BOOK VL PROP, ix. xv.] MATH EM ATIC S. PLANE GEOMETRY. 



687 



PROPOSITION IX. PROBLEM. 



From a given straight line (A B) to cut off any part 

 required. 



From A draw a straight line A C, making any angle 

 with A B ; and in it take any point D ; and take A C, the 

 same multiple of A 1), that A B is of the 

 part to be cut off. Draw B C, and D E 

 311. parallel to it, * then A E shall be 

 the part required to be cut off. Because 

 D E D is parallel to B C, a side of the tri- 



angle ABC, .-. CD:DA::BE:EA; 

 T 13 v. . . A C ; A D : : A B : A E ;t but 

 Const. AC is a multiple of AD,*.-. 

 A B is the same multiple of A E ; that is, 

 whatever part of A D is of A C, the same 

 c f 7 v. Cor. part is A E of A B,f . ' . from 

 A B, the part required, A E is cut off. Which was to be 

 done. 



PROPOSITION X. PROBLEM. 



To divide a given straight line (A B) similarly to a given 

 divided straight line (A C). 



Let A C, placed so as to make any angle with A B, be 

 divided in the points D, E. Draw B C ; and through the 

 points D, E draw D F, E G parallels to it ; A B shall be 

 divided in F, G similarly, to A C. 

 Through D draw D H K, parallel to 

 A B : then each of the figures F H 

 H B is a parallelogram, . ' . D H = 

 MI. FG,* and HK = GB; 



O/ H/ AS and because H E is parallel to K C, 



a side of the triangle DKC, . . CE : 

 , i 2 vi. E D : : K H : H D ; + but 



CKH = BG, and HD = GF ; .-. 



CE:ED::BG:GF. Again, because F D is parallel 

 to G E, a side of the triangle AGE,.-. ED:DA:: 

 G F : F A : and it was before proved that C E : E D : : 

 B G : G F ; . . the given straight line is divided similarly 

 to AC. Which was to be done. 



PROPOSITION XI. PROBLEM. 



To find a third proportional to two given straight lines 

 (AB.AC). 



Let the lines be placed so as to make 

 any angle A, and produce them to 

 D, E. Make B D = A C ; draw B C, 

 and through D draw D E parallel to it :* 



\C 11 1. C E shall be a third proportional 



to AB, A C. Because BC is parallel to 

 DE, .-.AB:BD::AC:CE; but BD 

 + Const. = A C ; t .'. A B : A C : : A C : 

 C E ; . . to the two given straight lines A B, 

 A C, a third proportional C E is found. 

 Which was to be done. Q. E. D. 



PROPOSITION XII. PROBLEM. 

 To find a fourth proportional to three given straight lines 



(A, B, C). 



Take two straight lines D E, D F, making any angle 

 D ; and upon these take D G = 

 A, G E = B, and D H = C. 

 Draw G H, and E F parallel to 

 it : F H shall be a fourth pro- 

 . portional to A, B, C. Because 

 G H is parallel to E F, . . D G 

 :GE::DH:HF;butDG = 

 A, G E = B, and D H = C,* 

 A : B : : C : H F, 



I F 



. to the three given straii/ht lines A, B, C, a fourth pro- 

 portional H F is found. Which was to be done. 



PROPOSITION XIII. PROBLEM. 



To find a mean proportional between two giten straight 

 lines. 



Let the parts A B, B C be equal to the two given 

 straight lines : it is required to find a mean proportional 



between them. Upon AC de- 

 scribe the semicircle ADC, and 

 from B draw B D perpendicular 

 m. to AC;' BD shall be 

 a mean proportional between 

 AB.BC. 



c Draw AD, CD. Then because 

 4 31 1. the angle A D C, in a semicircle is a right angle, f 

 and that in the right-angled triangle A D C, D B is drawn 

 from the vertex of the right angle perpendicular to the 

 base, DB is a mean proportional between AB, BC,* 

 _ 8 VI. Cor. .'. between A B, B C, o mean proportional D B, 

 is found. Which was to be done. 



PROPOSITION XIV. THEOREM. 



Equal parallelograms (A B, B C) which have an angle (B) 

 of the one equal to an angle of the. other, have their sides 

 about the equal angles reciprocally proportional: and 

 parallelograms that have an angle of the one equal to an 

 angle of the other, and the sides about those angles 

 reciprocally proportional, are equal to one another. 



Let the sides D B, BE, be placed in the same straight 



line ; then because FBD+FBE= two right angles, 



and that FBD= GBE (hyp.), /. GBE+FBE = 



^ if two right angles ; .". F B, 



\v v B G are in one straight 

 \ \ line. It is to be proved 

 \__\-, that DB : BE : : GB : BF. 



Complete the parallelogram 

 F E. Then because A B = 

 BC, AB:FE::BC:FE; 

 ,, but AB :FE : :DB:BE * 

 ivi. andBC:FE:: 

 GB : BF, /. DB : BE :: GB : BF;f .'. 

 the sides of the equal parallelograms A B, B C, about their 

 equal angles are reciprocally proportional. 



Next, let the sides about the equal angles be recipro- 

 cally proportional ; namely :DB:BE::GB:BF; 

 then is AB = B C. Because D B : BE : : GB : B F, 

 and DB : BE :: AB : FE, and GB : BF : : BC : FE,* 



ivi. .'. AB :FE : :BC :FE,t .'. AB = B C, .-. 

 + 2 V. equal parallelograms, &c. Q. E. D. 



PROPOSITION XV. THEOREM. 



Equal triangles (ABC, AD E), which have an angle (A) 

 of the one equal to an angle of the other, have their sides 

 about the equal angles reciprocally proportional ; and 

 triangles which have an angle of tlie one equal to an 

 angle of the other, and their sides about those angles re- 

 ciprocally proportional, are equal to one another. 



Let the triangles be placed so that their sides, C A, 

 A D may be in one straight line ; then it may be proved, 

 as in last proposition, that E A, A B are in one straight 

 line. It is to be proved that CA:AD::EA:AB. 



Draw B D. Then because the triangles ABC, A D E 

 are equal, .'. triangle ABC: triangle A B D : : triangle 

 A D E : triangle A B D ; but tri- B D 



angle ABC : triangle ABD : : CA 



. i vi. : AD,* and triangle ADE : 

 triangle A B D : : E A : A B, con- 

 2 v. sequentlyt C A : A D 

 E A : A B ; . . the sid-x of the 

 equal triangles ABC, ADE, about 

 the equal angles are reciprocally c 

 oroportioncl. 



Next, let the sides of the triangles, ABC, ADE, 

 about the equal angles be reciprocally proportional ; 

 namely, CA:A1) :: EA : AB; then is triangle ABC 

 triangle ADE. 



Draw BD as before. Then, because C A : A D : : E A : 

 A B, and C A : A D : : triangle ABC: triangle ABD,* 



i vi. and E A : A B : : triangle ADE: triangle 

 A B D; .'. triangle ABC: triangle ABD:: triangle 



t 2 v. ADE: triangle A B D ;t and the consequents 



9 V. Cor. being equal, the antecedents are equal,* 

 triangle A B C = triangle ADE; .'. equal triangles, 



&c. Q. E. D. 



