDF).

But the pyramid B-A C F may be regarded as having the base ABC and the vertex F.

.'.the truncated triangular prism A B C-DEFis equivalent to the three pyramids E-A B C, DA B C, and F-A B C.

Q. E. D F

582. Corollary 1. The volume of a truncated right tri angular prism is equal to the product of its base by one-third the sum of its lateral edges. For the lateral edges D A> EB, FC, being perpendicular to the base, are the altitudes of the three pyramids whose sum is equivalent to the truncated prism. And, since the volume of a pyramid is one-third the product of its base by its altitude, the sum of the volumes of these pyramids = ABCXi(DA + EB + FC).

583. Cor. 2. The volume of any truncated triangular prism is equal to the product of its right section by one-third the sum of its lateral edges.

For let A B C-Al B' C be any truncated triangular prism. Then the right section D E F divides it into two truncated right prisms whose volumes are D E F X J (A D + B E + C F) and DEFX £ (A'D + B'E-b C F).

Whence their sum is D EF X J (A A' + B B' + C C).

316 GEOMETRY. — BOOK VIE.

Exercises.

1. Given a pyramid whose base is a rectangle 80 feet by 60 feet, and whose lateral edges are each 1 30 feet ; find its volume, and its entire surface.

2. Given the frustum of a pyramid whose bases are hepta- gons ; each side of the lower base being 10 feet, and of the upper base 6 feet, and the slant height 42 feet ; find the convex surface in square yards.

3. Given a stick of timber 30 feet long, the greater end being 18 inches square, and the smaller end 15 inches square; find its volume in cubic feet.

4. Given a stone obelisk in the form of a regular quadrangular pyramid, having a side of its base equal to 25 decimetres, and its slant height 12 metres. The stone weighs 2.5 as much as water. What is its weight in kilogrammes 1

5. Given the frustum of a pyramid whose bases are squares j each side of the lower base being 35 decimetres, each side of the upper base 25 decimetres, and the altitude 15 metres ; find its volume in steres.

6. Given a right hexagonal pyramid whose base is inscribed in a circle 30 feet in diameter, and whose altitude is 20 feet ; find its convex surface, and its volume.

7. Given a right pentagonal pyramid whose base is inscribed in a circle 20 feet in diameter, and whose slant height is 30 feet ; find its convex surface, and its volume.

8. Find the difference between the volume of -the frustum of a pyramid, and the volume of a prism of the same altitude whose base is a section of the frustum parallel to its bases and equidis- tant from them.

9. Given a stick of timber 32 feet long, 18 inches wide, 15 inches thick at one end, and 12 inches at the other; find the number of cubic feet, and the number of feet board measure it contains. Find equivalents for the results in the metric system.

SIMILAR POLYHEDRONS.

317

On Similar Polyhedrons.

584. Def. Similar polyhedrons are polyhedrons which have the same form. They have, therefore, the same number of faces, respectively similar and similarly placed, and their corre- sponding polyhedral angles equal.

585. Def. Homologous faces, lines, and angles of similar

polyhedrons are faces, lines, and angles similarly placed.

8

I. The homologous edges of similar polyhedrons are pro- portional.

Since the faces SAB, SA C, SB C and A B C are similar respectively to S' A' B', S1 A' C, S' B1 C and A' B' C, we have

SA _SB^_^ etc ~SrA'~SrB'~ArB'1 6C*

§278

II. Any two homologous faces of similar polyhedrons are proportional to the squares of any two homologous edges.

SAB

Thus' S'A'B'

S A2

S' A'2

SAC SC2 SBC

S'A/C'~SrC<2~S'B'C'' §342

III. The entire surfaces of two similar polyhedrons are pro- portional to the squares of any two homologous edges.

™ . SAB SAC .

Inus, since . , = , etc.,

S'A'B' S'A'C

SAB + SAC, etc. SAB -$-#

S' A' B' + S' A' C, etc. S' A' B' ^p '

266

318

GEOMETRY.

BOOK VII.

Proposition XXII. Theorem.

586. Two similar polyhedrons may be decomposed into the same number of tetrahedrons similar, each to each, and similarly placed.

Let ABCDEOPQRS and A1 B' C D' E'-O' F Q' R' & be two similar polyhedrons of which P and P' are homologous vertices.

We are to prove that A BCDE-OPQRS and A'B'G'D'E'- 0' P' Q1 R' S' can be decomposed into the same number of tetrahe- drons similar and similarly placed.

Place two homologous faces A BCD and A'B'O'D' in the same plane, having two homologous edges AB and A' B' II and lying in the same direction.

On any two corresponding faces not adjacent to P and P'> as ABODE and A' B' C I> E', from two homologous vertices, as E and E', draw diagonals dividing these faces into A, similar and similarly placed.

From the homologous vertices P, P' of the polyhedrons draw straight lines to the vertices of these A.

Repeat this construction for each of the faces not adjacent to P, P>.

Then the polyhedrons will be divided into the same number of tetrahedrons ;

that is, into as many tetrahedrons as there are A in these faces.

SIMILAR POLYHEDRONS. 319

Now, any two corresponding tetrahedrons, as P-A B E and P'-A1 B' E', are similar ;

for the faces E A B and P A B are similar respectively to the faces E' A' B1 and P' A1 B', § 294

(being similarly situated & of similar polygons).

In the A PBE and P' B' E'

PB\s\\ to P' B', and B E to B' E't

(since they make equal A respectively with the II lines A B and A' B') ;

.\Z PBE = Z P'B'E't § 462

(two A not in the same plane having their sides II and lying in the same direction are equal) ;

and **=(A*\-M.. §278

P'B> \A'B'1 B'E' *

.*. face PBE is similar to face P' B> E'. § 284

Also, in the A P A E and P' A1 E'

PE (PB\ PA (AB\ AE . 27g

P'E'~\P'B') P' A'~\A' B')~ A' E'* S (being homologous sides of similar A ).

.'. face P A E is similar to face P' A' E'. § 282

Moreover, since any two corresponding trihedral A of these

tetrahedrons are formed by three plane A which are equal, each

to each, and similarly situated, they are equal. § 492

.'. P-A BE and P'-A' B' E' are similar. § 584

In like manner we may show that any other two tetrahe- drons similarly situated are similar.

That is, the two similar polyhedrons have the same number of tetrahedrons similar each to each, and similarly situated.

Q. E. D.

587. Corollary. Any two homologous lines in two similar polyhedrons have the same ratio as any two homologous edges.

320

GEOMETRY. BOOK VII.

Proposition XXIII. Theorem. 588. Similar tetrahedrons are to each other as the cubes of their homologous edges.

Let S-B CD and S'-B' CD' be two similar tetrahedrons having for bases the similar faces BCD and B1 CD', and for altitudes S 0 and S' 0'.

We are to prove

SBC D

BCZ S'-B'C'D'~ W(j»'

Apply the tetrahedron S'-B' C D' to the tetrahedron S-B C D, so that the polyhedral S' shall coincide with S.

Then the base B' C D' will be II to the face BCD, {since their planes make equal A with the face SB 0),

and the J_ S 0, _L to B C D, will also be _L to B' C D'.

SO' will be the altitude of the tetrahedron S-B' C D'.

Now

S-BCD BCDXSO

^x^,§575 * SO' *

S-B' CD' B'C'D'XSO' B' C D'

(any two tetrahedrons are to each other as the products of their bases and

altitudes).

Since the bases are similar,

BCD B~&

B'CD' ^C1

§343

SIMILAR POLYHEDRONS. 321

Also, = , § 587

SO' B'G' *

(in two similar polyhedrons any two homologous lines are in the same ratio as any two homologous edges).

. S-B C D BC2 BC _ BCZ

s-B'c iy~wc^x B'tf^WW*'

Q. E. D.

589. Corollary 1. Two similar polyhedrons are to each other as the cubes of any two homologous edges.

For, two similar polyhedrons may be decomposed into tetra- hedrons similar, eacli to each, and similarly placed, of which any two homologous edges have the same ratio as any two homolo- gous edges of the polyhedrons. And, since any pair of the simi- lar tetrahedrons are to each other as the cubes of any two homologous edges, the entire polyhedrons are to each other as the cubes of any two homologous edges. § 266

590. Cor. 2. Similar prisms or pyramids are to each other as the cubes of their altitudes ; and similar polyhedrons are to each other as the cubes of any two homologous lines.

Ex. 1. The portion of a tetrahedron cut off by a plane parallel to any face is a tetrahedron similar to the given tetrahedron.

Ex. 2. Two tetrahedrons, having a dihedral angle of one equal to a dihedral angle of the other, and the faces including these angles respectively similar, and similarly placed, are similar.

Ex. 3. Given two similar polyhedrons, whose volumes are 125 feet and 12.5 feet respectively j find the ratio of two homologous edges.

322 GEOMETRY. BOOK VII.

On Eegular Polyhedrons.

591. Def. A Regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhe- dral angles are equal.

The regular polyhedrons are the tetrahedron, octahedron and icosahedron, all of whose faces are equal equilateral triangles; the hexahedron, or cube, whose faces are squares ; the dodecahe- dron, whose faces are regular pentagons.

Only these five regular polyhedrons are possible, for a poly- hedral angle must have at least three face angles, and must have the sum of its face angles less than four right angles, (§ 488). Hence :

I. If the faces be equilateral triangles, polyhedral angles may be formed of them in groups of 3, 4, or 5 only, as in the tetrahedron, octahedron and icosahedron. Since each angle of an equilateral triangle is two-thirds of a right angle, the sum of six such angles is four right angles, and therefore greater than a convex polyhedral angle.

II. If the faces be squares, polyhedral angles may be formed of them in groups of three only, as in the regular hexahedron, or cube ; since four such angles would be four right angles.

III. If the faces be regular pentagons, polyhedral angles may be formed of them in groups of three only, as in the regular dodecahedron ; since four such angles would be greater than four right angles.

IV. "We can proceed no farther ; for a group of three angles of regular hexagons would equal four right angles, and of regular heptagons, etc., would be greater than four right angles.

REGULAR POLYHEDRONS.

323

450

Proposition XXIY. Problem.

592. Given an edge, to construct the five regular poly- hedrons.

Let A B be the given edge.

I. Upon AB to construct a regular tetrahedron.

D

Upon A B construct the equilateral A ABC. § 232

Find the centre 0 of this A, § 238 and erect 0 D _L to the plane ABC. Take the point D so that A D = AB. Draw DA,DB,DC. ABC D is the regular tetrahedron required. For, the edges are all equal, and hence the faces are equal equilateral A. and its polyhedral A are all equal. § 492

q II. To construct a regular hexahedron.

Upon the given edge AB construct the square ABCDy

and upon the sides of this square con- C struct the squares E B, FC, G D, HA _L to the plane ABC D.

Then A G is the regular hexahedron required.

III. To construct a regular octahedron.

Upon the given edge A B construct the square ABC D.

Through its centre 0 pass a J_ to its plane ABC D.

In this _L take two points E and F, one above and the other below the plane,

so that A E and A F are each equal toAB.

Join E and F to each of the vertices of the square.

Then E ABC D F is the regular octahedron required.

For, the edges are all equal,

and hence the faces are equal equilateral A.

And, since the A D EF and D A C are equal, § 108

D EBF is a square and the pyramid A-D EB F is equal in all its parts to the pyramid E-A BCD.

Hence, the polyhedral A A and E are equal.

In like manner all the polyhedral A of the figure are equal.

E

A

D

§ 450

324

GEOMETRY. BOOK VII.

IV. To construct a regular dodecahedron.

Upon A B construct the regular pentagon ABODE. § 395

On each side of this pentagon construct an equal pentagon, so inclined that trihedral A shall be formed at A, B, 0, D, E.

The convex surface thus formed is composed of six regular pentagons.

In like manner, upon an equal pentagon A' B' C D' E' con- struct an equal convex surface.

Apply one of these surfaces to the other, with their convexi- ties turned in opposite directions, so that P' 0' and P' Q1 shall fall upon P 0 and P Q.

Then every face Z of the one will, with two consecutive face A of the other, form a trihedral Z.

The solid thus formed is the regular dodecahedron required.

For, the faces are all regular pentagons, Cons,

and the polyhedral A are all equal. § 499

D D'

G G'

V. To construct a regular icosahedron.

Upon A B construct the regular pentagon ABODE. § 395 At its centre 0 erect a _U to its plane. In this X take P so that PA = A B.

REGULAR POLYHEDRONS.

325

Join P with each of the vertices of the pentagon ;

thus forming a regular pentagonal pyramid whose vertex is P, and whose dihedral A formed on the edges PA, P JB, PC, etc. are all equal. § 571

Taking A and B as vertices, construct two pyramids each equal to the first, and having for bases BPEFGwAAGECP respectively.

There will thus be formed a convex surface consisting of ten equal equilateral A.

In like manner upon an equal pentagon A' B' C D1 E' con- struct an equal convex surface.

Apply one of these surfaces to the other with their convexi- ties turned in opposite directions, so that every combination of two face A of the one, as P' D' C, P* D' E', shall with a combi- nation of three face A of the other, as BCH, BCP, PCD, form a pentahedral Z.

The solid thus formed is the regular icosahedron required.

For, the faces are all equal ; Cons.

and the polyhedral A are all equal, § 571

Q. E. D.

TETRAHEDRON.

		HEX AH	EDRON.

ICOSAHEDRON.

DODECAHEDRON.

593. Scholium. The regular polyhedrons can be formed thus :

Draw the above diagrams upon card-board. Cut through the exterior lines and half through the interior lines. The fig- ures will then readily bend into the regular forms required.

326 geometry. book vii.

Supplementary Propositions. Proposition XXV. Theorem. (Euler's.) 594. In any polyhedron the number of its edges in- creased by two is equal to the number of its vertices increased by the number of its faces.

Let E denote the number of edges of any polyhedron; V the number of its vertices, F the number of its faces.

We are to prove E + 2 = V + F.

S Beginning with one face ABODE,

we have E = V.

Annex a second face SAB by ap- plying one of its edges to an edge of the first face.

There is formed a surface having one edge A B, and two vertices A and )D B common to both faces.

.*. whatever the number of the B C sides of the new face, the whole num-

ber of edges is now one more than the whole number of ver- tices.

.-.for 2 faces E= V+ 1.

Annex a third face, SBC, adjacent to each of the former.

The new surface will have two edges SB and B C,

and three vertices S, B and C, in common with the preced- ing surface.

.*. the increase in the number of edges is again one more than the increase in the number of vertices.

According to the same law, for an incomplete surface of F—\ faces

E= V+ F-2.

When we add the last face SEA, necessary to complete the surface,

its edges SE, SA and A E, and its vertices S, E and A will be in common with the preceding surface.

.*. in a polyhedron of F faces E — V + F — 2. ,-.E+ 2= V+ F.

Q. E. D.

POLYHEDRONS. 327

Proposition XXVI. Theorem.

595. The sum of all the angles of the faces of any poly- hedron is equal to four right angles taken as many times as the polyhedron has vertices less two.

Let E denote the number of edges, V the number of vertices, F the number of faces, and S the sum of all the angles of the faces of any polyhedron.

We are to prove S = 4 rt. A X ( V— 2).

Since E denotes the number of the edges of the polyhedron,

2 E will denote the whole num- ber of sides of all its faces, con- sidered as sides of independent poly- gons.

A{

And since the sum of all the interior and exterior A of each poly- B c

gon is equal to 2 rt. A taken as many times as it has sides,

the sum of the interior and exterior A of all the faces is equal to 2 rt. A X 2 E.

And since the sum of the exterior A of each face is 4 rt. A, § 159

the sum of the exterior A of all the faces is equal to 4 rt. A X F.

.\ S+ 4 rt A X F*=* 2 it A X 2 E.

That is, . S = 4 rt. A X (E — F).

Since E + 2 = V + F, § 594

E- F= F-2,

.'. £=4rt. A X (F-2). q. E. d.

328 geometry. book vii.

On the Cylinder.

596. Def. A Cylindrical surface is a curved surface gen- erated by a moving straight line which continually touches a given curve and in all its positions is parallel to a given fixed straight line not in the plane of the curve.

Thus, the surface ABC ' D, generated by the moving line A D continually touching the curve ABC and always parallel to a given straight line M, is a cylindrical surface.

597. Def. The moving line is called the Generatrix; the curve which directs the motion of the generatrix is called the Directrix ; the generatrix in any position is called an Element of the surface.

The generatrix may be indefinite in extent, and the direc- trix a closed or an open curve. In elementary geometry the directrix is considered a circle.

598. Def. A Cylinder is a solid bounded by a cylindrical surface and two parallel planes.

599. Def. The Bases of a cylinder are its plane surfaces.

600. Def. The Lateral surface of a cylinder is its cylindri- cal surface.

601. Def. The Axis of a cylinder is the straight line join- ing the centres of its bases.

CYLINDERS. 329

602. Def. The Altitude of a cylinder is the perpendicular distance between the planes of its bases.

603. Def. A Section of a cylinder is a plane figure whose boundary is the intersection of its plane with the surface of the cylinder.

604. Def. A Right section of a cylinder is a section per- pendicular to the elements.

605. Def. A Radius of a cylinder is the radius of the base.

606. Def. A Right cylinder is a cylinder whose elements are perpendicular to its bases. Any element of a right cylinder is equal to its altitude.

607. Def. An Oblique cylinder is a cylinder whose elements are oblique to its bases. Any element of an oblique cylinder is greater than its altitude.

608. Def. A Cylinder of Revolution is a cylinder generated by the revolution of a rectangle about one side as an axis.

609. Def. Similar cylinders of revolution are cylinders generated by similar rectangles revolving about homologous sides.

610. Def. A Tangent line to a cylinder is a straight line which touches the surface of the cylinder, but does not intersect it.

611. Def. A Tangent plane to a cylinder is a plane which embraces an element of the cylinder without cutting the sur- face. The element embraced by the tangent plane is called the Element of Contact.

612. Def. A prism is inscribed in a cylinder when its lateral edges are elements of the cylinder and its bases are in- scribed in the bases of the cylinder.

613. Def. A prism is circumscribed about a cylinder when its lateral faces are tangent to the cylinder and its bases are cir- cumscribed about the bases of the cylinder.

330

GEOMETRY. BOOK VII.

Proposition XXVII. Theorem.

614. Every section of a cylinder made by a plane pass- ing through an element is a parallelogram.

G

K?\

^^7

Let ABC D be a section of the cylinder A G, made by a plane passing through A D.

We are to prove the section A B G D a parallelogram.

The line B G, in which the cutting plane intersects the curved surface a second time, is an element ;

for, if through the point B a line be drawn II to A D,

it will be an element of the surface.

It will also lie in the plane A G.

This element, lying in both the cylindrical surface and plane surface, is their intersection.

Now A D is II to B C,

(being elements of the cylinder),

and A B is II to D G, § 465

(the intersections of two II planes by a third plane are II lines).

.-. the section ABGD is a O. §125

Q. E. D.

615. Corollary. Every section of a right cylinder embrac- ing an element is a rectangle.

CYLINDERS.

331

Proposition XXVIII. Theorem. 616. The bases of a cylinder are equal. £_ C

Let ABE and DGG be the bases of the cylinder A G. We are to prove A B E = D C G.

Any sections A G and A G, embracing A D, an element of

§614

D G and A E = D G.

BG is II to EG, (each being II to AD).

BC = EG,

.'. EG is a O.

.\EB = GG,

\AEAB = A GDG.

the cylinder, are UJ.

.'.AB Now

Also

§134 §459

§464 § 136 § 134 § 108

Apply the upper base to the lower base, so that D G will coincide with A B.

Then A GDC will coincide with A EAB, and point G will fall upon point E.

That is, any point G in the perimeter of the upper base will coincide with the point in the same element in the lower base. .*. the bases coincide, and are equal.

Q. E. D.

617. Corollary 1. Any two parallel sections ABC and A' B' C, cutting all the elements of a cylinder E F, are equal. For these sections are the bases of the cylinder A C.

618. Cor. 2. Any section of a cylinder parallel to the base is equal to the base.

332 GEOMETRY. BOOK VII.

Proposition XXIX. Theorem. 619. The lateral area of a cylinder is equal to the product of the perimeter of a right section of the cylinder by an element of the surface. ^

Let ABC D E be the base, and A A' any element of the cylinder A C ; and let the curve abcdebe any right section of its surface.

Denote the perimeter of the right section by P, and the lateral surface of the cylinder by & We are to prove S = P X A A'.

Inscribe in the cylinder a prism whose right section abcde

will be a polygon inscribed in the right section a b c d e of the

cylinder. § 604

Denote the lateral area of the prism by s,

and the perimeter of its right section by p.

Then s=pXAA', §524

{the lateral area of a prism is equal to the product of the perimeter of a right section by a lateral edge).

Now let the number of lateral faces of the inscribed prism be indefinitely increased,

the new edges continually bisecting the arcs in the right section.

Then p approaches P as its limit, and s approaches S as its limit. But, however great the number of faces, $=p X A A1. .'.S=PX AA\ §199

Q. E. D.

CYLINDERS.

333

II

IV

620. Corollary 1. The lateral area of a right cylinder is equal to the product of the perimeter of its base by its altitude.

621. Cor. 2. Let a cylinder of revolution be generated by the rectangle whose sides are R and H revolving about the side H.

Then R is the radius of the base of the cylinder, and H the altitude of the cylinder.

The perimeter of the base is 2 n- R ; § 381

hence, S = 2 it R X H.

The area of each base is tt R2 ; § 381

hence, the total area T of a cylinder of revolution is ex- pressed by

T=2irRXH+27rR2 = 27rR(H+R).

622. Cor. 3. Let S, S' denote the lateral areas of two simi- lar cylinders of revolution ;

T, T' their total areas ; R, R' the radii of their bases ; //, H' their altitudes.

Since the generating rectangles are similar, we have

#=^=#jfljff -266

H' R1 H' + R1 *

S _2irRII S'~2 7rR'H'

x^ = ^2

R*

H' H'* R'2 '

and —- 2^(#+-ft) _& (H+ R\_ H2 _ R2 T'~2it R' {H' + R')~R' \H' + R1) ~ E12 ~ R'2'

That is, the lateral areas, or the total areas, of similar cylin- ders of revolution are to each other as the squares of their altitudes, or as the squares of the radii of their bases.

334

GEOMETRY BOOK VII.

Proposition XXX. Theorem. 623. The volume of a cylinder is equal to the product of its base by its altitude.

Let V denote the volume of the cylinder A G, B its base, and H its altitude.

We are to prove

V=BX H.

Let V denote the volume of the inscribed prism A G, B' its base, and H will be its altitude.

Then

V' = B'X H,

§543

{the volume of a prism is equal to the product of its base by its altitude). Now, let the number of lateral faces of the inscribed prism be indefinitely increased, the new edges continually bisecting the arcs of the bases.

Then B' approaches B as its limit,

and V approaches V as its limit.

But however great the number of the lateral faces,

V' = B'X H.

.'.V=BXH. §199

Q. E. D.

CYLINDERS. 335

624. Corollary 1. Let Vbe the volume of a cylinder of revolution, R the radius of its base, and H its altitude.

Then the area of its base is rr R2, § 381

.'. V=7rE2X h.

625. Cor. 2. Let V and V be the volumes of two similar cylinders of revolution, R and R' the radii of their bases, H and H' their altitudes.

Since the generating rectangles are similar, we have

B_= ^. H' R'}

V ttR2H R2 H H* R3 and = = — v — = a .

V itR^H' R'* H' H'8 R'*

That is, the volumes of similar cylinders of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases.

Ex. 1. Required, the entire surface and volume of a cylin- der of revolution whose altitude is 30 inches, and whose base is a circle of which the diameter is 20 inches.

2. Eequired, the volume of a right truncated triangular prism the area of whose base is 40 inches, and whose lateral edges are 10, 12, and 15 inches, respectively.

3. Let E denote an edge of a regular tetrahedron ; show that the altitude of the tetrahedron is equal to E y/~|~; that the surface is equal to E'2 ^~3 ; and that the volume is equal to

4. Required, the number of quarts that a cylinder of revo- lution will contain whose height is 20 inches, and whose diame- ter is 12 inches.

5. Given S, the surface of a cube, find its edge, diagonal, and volume. What do these become when S = 54 1

336

GEOMETRY. BOOK VII.

Proposition XXXI. Problem.

626. Through a given point to pass a plane given c$

to a

Case I. — When the given point is in the curved surface of the cylinder.

Let AC be a given cylinder, and let the given point be' a point in the element A A'.

It is required to pass a plane tangent to the cylinder and em- bracing the element A A'.

Draw the radius 0 A, and A T tangent to the base ;

and pass a plane R T' through A A' and A T.

The plane R T1 is the plane required.

For, through any point P in this plane, not in the ele- ment A A',

pass a plane II to the base, intersecting the cylinder in the O MN,

and the plane R T> in MP.

Prom the centre of the O M N draw Q M.

MP and M Q are II respectively to A T and A 0, § 465 (the intersections of two II planes by a third plane are II lines) ;

r.ZPMQ = Z TAO, § 462

(two A not in the same plane, having their sides II and lying in the same direction, are equal).

CYLINDERS. 337

.'. P M is tangent to the O MN at M. § 186

.'. P lies without the O M N,

and hence without the cylinder.

.*• the plane R T' does not cut the cylinder, and is tangent to it.

Case IT. — When the given point is urithout the cylinder.

Let P be the given point.

It is required to pass a plane through P tangent to th% cylinder.

Through P draw the line P T II to the elements of the cylinder,

meeting the plane of the base at T.

From T draw TA and TC tangents to the base. § 240

Through P T and the tangent TA pass a plane R V.

Since A A1 is II to P T, Cons.

the plane R T', passing through P T and the point A will contain the element A A',

(two II lives He in the same plane).

And, since R V also contains the tangent A T,

it is a tangent plane to the cylinder.

In like manner, the plane T S', passed through P T and the tangent line T C,

is a tangent plane to the cylinder.

Q. E. F.

627. Corollary 1. The intersection of two tangent planes to a cylinder is parallel to the elements of the cylinder.

628. Cor. 2. Any straight line drawn in a tangent plane, and cutting the element of contact, is tangent to the cylinder.

338 geometry. book vii.

On the Cone.

629. Def. A Conical surface is a surface generated by a moving straight line continually touching a given curve and passing through a fixed point not in the plane of the curve.

Thus the surface generated by the mov- ing line A A' continually touching the curve ABC D, and passing through the fixed point S, is a conical surface.

630. Def. The moving line is called the Generatrix ; the curve which directs the motion of the generatrix is called the Di- rectrix ; the generatrix, in any position, is called an Element of the surface.

631. Def. A conical surface generated by an indefinite straight line consists of two portions, called Nappes, one the Lower, the other the Upper Nappe.

632. Def. A Cone is a solid bounded by a conical surface and a plane.

633. Def. The Lateral surface of a cone is its conical sur- face.

634. Def. The Base of a cone is its plane surface.

635. Def. The Vertex of a cone is the fixed point through which all the elements pass.

636. Def. The Altitude of a cone is the perpendicular dis- tance between its vertex and the plane of its base.

637. Def. A Section of a cone is a plane figure whose boundary is the intersection of its plane with the surface of the cone.

638. Def. A Right section of a cone is a section perpen- dicular to the axis.

639. Def. A Circular cone is a cone whose base is a circle.

640. Def. The Axis of a cone is the straight line joining its vertex and the centre of its base.

641. Def. A Right cone is a cone whose axis is perpen- dicular to its base. The axis of a right cone is equal to its altitude.

642. Def. An Oblique cone is a cone whose axis is oblique to its base. The axis of an oblique cone is greater than its altitude.

coxes. 339

643. Def. A Cone of Revolution is a cone generated by the revolution of a right triangle about one of its perpendicular sides as an axis.

The side about which the triangle re- volves is the axis of the cone ; the other per- pendicular generates the base, the hypotenuse generates the conical surface. Any position '»i' the hypotenuse is an element, and any element is called the slant height.

644. Def. Similar cones of revolution are cones generated by the revolution of simi- lar right triangles about homologous perpen- dicular sides.

645. Def. A Truncated cone is the portion of a cone included between the base and a section cutting all the elements.

646. Def. A Frustum of a cone is a truncated cone in which the cutting section is parallel to the base.

647. Def. The base of the cone is called the Lower base of the frustum, and the parallel section the Upper base.

648. Def. The Altitude of a frustum is the perpendicular distance between the planes of its bases.

649. Def. The Lateral surface of a frustum is the portion of the lateral surface of the cone included between the bases of the frustum.

650. Def. The Slant height of a frustum of a cone of revo- lution is the portion of any element of the cone included between the bases.

651. Def. A Tangent line to a cone is a line having only one point in common with the surface.

652. Def. A Tangent plane to a cone is a plane embracing an element of the cone without cutting the surface. The element embraced by the tangent plane is called the Element of Contact.

653. Def. A pyramid is inscribed in a cone when its lat- eral edges are elements of the cone and its base is inscribed in the base of the cone.

654. Def. A pyramid is circumscribed about a cone when its lateral faces are tangent to the cone and its base is circum- scribed about the base of the cone.

340

GEOMETRY. BOOK VII.

Proposition XXXII. Theorem. 655. Every section of a cone made by a plane passing

through its vertex is a triangle.

S

Let SBD be a section of the cone S-ABG through the vertex S.

We are to prove the section SB D a triangle.

The straight lines joining S with B and D are elements of the surface. § 630

They also lie in the cutting plane, (for their extremities lie in tJie plane).

Hence, they are the intersections of the conical surface with the plane of the section.

B D is also a straight line, § 446

(the intersection of two planes is a straight line).

.'. the section SBD is a A.

Q. E. D

coxes. 341

Proposition XXXIII. Theorem.

656. Every section of a circular cone made by a plane parallel to the base is a circle.

Let the section a b c of the circular cone S-A B C be parallel to the base.

We are to prove that a b c is a circle.

Let 0 be the centre of the base, and let o be the point in which the axis S 0 pierces the plane of the II section.

Through SO and any number of elements, SA, SB, etc., pass planes cutting the base in the radii 0 A, OB, etc.,

and the section a b c in the straight lines o a, ob, etc.

Now o a and o b are II respectively to 0 A and 0 B, § 465 {the intersections of two II planes by a third plane are II lines). /.the A So a and Sob are similar respectively to the ASOAandSOB, . §279

and their homologous sides give the proportion

oa / 'S ' o\ ob OA = \S0) = OB '

But OA = OB; §163

.". o a = o b. That is, all the straight lines drawn from o to the perimeter of the section are equal.

.'. the section a b c is a O.

Q. E. D.

657. Corollary. The axis of a circular cone passes through the centres of all the sections which are parallel to the base.

842

GEOMETRY.

BOOK vn.

Proposition XXXIV. Theorem.

658. The lateral area of a cone of revolution is equal to one-half the product of the circumference of its base by the slant height.

Let A-E F G H K be a cone generated by the revolution of the right triangle A OE about AO as an axis, and let S denote its lateral area, G the circumference of its base and L its slant height.

We are to prove S = \ G X L.

Inscribe on the base any regular polygon E FG H K,

and upon this polygon as a base construct the regular pyra- mid A-E F G UK inscribed in the cone.

Denote the lateral area of this pyramid by s, the perimeter of its base by p, its slant height by I,

Then

s = \p X I,

569

{the lateral area of a regular pyramid is equal to one-half the product of the perimeter of its base by the slant height).

Now, let the number of the lateral faces of the inscribed pyramid be indefinitely increased,

cones. 343

the new edges continually bisecting the arcs of the base.

Then p, s and I approach C, 8 and L respectively as their limits.

But however great the number of lateral faces of the pyramid,

s = \p X I.

>\J3m*iOxL. §199

Q. E. D.

659. Corollary 1. If R be the radius of the base, we have C=27ri?(§381). Therefore S=±(2 vR X L) = vRL. Also, since the area of the base is ttR7, the total area Tof the cone is expressed by

T= ttRL + ttR2 = irR(L + R).

660. Cor. 2. Let S and S1 denote the lateral areas of two Bimilar cones of revolution, T and T their total areas, R and R' the radii of their bases, iiT and W their altitudes, L and V their slant heights. Since the generating triangles are Bimilar, we have

L H R R + L

L> H> R' W + L' '

irRL _R L L2 R* H*

~~~ ~z~. X — r~. '

206

£' ttR'L' R' L' L'2 R'* Hri'

T n wRX(L+R) _R L + R _ Z2 _ R* _ IP T1 ir R' X {L1 + R') R' L' + R' T* R* IF2'

That is : the lateral areas, or total areas, of similar cones of revolution are to each other as the squares of their slant heights, the squares of their altitudes, or the squares of the radii of their bases.

3U

GEOMETRY.

BOOK VII.

Proposition XXXV. Theorem.

661. The lateral area of the frustum of a cone of revo- lution is equal to one-half the sum of the circumferences of its buses multiplied by the slant height.

het II BC-E FG be the frustum of a cone of revolution, and let S denote its lateral area, G and c the cir- cumferences of its lower and upper bases, R and r the radii of the bases, and L the slant height.

We are to prove S = \ (G + c) X L.

Inscribe in the frustum of the cone the frustum of the reg- ular pyramid HBC-EFG,

and denote the lateral area of this frustum by s, the peri- meters of its lower and upper bases by P and p respectively, and its slant height by /.

Then s = \ (P + p) I, § 570

{the lateral area of the frustum of a regular pyramid is equal to one-half the sum of the perimeters of its bases multiplied by the slant height).

Now, let the number of lateral faces be indefinitely in- creased, the new elements constantly bisecting the arcs of the bases.

CONES.

345

Then P, p, and /, approach C, c, and L, respectively as their limits.

But, however great the number of lateral faces of the frus- tum of the pyramid,

• - J (# + p) X I

X = b(C+ c)X L.

§ 199

Q. E. D.

662. Corollary. The lateral area of a frustum of a cone of revolution is equal to the circumference of a section equidistant from its bases multiplied by its slant height.

For the section of the frustum equidistant from its bases cuts the frustum of the regular inscribed pyramid equidistant from its bases.

Therefore the perimeter / LK = J the sum of the perim- eters II B C and EFG. § 142

And this will always be true, however great the number of the lateral faces of the frustum of the pyramid.

Hence, circumference ILK = J the sum of the circumfer- ences HB G and EFG. § 199

346

GEOMETRY. — BOOK VII.

Proposition XXXVI. Theorem.

G63. Any section of a cone parallel to the base is io the base as the square of the altitude of the part above the section is to the square of the altitude of the cone.

Let B denote the base of the cone, H its altitude, b a section of the cone parallel to the base, and h the altitude of the cone above the section.

We are to prove B : b : : IT2 : Jr.

Let J51 denote the base of an inscribed pyramid, b1 the base of the pyramid formed in the section of the cone.

Then B' : V : : IP : k\ § 566

(any section of a pyramid II to its base is to the base as tJie square of the JL from the vertex to tlie plane of the section is to tlie square of tJie altitude of the pyramid).

Now let the number of lateral faces of the inscribed pyiv mid be indefinitely increased,

the new edges continually bisecting the arcs in the base of the cone.

Then B' and b' approach B and b respectively as their limits.

But however great the number of lateral faces of the pyra- mid,

B' :b' ::H2: h\
.B:b ::H*:h*,	§199
	a e. d.

CONES. 347

Proposition XXXVII. Theorem. 664. The volume of any cone is equal to the product of one-third of its base by its altitude.

Let V denote the volume, B the base, and H the al- titude of the cone.

We are to prove V = \ B X II.

Let the volume of an inscribed pyramid AC DEFG be denoted by F, and its base by B'.

II will also be the altitude of this pyramid.

Then V' = %B'XH, §574

Now, let the number of lateral faces of the inscribed pyra- mid be indefinitely increased, the new edges continually bisect- ing the arcs in the base of the cone.

Then V approaches to V as its limit, and B' to B as its limit.

But however great the number of lateral faces of the pyramid,

p-j	B' X H.
. r— \	B X H.		§ 199 Q. E. D.
If the	cone be a cone	of ]	'evolution,
of the	base, then B =	irtf	1 (§381);

665. Corollary 1. and R be the radius .-. V = \ttR2X H.

666. Cor. 2. Similar cones of revolution are to each other as the cubes of their altitudes, or as the cubes of the radii of their bases. For, let R and R' be the radii of two similar cones of revolution, H and II' their altitudes, V and V their volumes. Since the generating triangles are similar, we have

Hill' :: R : R'.

V'~~hv K/2X H'~ Rh H'~~ H'*~~ R'z'

348 GEOMETRY. — BOOK VII.

Proposition XXXVIII. Theorem. 667. A frustum of any cone is equivalent to the sum of three cones whose common altitude is the altitude of the frus- tum and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum.

h j

Let V denote the volume of the frustum, B its lower base, b its upper base, and H its altitude.

We are to prove V=±H(B+b + y/ B X b).

Let V denote the volume of an inscribed frustum of a pyra- mid, B' its lower base, b' its upper base. Its altitude will also be H.

Then, V = J H {B' + b' + y/ B> X b'\ § 578

(a frustum of any pyramid is ^ to the sum of three pyramids whose common altitude is the altitude of tfie frustum, and whose bases are the lower base, the upper base, and a mean proportional between the bases of the frustum).

Now, let the number of lateral faces of the inscribed frus- tum be indefinitely increased,

the new edges continually bisecting the arcs in the bases of the frustum of the cone.

But however great the number of lateral faces of the frus- tum of the pyramid,

V = \H(B' + V + y/ B' X V.

A V=$H(B+ b+ s] BXb). § 199

Q. E. D.

668. Corollary. If the frustum be that of a cone of revo- lution, and R and r be the radii of its bases, we have B = it R2, and b = n r2,

and \/ BXb = 7rJRr.

.'. V= J *•#(/?»+ r2+ Rr).

BOOK VIII.

THE SPHERE.

On Sections and TANGENTa

669. Def. A Sphere is a solid bounded by a surface all points of which are equally distant from a point within called the centre. A sphere may be generated by the revolution of a semicircle about its diameter as an axis.

670. Def. A Radius of a sphere is the distance from its centre to any point in the surface. All the radii of a sphere are equal.

671. Def. A Diameter of a sphere is any straight line passing through the centre and having its extremities in the surface of the sphere. All the diameters of a sphere are equal, since each is equal to twice the radius.

672. Def. A Section of a sphere is a plane figure whose boun- dary is the intersection of its plane with the surface of the sphere.

673. Def. A line or plane is Tangent to a sphere when it has one, and only one, point in common with the surface of the sphere.

674. Def. Two spheres are tangent to each other when their surfaces have one, and only one, point in common.

675. Def. A polyhedron is circumscribed about a sphere when all of its faces are tangent to the sphere. In this case the sphere is inscribed in the polyhedron.

676. Def. A 2Joli/hedron is inscribed in a sphere when all of its vertices are in the surface of the sphere. In this case th .■ sphere is circumscribed about the polyhedron.

677. Def. A Cylinder or cone is circumscribed about a sphere when its bases and cylindrical surface, or its base and conical surface, are tangent to the sphere. In this case the sphere is inscribed in the cylinder or cone.

350

GEOMETRY.

BOOK VIII.

Proposition I. Theorem. 678. Every section of a sphere made by a plane is a circle.

Let the section ABC be a, plane section of a sphere whose centre is 0.

We are to prove section ABC a circle.

From the centre 0 draw 0 D _L to the section, and draw the radii 0 A, OB, 0 C, to different points in the boundary of the section.

In the rt. A 0 D A, 0 D B and ODC,

0 D is common, and 0 A, 0 B and 0 C are equal,

(being radii of the sphere).

.". the rt. AODA,ODB and ODC are equal, § 109

(two rt. A are equal when they have a side and hypotenuse of the one equal respectively to a side and hypotenuse of the other).

.'. DA, D B and D C are equal, (being homologous sides of equal &).

.*. the section A B C is a circle whose centre is D.

Q. E. D.

679. Corollary 1. The line joining the centres of a sphere and a circle of a sphere is perpendicular to the circle.

680. Cor. II. If K, r and p, respectively, denote the radius of a sphere, the radius of a circle of a sphere, and the per- pendicular from the centre of the sphere to the circle, then r = y R2 — p*. Therefore all circles of a sphere equally distant from the centre are equal, and of two circles unequally distant from the centre of the sphere the more remote is the smaller.

Again, if p — 0, then r = R, and the centre of the sphere and the centre of the circle coincide ; such a section is the greatest possible circle of the sphere.

THE SPHERE.

351

G81. Def. A Great circle of a sphere is a section of the sphere made by a plane passing through the centre.

682. Def. A Small circle of a sphere is a section of the sphere made by a plane not passing through the centre.

683. Def. An Axis of a circle of a sphere is the diameter (if the sphere perpendicular to the circle; and the extremities of the axis are the Poles of the circle.

684. Every great circle bisects the sphere. For, if the parts be separated and placed with their plane sections in coinci- dence and their convexities turned the same way, their convex surfaces will coincide ; otherwise there would be points in the spherical surface unequally distant from the centre.

685. Any two great circles, ABC D and AECF, bisect each other. For the intersection A C of their planes passes through the centre of the sphere, and is a diameter of each circle.

686. An arc of a great circle may be drawn through any two given points A and E in the surface of a sphere. For the two points A and E, and the centre 0, determine the plane of a great circle whose circumference passes through A and E. §443

If, however, the two given points are the extremities A and C of the diameter of the sphere, the position of the circle is not determined. For, the points A, O and C, being in the same straight line, an infinite number of planes can pass through

them.

§441

687. One circle, and only one, may be drawn through any three given points on the surface of a sphere. For, the three points determine the plane which cuts the sphere in a circle.

352

GEOMETRY. BOOK VIIT.

Proposition II. Theorem.

688. A plane perpendicular to a radius at its extremity is tangent to the sphere.

Let 0 be the centre of a sphere, and M N a plane per- pendicular to the radius 0 P, at its extremity P.

We are to prove M N tangent to the sphere.

From 0 draw any other straight line 0 A to the plane M N.

OP<OA, §448

(a ± is the shortest distance from a point to a plane).

.*. point A is without the sphere.

But 0 A is any other line than 0 P,

.'. every point in the plane M N is without the sphere, except P.

.'. M N is tangent to the sphere at P. § 673

Q. E. D.

689. Corollary 1. A plane tangent to a sphere is perpen- dicular to the radius drawn to the point of contact.

690. Cor. 2. A straight line tangent to a circle of a sphere lies in a plane tangent to the sphere at the point of contact.

691. Cor. 3. Any straight line in a tangent plane through the point of contact is tangent to the sphere at that point.

692. Cor. 4. The plane of any two straight lines tangent to the sphere at the same point is tangent to the sphere at that point.

THE SPHERE. 353

Proposition III. Problem. 693. Given a material sphere to find its diameter.

P

P

C> ^.

<r

d\ --■>& /f...%

\ \ / /

A' ' pi P'

Let P B P' G represent a material sphere. It is required to find its diameter.

From any point P of the given surface, with any opening of the compasses, describe the circumference A B G on the surface.

Then the straight line P B, being the opening of the compasses, is a known line.

Take any three points A, B and G in this circumference, and with the compasses measure the rectilinear distances A B, B Guild GA.

Construct the A A' B' C, with its sides equal respectively to A B, B G and G A. §232

Circumscribe a circle about the A A' B' G'. § 239

The radius D' B' of this O is equal to the radius of O A BG.

Construct the rt. Abdp, having the hypotenuse b p=B P, and one side b d = B' D'.

Draw b p1 J_ to b p, and meeting p d produced in p'.

Then p p' is equal to the diameter of the given sphere.

For, if we bisect the sphere through P and B, and in the section draw the diameter P P' and chord BP', the A bpp', when applied to A B P P', will coincide with it.

Q. E. F.

354

GEOMETRY. — BOOK VIII.

Proposition IV. Theorem.

694. Through any four points not in the same plane, one spherical surface can be made to pass, and but one.

Let A, B, C, D, be four points not in the same plane.

We are to prove that one, and only one, spherical surface can be made to pass through A, B, C, D.

Construct the tetrahedron A BC D, having for its vertices A, B, C, D.

Let E be the centre of the circle circumscribed about the face ABC.

Draw EM J_ to this face.

Every point in E M is equally distant from the points A, B and C, § 450

(oblique lines drawn from a point to a plane at equal distances from tlie foot of the _L are equal).

Also, let F be the centre of the circle circumscribed about the face BC D ;

and draw F K _L to this face.

Let H be the middle point of B C.

Draw EH and F H.

Then E H and FH are JL to B C § 184

THE SPHERE. 355

.'.the plane passed through EH and FH is 1_ to BC, § 449 (if a straight line be JL to two straight lines drawn through its foot in a plane, it is A. to the plane, and in this case the plane is _L to the line).

Hence, this plane is also _L to each of the faces ABC and BOB, §471

(if a straight line be ± to a plane, every plane ixisscd through that line is _L to the plane).

.'. the J« E M and FK lie in the plane EHF.

Hence they must meet unless they be parallel.

But if they were II, the planes BCD and ABC would be one and the same plane, which is contrary to the hypothesis.

Now 0, the point of intersection of the J* E M and FK, is equally distant from A, B and C ; and also equally distant from B, C and D ;

.*. it is equally distant from A, B, C and D.

Hence, a spherical surface, whose centre is 0, and radius 0 A, will pass through the four given points.

Only one spherical surface can be made to pass through the points A, B, C and D.

For the centre of such a spherical surface must lie in both the J*E M and FK.

And, since 0 is the only point common to these J», 0 is the centre of the only spherical surface passing through A, B, C and D.

Q. E. D.

695. Corollary 1. The four perpendiculars erected at the centres of the faces of a tetrahedron meet at the same point.

696. Cor. 2. The six planes perpendicular to the six edges of a tetrahedron at their middle point will intersect at the same point.

356 GEOMETRY. — BOOK VIII.

Proposition V. Theorem. 697. A sphere may be inscribed in any given tetrahedron. D

B Let ABCB be the given tetrahedron.

We are to prove that a sphere may be inscribed in ABC B.

Bisect the dihedral A at the edges A B, B C and A C by the planes 0 A B, 0 B C and 0 AC respectively.

Every point in the plane 0 A B is squally distant from the faces ABC and ABB, § 477

For a like reason, every point in the plane 0 B C is equally distant from the faces ABC and BBC;

and every point in the plane 0 A C is equally distant from the faces A B C and A B C.

.*. 0, the common intersection of these three planes, is equally distant from the four faces of the tetrahedron.

.*. a sphere described with 0 as a centre, and with the radius equal to the distance of 0 to any face, will be tangent to each face, and will be inscribed in the tetrahedron. § 673

Q. E. D.

698. Corollary. The six planes which bisect the six dihe- dral angles of a tetrahedron intersect in the same point.

On Distances Measured on the Surface of the Sphere.

699. Def. The distance between two points on the surface of a sphere is understood to be the arc of a great circle joining the points, unless otherwise stated.

700. Def. The distance from the pole of a circle to any point in the circumference of the circle is the Polar distance *>f the circle.

THE SPHERE.

357

Proposition VI. Theorem.

701. The distances measured on the surface of a sphere from all points in the circumference of a circle to its pole are equal. p

A'

A^lr	/ ;i**^ n
/ \S*»»«-_ V
K^jp	JQL_/ \
\ * 1 '<	if j

Let P,F be the poles of the circle ABC.

We are to prove arcs PA, PB, PC equal.

The straight lines PA, PB and PC are equal, § 450 {oblique lines drawn from a point to a plane at equal distances from the foot of the ± are equal) ;

."• the arcs P A, P B and P C are equal, § 182

(in equal (D equal chords subtend equal arcs).

In like manner arcs Pf A, P' B and P' C are equal.

Q. E. D.

702. Corollary 1. The polar distance of a great circle is a quadrant. Thus, arcs PA', PB', P' A', P' B', polar distances of the great circle A' B' C D', are quadrants ; for they are the meas- ures of the right angles A' OP, B' 0 P, A' 0 P', B' 0 P', whose vertices are at the centres of the great circles PA'P'C, PB'P'B'.

703. Scholium. Every point in the circumference of a small circle is at unequal distances from the two poles of the circle.

358

GEOMETRY.

BOOK VIII.

Proposition VII. Problem.

704. To pass a circumference of a great circle through any two points on the surface of a sphere.

Let A and B be any two points on the surface of a

sphere.

It is required to pass a circumference of a great circle through A and B.

Prom iasa pole, with an arc equal to a quadrant, strike an arc a b,

and from B as a pole, with the same radius, describe an arc c d, intersecting a b at P.

Then a circumference described with a quadrant arc, with P as a pole, will pass through A and B and be the circumference of a great circle.

Q. E. F.

705. Corollary. Through any two points on the surface of a sphere, not at the extremities of the same diameter, only one circumference of a great circle can be made to pass.

706. Scholium. By means of poles arcs of circles may be drawn on the surface of a sphere with the same facility as upon a plane surface, and, in general, the methods of construction in Spherical Geometry are similar to those of Plane Geometry. Thus we may draw an arc perpendicular to a given spherical arc, bisect a given spherical angle or arc, make a spherical angle equal to a given spherical angle, etc., in the same way that we make analogous constructions in Plane Geometry.

THE SPHERE. 359

Proposition VIII. Theorem.

707. The shortest distance on the surface of a sphere between any two points on that surface is the arc, not greater than a semi-circumference, of the great circle which joins them.

Let A B be the arc of a great circle which joins any two points A and B on the surface of a sphere ; and let A C PQB be any other line on the surface between A and B.

We are to prove arc ABKACPQB.

Let P be any point in A C P Q B.

Pass arcs of great circles through A and P, and P and B. § 704

Join A, P and B with the centre of the sphere 0. The A A OB, AOP and POB are the face A of the tri- hedral A whose vertex is at 0.

The arcs A B, A P and P B are measures of these A. § 202

NowZAOB<A AOP + A POB, §487

(the sum of any two face A of a trihedral is > the third Z.).

.'. arc A B < arc A P + arc P B.

In like manner, joining any point in A C P with A and P by arcs of great (D, their sum would be greater than arc A P ;

and, joining any point in P Q B with P and B by arcs of great (D, the sum of these arcs would be greater than arc P B.

If this process be indefinitely repeated the distance from A to B on the arcs of the great © will continually increase and approach to the line A C P Q B.

.\a.vcAB<ACPQB.

Q. E. D.

360

GEOMETRY.

BOOK VIII.

Proposition IX. Theorem.

708. Every point in an arc of a great circle bisects a given arc at right angles is equally distant from the extremities of the given arc.

Let arc CD bisect arc A £ at right angles.

We are to prove any point 0 in G D is equally distant from A and B.

Since great circle CDE bisects arc A B at right angles, it also bisects chord A B at right angles.

Hence, chord A B is _L to the plane (7 Z>^ at K.

.'. 0 K is J_ to chord A B at its middle point.

.'.straight lines 0 A and OB are equal.

.'.arcs 0 A and OB are equal.

§430

§58

§182

Q. E. D.

Proposition X. Problem.

709. To pass the circumference of a small circle through any three points on the surface of a sphere.

Let A, B and C be any three points on the surface of a sphere.

It is required to pass the circum- ference of a small circle through the points A, B and C.

Pass arcs of great circles through A and B, A and C, B and 0. § 704

Arcs of great circles a o and b o J_ to A C and B C at their middle points intersect at o.

Then o is equally distant from A, B and C. § 708

.*. the circumference of a small circle drawn from o as a

pole, with an arc o A will pass through A, B and C, and be the

circumference required.

Q. E. D.

THE SPHERE. 361

On Spherical Angles.

710. Def. The angle of two curves which have a common point ia the angle included by the two tangents to the two curves at that point.

711. Def. A spherical angle is the angle included between two arcs of great circles.

Proposition XI. Theorem.

712. The angle of two curves which intersect on the sur- face of a sphere is equal to the dihedral angle between the planes passed through the centre of the sphere, and the tan- gents of the two curves at their point of intersection.

Let the curves A B and A C intersect at A on the sur- face of a sphere whose centre is 0 ; and let A T and A S be the tangents to the two curves re- spectively.

We are to prove Z T AS equal to the dihedral angle formed by the planes OAT and 0 A S.

Since A T and A S do not cut the curves at A, they do not cut the surface of the sphere,

and are therefore tangents to the sphere. .'.AT and A S are J_ to the radius 0 A, drawn to the point of contact. § 186

.*. Z T AS measures the dihedral Z of the planes OAT and 0 A S, passed through the radius 0 A and the tangents A T and AS. § 470

But Z TA S is the Z of the two curves A B and A C. § 710

.'. the Z of the two curves A B and AC = the dihedral Z of the planes 0 A T and 0 A S.

Q. E. D.

362

GEOMETRY.

BOOK VIII.

Proposition XII. Theorem.

713. A spherical angle is equal to the measure of the dihedral angle included by the great circles whose arcs form the sides of the angle.

P

Let BPC be any spherical angle, and B P D P' and C P E P' the great circles whose arcs B P and C P include the angle.

We are to prove /.BPC equal to the measure of the dihe- dral Z C-PP'-B.

Since two great © intersect in a diameter, P P' is a diameter. § 685

Draw P T tangent to the O BPDP1.

Then P T lies in the same plane as the O B P D P', and is

_L to PP< at P.

In like manner draw P T' tangent to the O CPEP'. Then P T' lies in the same plane as the O C P EP', and is

-L to PP' at P.

.'. Z TPT is the measure of the dihedral Z C-PP'-B. § 470 But spherical Z B P C is the same as plane ZTPT'; § 7 1 0 .*. spherical Z BPC is equal to the measure of dihedral

Z C-PP'-B.

Q. E. D.

714. Corollary. A spherical angle is measured by the art of a great circle described about its vertex as a pole and intercepted by its sides (produced if necessary). For, if B C be the arc of a great circle described about the vertex P as a pole, P B and P C are quadrants. Hence, B 0 and C 0 are perpendicular to P P'. Therefore BO C measures the dihedral angle B-P O-C, and, hence, the spherical angle BPC. Therefore, arc B C, which measures the angle BO C, measures the spherical angle BPC.

THE SPHERE. 363

On Spherical Polygons and Pyramids.

715. Def. A spherical Polygon is a portion of a surface of a sphere bounded by three or more arcs of great circles.

The sides of a spherical polygon are the bounding arcs ; the angles are the angles included by consecutive sides; the vertices are the intersections of the sides.

716. Def. The Diagonal of a spherical polygon is an arc of a great circle dividing the polygon, and terminating in twG vertices not adjacent.

The planes of the sides of a spherical polygon form by their intersections a polyhedral angle whose vertex is the centre of the sphere, and whose face angles are measured by the sides of the polygon.

717. Def. A spherical Pyramid is a portion of a sphere bounded by a spherical polygon and the planes of the sides of the polygon.

The spherical polygon is the base of the pyramid, and the centre of the sphere is its vertex.

718. Def. A spherical Triangle is a spherical polygon of three sides.

A spherical triangle, like a plane triangle, is right, or oblique ; scalene, isosceles or equilateral.

719. Def. Two spherical triangles are equal if their suc- cessive sides and angles, taken in the same order, be equal each to each.

720. Def. Two spherical triangles are symmetrical if their successive sides and angles, taken in reverse order, be equal each to each.

721. Def. The Polar of a spherical triangle is a spherical triangle, the poles of whose sides are respectively the vertices of the given triangle.

Since the sides of a spherical triangle are arcs, they may be expressed in degrees and minutes.

364

GEOMETRY. BOOK VIII.

Proposition XIII. Theorem.

722. Any side of a spherical triangle is less than the of the other two sides.

Let ABC be any spherical triangle. We are to prove BG < B A + AG. Join the vertices A, B and G with the centre 0 of the sphere.

Then, in the trihedral A O-A BG thus formed, the face A A 0 G, AOB and BOG are measured, respectively, by the sides A G, A B and B G. § 202

Now, BOG<BOA + AOG, § 487

any two A of a trihedral is greater than the third Z. ).

.'.BG<BA + AG.

Q. E. D.

sum of

723. Corollary. Any side of a spherical polygon is less than the sum of the other sides.

Ex. 1. Given a cone of revolution whose side is 24 feet, and the diameter of its base 6 feet ; find its entire surface, and its volume.

2. Given the frustum of a cone whose altitude is 24 feet, the circumference of its lower base 20 feet, and that of its upper base 16 feet; find its volume.

3. The volume of the frustum of a cone of revolution is 8025 cubic inches; its altitude 14 inches; the circumference of the lower base twice that of the upper base. What are the cir- cumferences of the bases 1

4. The frustum of a cone of revolution whose altitude is 20 feet, and the diameters of its bases 12 feet and 8 feet respec- tively, is divided into two equal parts by a plane parallel to its bases. What is the altitude of each part 1

THE SPHERE.

365

Proposition XIV. Theorem.

724. The sum of the sides of a spherical polygon is less than the circumference of a great circle.

Let ABC BE be a spherical polygon.

We are to prove AB + B G etc. less than the circumference of a great circle.

Join the vertices A, B, G etc., with 0 the centre of the sphere.

The sum of the face A A 0 B, BOG etc., which form a polyhedral Z at 0, is less than four rt. A . § 488

.*. the sum of the arcs A B, B G etc., which measure these face A , is less than the circumference of a great circle.

Q. E. D.

725. Corollary. If we denote the sides of a spherical tri- angle by a, b and c, then a + b + c < 360°.

Ex. 1. Tho surface of a cone is 540 square inches; what is the surface of a similar cone whose volume is 8 times as great 1

2. The lateral surface of a cone is S ; what is the lateral surface of a similar cone whose volume is n times as great ]

366

GEOMETRY. BOOK VIII.

Proposition XV. Theorem.

726. A point upon the surface of a sphere, which is at the distance of a quadrant from each of two other points, is one of the poles of the great circle which passes through these points.

Let P be a point at the distance of a quadrant from each of the two points A and B.

We are to prove P a pole of the great circle which passes through A and B.

Since PA and P B are quadrants,

A POA and P 0 B me it A.

.'. P 0 is _L to the plane of the O ABC, § 449

(a straight line _L to two straight lines drawn through its foot in a plane is _L to the -plane).

P is a pole of the O A B G.

§ 683

Q. E. D.

Ex. 1. Show that two symmetrical polyhedrons may be de- composed into the same number of tetrahedrons symmetrical each to each.

2. Show that two symmetrical polyhedrons are equivalent.

3. Show that the intersection of two planes of symmetry of a solid is an axis of symmetry.

4. Show that the intersections of three planes of symmetry of a solid are three axes of symmetry; and that the common intersection of these axes is the centre of symmetry.

THE SPHERE.

367

Proposition XVI. Theorem.

727. If, from the vertices of a given spherical triangle as poles, arcs of great circles be described, another triangle is formed, the vertices of which are the poles of the sides of the given triangle.

Let A B G be the given triangle; and, from its vertices A, B and G as poles, let the arcs B'C, A' C and A' B' respectively be described.

We are to prove vertices A', B' and C poles respectively of arcs BC, A G and A B.

Since B is the pole of the arc A' C', and G the pole of the oxcA'B',

A' is at a quadrant's distance from each of the points B and G.

,'J'isa pole of the arc B G, § 726

(a point upon the surface of a sphere which is at a quadrant's distance from each of two other points is one of the poles of the great circle which passes through those points).

In like manner, it may be shown that B' is a pole of the arc A G, and G' a pole of the arc A B.

Q. E. D.

728. Scholium 1. A A1 B' G' is the polar of A A B G, and, reciprocally, A A B C is the polar of A A' B' G'.

729. Sch. 2. The arcs of great circles described about A, B and C as poles will, if produced, form three triangles exterior to the polar. The polar triangles are distinguished by having their homologous vertices A and A' on the same side of B G and B1 G', B and B' on the same side of A G and A' G', and G and G' on the same side of A B and A' B'.

368 GEOMETRY. BOOK VIII.

Proposition XVII. Theorem. 730. In two polar triangles each angle of either is the supplement of the side lying opposite to it in the other.

Let ABG and A' B' C be two polar triangles.

We are to prove A A, B and G respectively the supplements of the sides B' C, A1 C and A' B'.

Let the sides A B and A G, produced if necessary, meet the side B' G' in the points b and c.

Since the vertex A is a pole of the arc B' C', §721

A A is measured by b c, § 714

(a spherical Z is measured by the arc of a great circle described about its vertex as a pole and intercepted by its sides).

Now, since B' is the pole of the arc Ac, B' c = 90°.

Since G' is the pole of the arc Ab,C'b = 90°.

.-. B'c+G/b = B,C,+ bc=l80°.

.'. Z A ( = b c) is the supplement of the side B' G'.

In like manner it may be shown that each A of either A is the supplement of the side lying opposite to it in the other.

Q. E. D.

731. Scholium. In two polar triangles each side of either is the supplement of the angle lying opposite to it in the other. If A, B and G denote the angles, and a, b and c the sides of a tri- angle, the angles of the polar triangle will be 180° — a, 180° — b and 180° — c; and the sides of the polar triangle will be 180° - A, 180° - B and 180° - G.

By reason of these relations polar triangles are often called supplemental triangles.

THE- SPHERE. 369

Proposition XVIII. Theorem.

73£. The sum of the angles of a spherical triangle is greater than two, and less than six, right angles.

Let ABC be a spherical triangle.

We are to prove ZA+ZB+ZC greater than 2, and less than 6, right angles.

Denote the sides of the polar A opposite the A A,B, C re- spectively, by a', b', cf.

Then Z A = 180° -a', Z B= 180° - V and Z C = 180° - c7, § 730

(in two polar A each Z of either is tJie sitjqrtement of the side lying opposite to it in the other.)

By adding, ZA + ZB + ZC = 540° - (a' + V + c').

But a' + V + d is less than 360°, § 724

(the sum of the sides of a spherical polygon is less than the circumference of a great circle).

.'.ZA + ZB + ZOI&00.

Also, since each Z is less than 2 rt. A,

their sum is less than 6 rt. A.

Q. E. D.

733. Corollary. A spherical triangle may have two, or even three right angles ; or two, or even three obtuse angles.

734. Def. A spherical triangle having one right angle is called rectangular; having two right angles, bi-rectangular ; having three right angles, tri-rectangular.

Each of the sides of a tri-rectangular triangle is a quadrant, and the triangle is called, when reference is had to its sides, tri- quadrantal.

370 GEOMETRY. BOOK VIII.

Proposition XIX. Theorem.

735. Each angle of a spherical triangle is greater than the difference between two right angles and the sum of the other two angles.

Let AA,B and C be the angles of the spherical tri- angle ABC.

We are to prove Z A greater than the difference between 1 80° and(ZB + ZC).

I. Suppose (Z B + Z C) < 180°.

Now ZA + ZB + ZC> 180°. § 732

By transposing, Z A > 180° - (Z B + Z C).

II. Suppose (Z B + Z C) > 180°.

Now of the three sides (180° - Z A), (180° - Z B), (180° — Z C), of the polar A, each is less than the sum of the other two, § 722

{cither side of a spherical A is less than the sum of the other two sides).

.'. (180° - Z B) + (180° - Z C) > 180° - Z A ; ' or, 360° - (Z B + Z C) > 180° - Z A.

By transposing, Z A>{ZB + ZC)~ 180°.

Q. E. D.

Ex. 1. The volume of a cone is 1728 cuhic inches; what is the volume of a similar cone whose surface is 4 times as great 1

2. The volume of a cone is V ; what is the volume of a simi- lar cone whose surface is n times as great 1

THE SPHERE.

371

736. Def. Equal spherical triangles are triangles which have their corresponding sides and angles equal each to each and arranged in the same order, so that when applied to each other they will coincide. Thus in Fig. 1, ABC and A' B' C are equal spherical triangles.

Fig. 1. Fig. 2.

737. Def. Symmetrical spherical triangles are triangles which have their corresponding sides and angles equal each to each, but arranged in reverse order.

Thus, in Fig. 2, A B C and A' B' C are symmetrical spheri- cal triangles. For, since the face angles of the two trihedrals are equal respectively, but are arranged in reverse order, the sides of the spherical triangles, which measure these face angles, are equal, each to each, and are arranged in reverse order ; and since the dihedral angles of the two trihedrals are equal respec- tively, but are arranged in reverse order, the angles of the spherical triangles, which are equal to these trihedrals, are equal, each to each, and are arranged in reverse order.

In like manner we may have symmetrical spherical poly- gons of any number of sides, and corresponding symmetrical spherical pyramids.

Two symmetrical spherical triangles cannot be made to coincide. For, if their convexities lie in opposite directions, they evidently will not coincide ; and if their convexities lie in the same direction, and we apply A B to A' B', the vertices G and C will lie on opposite sides of A1 B'.

372 GEOMETRY. BOOK VIII.

738. There is, however, one exception. Two symmetrical isosceles spherical triangles can be made to coincide.

Thus, if A B C be an isosceles spherical triangle, AB = AO and in its symmetrical triangle A1 B' = A' C. Hence A B = A' C and AC = A' B'. And, since A A and A! are equal, if A B be placed on A' C, A G will fall on its equal A' B'.

In consequence of the relations established between poly- hedral angles and spherical polygons, from any property of poly- hedral angles, we may infer a corresponding property of spherical polygons. Reciprocally, from any property of spherical polygons, we may infer a corresponding property of polyhedral angles.

Ex. 1. The altitude of a cone of revolution is 12 inches ; at what distances from the vertex must three planes be passed par- allel to the base of the cone, in order to divide the lateral surface into four equal parts 1

2. The altitude of a given solid is 2 inches, its surface 24 square inches, and its volume 8 cubic inches ; find the altitude and surface of a similar solid whose volume is 512 cubic inches.

3. The volumes of two similar cones of revolution are 6 cubic inches and 48 cubic inches respectively, and the slant height of the first is 5 inches ; find the slant height of the second.

THE SPHERE.

373

Proposition XX. Theorem. 739. Two symmetrical spherical triangles are equivalent.

Let ABC and A1 B' C be two symmetrical spherical triangles, having A B, A C and B C equal respectively to A' B>, A'C andB'C.

We are to prove A ABC ^ A A' B'C.

Let P and P' be poles of small circles which pass through A, B, C and A', B', C.

Now, since the arcs A B, A C and B C = A' B', A' C and B' C respectively, the chords of the arcs AB, AC and B C = chords of the arcs A' B'} A' C and B' C respectively. § 181

.*. the plane A formed by the chords of these arcs are equal. § 108

.*. ©ABC and A1 B' C which circumscribe these equal plane A are equal.

.*. the six spherical distances PA, P B, P' A' etc. are equal, {being polar distances of equal (D on tlie same sphere).

,' . A P A B, P' A' B' are symmetrical and isosceles.

So likewise are A P B C, P' B' C and A PAC, P'A' C.

.'. A P AB may be applied to A P' A1 B' and will coincide with it. § 738

So likewise A PBC with A P' B' C and A PAC with A P' A' C.

.'. APAB + PBC-PAC^AP'A'B'+ P< B1 C - P'A'C.

.'.A ABC- A A' B'C

Q. E. D.

740. Corollary. Two symmetrical spherical pyramids are equivalent.

374

GEOMETRY. BOOK VIII.

Proposition XXI. Theorem. 741. On the same sphere, or equal spheres, two triangles are either equal, or symmetrical and equivalent, if two sides and the included angle of the one be respectively equal to two sides and the included angle of the other.

In the AABG and B E F, let Z A = Z B, and the sides A B and A G equal respectively the sides BE and D F.

We are to prove A A B C and D E F equal, or symmetrical and equivalent.

I. When the parts of the two A are in the same order as in A

ABC and BE F,

A A B G can be applied to A B E F, as in the corre- sponding case of plane A, and will coincide with it. § 106

II. When the parts are in reverse order, as in A A B G and

B' E1 F,

construct the A BE Asymmetrical with respect to A B'E'F.

Then A B E F will have its A and sides equal respectively to those of the A B'E'F. § 737

Now in the A A B G and B E F,

Z A=Z B, AB = BE and A G = B F,

and these parts are arranged in the same order.

.'. A A B G = A B EF. Case I.

But A B'E'F- A BEF, § 739

.'.AABG^AB'E'F.

Q. E. D.

THE SPHERE.

m

Proposition XXII. Theorem.

742. Two triangles on the same sphere, or equal spheres, are either equal, or symmetrical and equivalent, if a side and two adjacent angles of the one be equal respectively to a side and two adjacent angles of the other.

For one of the A may be applied to the other, or to its sym- metrical A, as in the corresponding case of plane A. § 107

Q. E. D.

Proposition XXIII. Theorem.

743. Two mutually equilateral triangles on the same sphere, or equal spheres, are mutually equiangular, and are either equal, or symmetrical and equivalent.

For the face A of the corresponding trihedral angles at the centre of the sphere are equal respectively, § 202

(since they arc measured by equal sides of the A). .*. the corresponding dihedral A are equal. § 492

.*. the A of the spherical A are respectively equal. .'. the A are either equal, or symmetrical and equivalent, according as their equal sides are arranged in the same, or reverse order.

Q. E. D.

376

GEOMETRY. BOOK VIII.

Proposition XXIV. Theorem.

744. Two mutually equiangular triangles on the same sphere, or equal spheres, are mutually equilateral, and are either equal, or symmetrical and equivalent.

Let the spherical triangles ABC and D E F be mutually equiangular.

We are to prove A A B C and DBF mutually equilateral, and equal, or symmetrical and equivalent.

Let A A' B' C and D' E' F be the polar A of A A B 0 and D E F respectively.

Then the A A' B' C and D' E' F are mutually equilat- eral, § 731

(in two polar A each side of the one is the supplement of the A lying opposite to it in the other).

.'. A A'B'C and D E' F are mutually equiangular, § 743 (two mutually equilateral A on equal spheres are mutually equiangular).

.'. A A B C and DEF are mutually equilateral ; § 731

hence A A B C and D E F are either equal, or symmetri- cal and equivalent, § 743 (two mutually equilateral A on equal spheres are either equal, or symmetrical

and equivalent).

Q. E. D.

THE SPHERE. 377

Proposition XXY. Theorem.

745. The angles opposite equal sides of an isosceles spherical triangle are equal.

In the spherical A A B C, let A B = AC. We are to prove Z. B = Z C.

Draw arc A D of a great circle, from the vertex A to the middle of the base B C.

Then A A B D and A C D are mutually equilateral.

.'. A A B D and A CD are mutually equiangular, § 743 {two mutually equilateral & on the same sphere are mutually equiangular).

.-.ZB = ZC,

(since tliey are lwmologous A of symmetrical &).

Q. E. D.

746. Corollary. The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle of the base bisects the vertical angle, is perpendicular to the base, and di- vides the triangle into two symmetrical triangles.

378

GEOMETRY. BOOK VIII.

Proposition XXVI. , Theorem.

747. If two angles of a spherical triangle be equal, the sides opposite these angles are equal, and the triangle is

In the spherical A A B C, let Z B = Z C.

We are to prove A C = A B.

Let A A' B' a be the polar AofAi.BC.

Since Z B = ZC,

.\A/C, = A'Bf, (in two polar A each side of one is the supplement of the Z it in the other).

Hyp. §731

opposite to

.-. Z B' = Z C,

§745

(in an isosceles spherical A, the A opposite the equal sides are equal).

AC = AB.

§731

Q. E. D.

THE SPHERE. 379

Proposition XXVII. Theorem.

748. In a spherical triangle the greater side is opposite the greater angle ; and, conversely, the greater angle is oppo- site the greater side.

I. In the A ABC, let Z ABO Z 0. We are to prove A C > A B.

Draw the arc BDofa great circle, making Z 0 B D = Z C.

Then DC=DB, §747

(if two A of a spherical A be equal the sides opposite these A are equal).

Add D A to each of these equals ;

then DC + DA =DB + DA.

But DB + DA> AB, §722

(the sum of two sides of a splierical A is greater tlian the third side).

.\DC+DA>AB,otAOAB.

II. Let AC > A B. We are to prove Z A B C> Z C.

If ZABC = Z C, AC = AB, §747

andif Z ABC<Z C,AC<AB. Case I.

But both of these conclusions are contrary to the hypothesis. .*. Z A B C> Z C.

Q. E. D.

380 . GEOMETRY. BOOK VIII.

Proposition XXVIII. Theorem.

749. On unequal spheres mutually equiangular triangles are similar.

From 0, the common centre of two unequal spheres, draw the radii 0 A, 0 B and 0 C cutting the sur- face of the smaller sphere in a, b and c. Draw arcs of great circles, AB, AC, BC, ab, a c, be. We are to prove A AB C similar to A ab c.

A A, B, C are equal respectively to A a, b, c, (since the corresponding dihedrals in each case are the same).

In the similar sectors A 0 B and a Ob,

AB :ab : : A 0 :aO; §385

and in the similar sectors A 0 G and aOc,

AC :ac::AO :aO. § 385

.*. A B : ab :: A C : ac. In like manner, AB : ab : : B C :bc.

That is, the homologous sides of the two A are proportional, and their homologous A are equal.

.'.A A B C is similar to A ab c.

Q. E. D.

750. Scholium. The statement that mutually equiangular spherical A are mutually equilateral, and equal, or symmetrical and equivalent, is true only when limited to the same sphere, or equal spheres. But when the spheres are unequal, the spherical A are similar, but not equal. Hence, to compare two similar spherical A, it is necessary to know the linear extent of two homologous sides ; or, what is equivalent, to know the radii of the spheres. And, as in the case of plane A, two similar spheri- cal A have the same ratio as the squares of the linear measures of any two homologous sides, and therefore as the squares of the radii of the spheres.

THE SPHERE.

381

On Comparison and Measurement of Spherical Surfaces.

751. Def. A Lune is a part of the surface of a sphere in- cluded between two semi-circumferences of great circles.

752. Def. The Angle of a lune is ^ the angle included by the semi-circum- ferences which forms its boundary. Thus Z. CAB is the angle of the lune.

753. Def. A Spherical Ungula, or Wedge, is a part of a sphere bounded by a lune and two great semicircles.

754. Def. The Base of an ungula is the bounding lune.

755. Def. The Angle of an ungula is the dihedral of its bounding semicir- cles, and is equal to the angle of the bounding lune.

756. Def. The Edge of an ungula is the edge of its angle.

757. Def. The Spherical Excess of a spherical triangle is the excess of the sum of its angles over two right angles.

C 758. Def. Three planes which

pass through the centre of the sphere, each perpendicular to the other two, divide the surface of the sphere into eight tri-rectangular triangles. Thus 1 5 the three planes 0 A D B, CEDE and AEBF divide the surface o£ the sphere into the eight tri-rectangular triangles C E B, D E B, 0 B E, DB F, etc.

As in Plane Geometry the whole angular magnitude about any point in a plane is divided by two straight lines perpendicular to each other into four right angles, and each right angle is measured by a quadrant, or fourth part of a circumference described about that point as a centre with any given radius ; so, if, through a point in space, three planes be made to pass perpendicular to one another, they will divide the whole angular magnitude about that point into eight solid right angles, each of which is measured by an eighth part of the surface of a sphere described about that point with any given radius.

And, as in Plane Geometry, each quadrant which measures a right angle is divided into 90 equal parts called degrees, so each of the eight tri-rectangular spherical triangles is divided into 90 equal parts called degrees of surface. Hence, the whole surface of the sphere is divided into 720 degrees of surface.

382

GEOMETRY.

BOOK VIII.

Proposition XXIX. Lemma. 759. The area of the surface generated by the revolution of a straight line about another line in the same plane with it as an axis, is equal to the product of the projection of the line on the axis by the circumference whose radius is perpendicular to the revolving line erected at its middle point arid termi- nated by the axis.

Let the straight line A B revolve about the axis Y T in the same plane ; let E F be its projection on the axis; and G 0 the perpendicular to A B at its middle point G, and terminated in the axis. We are to prove area A B = E F X 2 it 0 G. The surface generated by A B is the lateral surface of the frustum of a cone of revolution.

Draw GH±, and A D II, to YY.

Then area A B = A B X 2 rr C H, §662

(the lateral area of a frustum of a cone of revolution is equal to the slant height multiplied by the circumference of a section equidistant from its bases).

The A ABD and G 0 H are similar ; § 287

.'.AD :AB :: GH : GO.

ButCH :GO ::2ttGH :2ttCO, §375

(circumferences of © have the same ratio as their radii).

.'.AD :AB ::2,rGH'.2irGO.

.•.ADX2ttGO = ABX 2 7rGff.

.'. area ofAB = ADX27rCO.

Now AD = EF. § 135

.'.2Lre&AB = EFX 2# GO.

Q. E. D.

760. Scholium. If either extremity of A B he in the axis YY', A B generates the lateral surface of a cone of revolution ; and if A B be parallel to the axis Y Y', it generates the lateral area of a cylinder of revolution. In either case the formula holds good

THE SPHERE.

Exercises.

1. If, from the extremities of one side of a spherical triangle, two arcs of great circles be drawn to a point within the triangle, the sum of these arcs is less than the sum of the other two sides of the triangle.

2. On the same sphere, or on equal spheres, if two spherical triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

3. To draw an arc perpendicular to a given spherical arc, from a given point without it.

4. At a given point in a given arc, to construct a spherical angle equal to a given spherical angle.

5. To inscribe a circle in a given spherical triangle.

6. Given a spherical triangle whose sides are 60°, 80°, and 100° ; find the angles of its polar triangle.

7. The volume of a pyramid is 200 cubic feet ; find the vol- ume of a similar pyramid which is three times as high.

8. Find the centre of a sphere whose surface shall pass through three given points, and shall touch a given plane.

9. Find the centre of a sphere whose surface shall pass through three given points, and shall also touch the surface of a given sphere.

10. Find the centre of a sphere whose surface shall touch two given planes, and also pass through two given points which lie between the planes.

384

GEOMETRY. BOOK VIII.

Proposition XXX. Theorem.

761. The area of the surface of a sphere is equal to the product of its diameter by the circumference of a great circle. A A

Lei ABODE be the circumference of a great circle, and AD the diameter, and OA the radius of a sphere.

We are to prove surface of sphere = ADX2irOA.

Let the semicircle and any regular inscribed semi-polygon revolve together about the diameter A D.

The semi-circumference will generate the surface of the sphere,

and the semi-perimeter a surface equal to the sum of the surfaces generated by the sides A B, B 0, CD, etc.

Draw from the centre 0, _k 0 ff, 0 1 and 0 K to the chords AB,BG, CD, etc.

These J? bisect the chords and are equal ; .'. area AB = AP X 2 ir Off;

area BC = PR X 2 tt 01; and area CD=RDX2ttOK.

§185 § 759

THE SPHERE. 385

Adding, and observing that 0 H, 0 1 and 0 K are equal,

area, ABCD = (A P+PR + RD)X2nOH. .'.area ABC D = AD X 2 tt OH.

Now, if the number of sides of the regular inscribed semi- polygon be indefinitely increased, the surface generated by tho semi-perimeter will approach the surface of the sphere as its limit, and 0 H will approach 0 A as its limit.

.'.at the limit we have

surface of the sphere = ADX2nOA. §199

Q. E. D.

762. Corollary 1. If 7? denote the radius of the sphere, then A D will equal 2 R, and 0 A will equal R. Hence the surface of a sphere equals 2 R X 2 it R = 4 ir R2.

763. Cor. 2. Since the area of a great circle of a sphere is equal to n R2 (§ 381), and the area of the surface of a sphere is equal to 4 n R2, the surface of a sphere is equal to four great circles.

764. Cor. 3. If we denote the surfaces of two spheres by S and aS7, and their radii by R and R'} we have £ : S' : : 4 it R2 : 4 ir R'2, or S : S' : : R2 : R'2 ; that is, the surfaces of two spheres have the same ratio as the squares on their radii.

765. Cor. 4. Since S = 4 n R2 = *r (2 R)\ the surface of a sphere is equivalent to a circle whose radius is equal to the diameter of the sphere.

386

GEOMETRY.

BOOK VIII.

Proposition XXXI. Theorem.

766. A lune is to the surface of the sphere as the angle of the lune is to four right angles.

Let L denote the lune AB EG whose angle is A; S, the surface of the sphere; and B G D F, a great circle whose pole is A.

L A

S~ 4 rt. A'

We are to x>rove

Now the arc B G measures the Z A of the lune ; § 714

and the circumference B G D F measures 4 rt. A .

Case I. — If BO and B CDF be commensurable.

Find a common measure of B C and BC D F.

Suppose this common measure to be contained in BG 3 times,

and m BGDF 25 times.

Then

4 rt. A

/ BG \ _ 3

\BGD F/~ 25'

Pass arcs of great © through A and these points of division. The entire surface will be divided into 25 equal lunes, of which lune L will contain 3.

" S ~ 25 * A ^ . L A

4 rt. A ~ 25 ' " S ~ 4 rt. A '

But

THE SPHERE. 387

Case II. — If BO and B CD F be incommensurable,

the proposition can be proved by the method of limits, as employed in § 201.

Q. E. D.

767. Corollary. If we denote the surface of the tri-rectan- gular triangle by T, the surface of the whole sphere will be 8 T (§ 758), aud if we denote the surface of the lune by L, and its angle by A, the unit of the angle being a right angle, we shall

have -—-„ = -• Therefore L = T X 2 A.

And if we take the tri-rectangular triangle as the unit of surface in comparing surfaces on the same sphere, we shall have L = 2 A. That is, if a right angle be the unit of angles and the tri-rectangular triangle be the unit of spherical surfaces, the area of a lune is expressed by twice its angle.

768. Scholium. We mag also obtain the area of a lune whose angle is known, on a given sphere, by finding the area of the sphere, and multiplying this area by the ratio of the angle of the lune, expressed in degrees, to 360°. Thus, if the angle of the lune be 60°, the area of the lune will be ^^ of the area of the sphere.

Ex. 1. Given the radius of a sphere is 10 feet; find the area of a lune whose angle is 30°.

2. Given the diameter of a sphere is 1 6 feet ; find the area of a lune whose angle is 75°.

3. Given the diameter of a sphere is 20 inches; find the entire surface of its circumscribed cylinder ; and of its circum- scribed cone, the vertical angle of the cone being 60°.

388 GEOMETRY. BOOK VIII.

Proposition XXXII. Theorem.

769. If two circumferences of great circles intersect on the surface of a hemisphere, the su?n of the opposite triangles thus formed is equivalent to a lune whose angle is equal to that included by the semi-circumferences.

Let the semi- circumferences BAD and GAE intersect at A on the surface of a hemisphere.

We are to p?we A A B G + A DAE equivalent to a lune whose angle is BAG.

The semi-circumferences produced intersect on the opposite hemisphere at A'.

Then each of the arcs A D and A' B is the supplement of AB,

(hvo great © bisect each other).

.-. AD = A' B.

In like manner, A E — A1 '0 and D M -■ B G.

.'. A A D E and A' BG are symmetrical and equiva- lent. § 743

.\AABG+AADE = AABG+ A A' B C = lune ABA'GA.

That is, A ABC + A ADE = lune whose Z is BA G.

Q. E. D.

770. Corollary. The sum of two spherical pyramids, the sum of whose bases is equivalent to a lune, is equivalent to a wedge whose base is the lune.

THE SPHERE.

Proposition XXXIII. Theorem. 771. The area of a spherical triangle is equal to the tri-rectangular triangle multiplied ly the ratio of the spherical excess of the given triangle to one right angle.

Let ABC be a spherical triangle, and T the area of the tri-rectangular triangle. We are to prove AABC = T(AA + B+C — 2).

Complete the circumference A B D E. Produce A C and B C to meet this circumference in D and E. Then A ABC + BCD (= lune A) = TX 2 Z A. § 767 AABC + ACE(= lune B) = T X 2 Z B, § 767

A A B C + I) C E (=luue C) (§ 769) = TX 2 Z C. §767 By adding these equalities, 2AABC+AABC+BCD+ACE+DCE = TX2(AA + B+C). But AABC + BCIJ + ACE+ D C E = \ T, §758

(the surface of a hemisphere is equal to 4 tri-rectcouju,lar &).

.-. 2 A ABC + 4 T= TX2 (AA+ B+ C);

.- . A A B C = T X (A A + B + C - 2).

Q. E. D.

772. Scholium 1. If Z A = 140°, Z^= 120° and Z C = 1 00°, a right angle being the unit,

then, A^C= W!ii° +i^! + 1^! - 2 W 2 ?'. \90° 90° 90° /

773. Scho. 2. To find the area of a spherical triangle on a given sphere, the angles of the triangle being given, we may multi- ply the area of the hemisp/iere by the ratio of the spheHcal excess to 360°.

Thus if Z A = 140°, Z B = 120° and Z C = 100°, since the hemisphere is 2 n R\ we have AABC = 2 n R2 X Z A + Z B + Z C- 180° )2 18<T _

360° " " 2 "" 7l>2 X 3605 ~" v ^

390

GEOMETRY. BOOK VIII.

Proposition XXXIV. Theorem. 774. The area of a spherical polygon is equal to ike tri-rectangular triangle multiplied by the ratio of the spherical excess to one right angle.

Let P denote the area, of the spherical polygon ; S the sum of its angles; n the number of its sides ; t, t', t" . . . the areas of the triangles formed by drawing diagonals from any vertex A ; s, s'} s" ... respec- tively the sums of the angles of these triangles; and T the tri- rectangular triangle.

We are to prove P = T [S — 2 (n — 2) ].

Now t = T (s - 2), § 771

(the area of a spherical A is equal to its spherical excess multiplied into the area of the tri-rectangular A).

t' = T (J - 2), § 771

and t" = T (s" — 2), . . .

By adding these equalities,

t + t' + t", . . . — T [s + s* + s" + . . . - 2 (n - 2) ].

But t + t' + t" + . . . = P-,

and s + s' + s" + . . . = 8.

,'.f = T[S-2 {n-2)].

THE SPHERE.

39J

775. Corollary. The volume of a spherical pyramid is to the volume of the tri-rectangular pyramid, as the base of the pyra- mid is to the tri-rectangular triangle. And, since the volume of the tri-rectangular pyramid is |- the volume of the sphere, and the area of the tri-rectangidar triangle is ^ of the surface of the sphere ; the volume of a spherical pyramid is to the volume of the sphere as its base is to the surface of the sphei*e.

776. Def. A Zone is the part of the surface of a sphere in- cluded between two parallel circles of the sphere ; as the surface included between the circles ABC and E FG.

777. Def. The Bases of a zone are the circumferences of the intercepting circles; as circumferences ABC and EFG. If the plane of one base become tangent to the sphere, that base becomes a point, and the zone will have but one base.

778. Def. The altitude of a zone is the perpendicular dis- tance between the planes of its bases.

779. Def. A Spherical Segment is a part of the sphere in- cluded between two parallel planes.

780. Def. The Bases of a spherical segment are the bound- ing circles.

One of the planes may become a tangent plane to the sphere. In this case the segment has but one base.

781. Def. The Altitude of a spherical segment is the per- pendicular distance between the planes of its bases.

392

GEOMETRY. BOOK VIII.

782. Def. A Spherical Sector is a part of a sphere gener- ated by a circular sector of the semicircle which generates the sphere ; as A 0 C K.

783. Def. The Base of a spherical sector is the zone gener- ated by the arc of the circular sector ; as AC K.

The other bounding surfaces of a spherical sector may be one conical surface, or two conical surfaces ; or one conical and one plane surface.

Thus, let A B be the diameter around which the semicircle AG B revolves to generate the sphere. The solid generated by the circular sector A 0 G will be a spherical sector having the zone AG K for its base, and for its other bounding surface the conical surface generated by CO.

The spherical sector generated by C 0 D has for its base the zone generated by G D, and for its other surfaces the concave conical surface generated by D 0, and the convex conical surface generated by G 0.

The spherical sector generated by E 0 F has for its base the zone generated by E Ft and for one surface the plane surface generated by E 0, and for the other surface the concave conical surface generated by FO.

THE SPHEKE.

393

Proposition XXXV. Theorem.

784. The area of a zone is equal to the product of its altitude by the circumference of a great circle.

Let A BC D E be the circumference of a great circle, BC any arc of this circumference, and 0 A the radius of the sphere. And, let PR be the altitude of the zone generated by arc B C.

We are to prove zone B C = P R X 2 tt 0 A.

If the semicircle A BC D revolve about the diameter A D as an axis, the semi-circumference ABC D will generate the sur- face of a sphere ; the arc B C, a zone,

and the chord B (7, a surface whose area is PR X 2 it 0 1. § 759

Now if we bisect the arc B C, and continue this process in- definitely, the surface generated by the chords of these arcs will approach the zone as its limit ;

the _L 0 1 will approach the radius of the sphere as its limit ;

while P R will remain constant.

.-. at the limit, zone BC = PRX2nOA.

Q. E. D.

785. Corollary 1. Zones on the same sphere, or equal spheres, have the same ratio as their altitudes.

786. Cor. 2. A zone is to the surface of the sphere as the altitude of the zone is to the diameter of the sphere.

787. Cor. 3. Let arc A B generate a zone of a single base. Then, zone AB_ = A P X 2 tt 0 A. Hence, zone AB = ir AP X AD = tt A~B2. (§ 307.) That is, a zone of one base is equiv- alent to a circle whose radius is the chord of the generating arc.

394

GEOMETRY.

BOOK VIII.

On the Volume of the Sphere.

Proposition XXXVI. Theorem.

788. The volume of a sphere is equal to the area of its ■nrface multiplied by one-third of its radius.

Let R be the radius of a sphere whose centre is 0, S its surface, and V its volume.

We are to prove V = S X ^ R.

Conceive a cube to be circumscribed about the sphere.

From 0, the centre of the sphere, conceive lines to be drawn to the vertices of each of the polyhedral AA,B,C,D, etc.

These lines are the edges of six quadrangular pyramids, whose bases are the faces of the cube, and whose common altitude is the radius of the sphere.

The volume of each pyramid is equal to the product of its base by £ its altitude. § 574

.*. the volume of the six pyramids, that is, the volume of the circumscribed cube, is equal to the surface of the cube mul- tiplied by \ R.

Now conceive planes drawn tangent to the sphere, cutting each of the polyhedral A of the cube.

We shall then have a circumscribed solid whose volume will be nearer that of the sphere than is the volume of the circum- scribed cube.

THE SPHERE. 395

From 0 conceive lines to be drawn to each of the polyhedral A of the solid thus formed, a, b, c, etc.

These lines will form the edges of a series of pyramids, whose bases are the surface of the solid, and whose common alti- tude is the radius of the sphere ;

and the volume of each pyramid thus formed is equal to the product of its base by J its altitude.

.'. the sum of the volumes of these pyramids, that is, the volume of this new solid, is equal to the surface of the solid mul- tiplied by J R.

Now, this process of cutting the polyhedral A by tangent planes may be considered as continued indefinitely,

and, however far this process is carried, it will always be true that the volume of the solid is equal to its surface multiplied

But the sphere is the limit of this circumscribed solid.

.\ V=SX}R. § 199

Q. E. D.

789. Corollary 1. Since £=4 * R2 (j 7C2), F=4Wi?2X l7?= + 7r/r8. If we denote the diameter of the sphere by

*.*-(T)-T"rTi»*

790. Cor. 2. Denote the radius of another sphere by R' and its volume by V : we have V'= 4 ir R/S. .'• -p-, = ^ — tt^== t^t-

J ' J I' $nR/a R'8

That is, spheres are to each other as the cubes of their radii.

791. Cor. 3. The volume of a spherical sector is equal to the product of the area of the zone which forms its base by one-third the radius of the sphere.

Let R denote the radius of a sphere, C the circumference of a great circle, H the altitude of the zone, Z the surface of the zone, and V the volume of the corresponding sector.

396	GEOMETRY. BOOK VIII.
Then	<7 = 2tt7?;	§ 381
	Z=0 X 11=2 w RX H;	§ 784
	V=lZXR = l-nR2XH.

792. Cor. 4. The volumes of spherical sectors of the same sphere, or equal spheres, are to each other as the zones which form their bases, or as the altitudes of these zones.

For, let V and V denote the volumes of two spherical sectors, Z and Z' the zones which form their bases, H and W the altitudes of these zones, and R the radius of the sphere.

Then

And since

V =	= Z X Z' X	\R_	Z'
z _	H
Z'	w
V _	_H
V	H'

§785

793. Cor. 5. The volume of a spherical segment of one base, less than a hemisphere, generated by the revolution of a semi-segment ABC about the diameter A D, may be found by subtracting the volume of the cone of revolution generated by 0 B G from that of the spherical sector A 0 B.

In like manner, the volume of a spherical segment of one base, greater than a hemisphere, generated by the revolution of

THE SPHERE. 897

A B'C may be found by adding the volume of the cone of revo- lution generated by 0 B' C to that of the spherical sector gener- ated by A 0 B'.

794. Cor. 6. The volume of a spherical segment of two bases, generated by the revolution of C B B' C about the diame- ter A D, may be found by subtracting the volume of the segment of one base generated by A B C from that of the segment of one base generated by A B' C

Exercises.

1. Given a sphere whose diameter is 20 inches; find the cir- cumference of a small circle whose plane cuts the diameter 4 inches from the centre.

2. Construct, on the spherical blackboard, spherical angles of 30°, 45°, 90°, 120°, 150° and 135°.

3. Construct, on the spherical blackboard, a spherical triangle, whose sides are 100°, 80° and 70° respectively. What is true of its polar triangle 1

4. Find the surface and volume of a sphere whose radius is 10 inches ; also find the area of a spherical triangle on this sphere, the angles of the triangle being 80°, 85° and 100° respectively.

5. If 7 equidistant planes cut a sphere, each perpendicular to the same diameter, what are the relative areas of the zones?

6. Given, two mutually equiangular triangles on spheres whose radii are 10 inches and 40 inches respectively ; what are their relative areas ?

7. Let V denote the volume of a spherical pyramid, S its base, E the spherical excess of its base, and R the radius of the sphere ; show that S = £ 7T R2 E, and V = £ *r R* E.

8. Given, the volume of a sphere 1728 inches : find its radius

GEOMETRY. BOOK VIII.

9. Find the ratio of the surfaces, and the ratio of the volumes, of a cube and of the inscribed sphere.

10. Find the ratio of the surfaces, and the ratio of the vol- umes, of a sphere and the circumscribed cylinder.

11. Let V denote the volume and // the altitude of the spher- ical segment of one base, and R the radius of the sphere ; show tb»t V=n IP (R - ( II). Also, find V when R = 12 and

12. Given, a sphere 2 feet in diameter; find the volume of a segment of the sphere included between two parallel planes, one at 3 and the other at 9 inches from the centre. (Two solutions.)

13. A sphere 4 inches in diameter is bored through the centre with a two-inch auger j find the volume remaining.

THE END.

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