MATHEMATICS. PLANE GEOMETRY. 



[BOOK i. pnor. v. s.. 



Ax. I. onpfc C 0* angle F.* And as the triangle 

 A I) C thai ooinci.lt, in every respect, with the tri- 

 angle D E F, Itn surfaces of On triangles must be 



At I tftal ; t .'. if tux, triangles, Ac. Q.E.D. 



PUol-osiTloN V. THBOEEK. 

 The angles (A B C, A C B) at the ban (B O of an isosceles 



triable (A B C) arc equal ; and if the equal tide* (A R, 



A C) be prolonged, the anafo (D B C, E C B) upon ike 



otter tide of the bate shall be equal. 



In B D take njr point, F, and from A E, the greater, 



Pr. . cut off A O - A F, the ten ; * and draw 

 FC, OB- 



I COM. Became F A -G A, t and AO-AB /. FA, 



ii. A C are equal to A, AB, each to each ; 

 and the angle A is common to 



both the triangles A F C, A O B ; 



trr.4. .'. FC-GB,t toe 

 angle A C F - the angle A B G,f 

 an<l the angle A F C the angle 

 A G B,t these being the angles 

 to which the equal sides are 

 opposite. 



Again : because the whole A F 

 the whole A G, and that the 

 parts A R, A C are equal, tho 

 remainders B F, C G are 



AX . equal ; * and it was 

 proved that F C = G B, .- . the 

 two sides B F, F C, are equal to 



the two C G, G B, each to each : it was also proved that 

 the angle B F C = the angle C G B, . . the angle F B C 

 t Pr. 4. = the angle G C B,t and the angle B C F = 

 the angle C B G. And since it was demonstrated that 

 the whole angle A B G, = the whole angle A C F, the 

 parts of which, the angles C B G, B C F, are also equal, 

 .. the remaining angle A B C = the remaining angle 



Ax. s. A C B.* These are the angles at the bate of 

 the isosceles triangle ABC, and the angles F B C, GOB, 

 before proved to be equal, are the angles on the other ride 

 of the bate. Therefore the angles at the base, <tc. Q. E.;D. 



COROLLARY. Every equilateral triangle is equiangular. 



PROPOSITION VI. THEOREM. 

 If two angles (A B C, AC B) of a triangle are equal, 



the side* (A C, A B) which tubtend, or are opposite to, 



the equal angles shall be. equal. 



For, if A B be not equal to A C, one of them, as A B, 

 must be tho greater. Let a part B D, equal to A C, the 



Pr. s. less, be cut off,* and draw 

 DC. Then, because in the triangles 

 DEC, A C B, D B = A C, and BO 

 common to both, the two sides D B, 

 B C are = the two sides AC, C B, 

 each to each ; and the angles DBG, 



Hyp. A C B are equal ; * . . the 

 triangle D B C = the triangle A C H,t 



t Pr. 4. a part to the whole, which 



is absurd. Therefore, A B, A C are 



not unequal that is, they are equal; .-. if two angles, 



ic. Q.E.D. 



COR. Every equiangular triangle is equilateral. 

 PROPOSITION VII. THEOREM. 



Upon the same bate (A R), and on the same side of if, 

 there cannot be two triangles (A C B, AD B), hating 

 their sides (A C, A U), which are terminated in one ex- 

 tremity of the base, equal to one another, and likewise 

 (B C, B D) those which are terminated in the other 

 estremiti/. 



Join C D : then, in the cose in which the vertex of 

 each triangle is without the other tri- 



Hyp. angle, because AC A ]),* 

 ,PT.J. the angle AC I)-ADC.t 



But the angle A C D is greater than 

 the angle BCD, which is only a 

 part of it ; . . the angle A I) C is 

 greater also than BCD: much 

 more, then, is the angle ft D 

 Jin>. greaterthanBCD. Again: because BC-BD,* 



1 Pr. i. the angle B D C - B C D ;t but it has been 

 di'tnonstrated to be greater than BCD, which is im- 

 possible, . . in this ease it is impossible that A C A 1), 

 and liktirite BC-BD. 



But if one of the vertices, as D, be in'Miu the other 

 triangle, prolong AC, AD to E, F. Then, because 



Hyp. A C A D* in the tri- 

 angle A C D, the angles BCD, 

 F D C, upon the other side of the 



t Pr. . base C D, are equal ;t 

 but the angle E C D is greater 



Ax.9. thanBCD,*.-.FDC is 

 likewise greater than BCD: much 

 more, then, is B D C greater than 

 BCD. Again: because BC - 



t Hyp. BD,t the angle B DC 



Pr. s. B C D ;* but B D C was A 



proved greater than BCD, which is impossible, .-.in 



this case also it is impossible that AC AD, and likewise 



BC-BD. 



The case in which the vertex of one triangle, is upon 

 t AX. . a side of the other, needs no demonstration. t 



Therefore, upon the same base, <fec. Q. E. D. 



PROPOSITION VIII. THEOREM. 



// two triangles (A B C, D E F) have two si<les (A B, AC) 

 of the one equal to two sides (D E, D F) of the other, 

 each to each, and have likewise their bases or third sides 

 (B C, E F) equal, then the angle (A) contained fry the 

 two sides of the one shall be equal to the angle (D) con- 

 tained by the two sides equal to them of the other. 

 For if the triangle A B C be applied to D E F, so that 

 the point B may be on E, and B C on E F, the point C 

 shall coincide with A. _ 



F, because B C - 



.nrp. EF.* 

 And BC thus coin- 

 ciding with E F, 

 B A and A C must 



coincide with E D 



and D F ; for if B C E r 



the base B C, coinciding with E F, the sides B A, A C 

 mlil fall otherwise than on E D, D F, and have different 

 situations, as E G, G F, then upon the same base E F, 

 and on the same side of it, there could be two triangles 

 having the sides E D, E G, terminated in one extremity 

 of the base, equal to one another, and likewise the sides 

 F D, F G terminated in the other extremity. But this 

 t Pr. 7. is impossible ;t ' if the bases coincide, the 

 remaining sides cannot but coincide, and . . the angle 

 A must coincide with the angle D, and be ecpud to 



Ax. 8. it. * Therefore, if two triangles, &c. Q. E. D. 



PROPOSITION IX. PROBLEM. 



To bisect 



a given rtctiline.nl angle (B A C) ; that is, to 



divide it into two equ/il angles. 

 Take any point D in A B, and from A C cut off A E = 



Pr. s A 1) :* draw D E, and 

 upon it describe an equilateral 



t Pr. l. triangle D E F, t so 

 that tho vertex F may be on Uia 

 opposite side of D E to the vertex 

 A. Draw A F, then A F shall 

 bisect the angle B A C. 



ront. Because A D A K, * 

 and that A F is common to the 

 two triangles D A K, 1C A K : the 

 two sides DA, A F are the 

 two sides E A, A F, each to each. 



the base E F ; . . the angle D A F 



Pr. - the angle E A F ;* . . the giren angle B A C 

 is bisected by the straight line A F. Which was to be done. 



PROPOSITION X. PROBLEM. 



To bisect a giren finite straight line (A B) ; that is, to 

 divide it into two equal parts. 



Pr. l. Upon AB describe an equilateral triangle A llf ',* 

 and bisect the angle A C U by the straight line 



and the base D F > 



