MATHEMATK - I'LANE GEOMETRY. [BOOK i. 



-PROP. IL. XLVL 



ame other line A G be 

 parallel to B K, and draw 

 OKF. The triangle A HC- 

 , Hrp. UietrianHeGl I ,t 

 as they are on equal base* 



__ - _^_ *" d between the ame paral- 



B OS r Jels. But the triangle ABC 



pr *, - the triangle I) K K, .'. triangle D K F - 

 triangle G 1-. 1 . though one only a part of the other, 

 which U impossible, .'. any other line A G is net parallel 

 to K v 1) u tt parallel to UP drawn from A, 



.-. eqwtl triangle*, Ac. Q. E. D. 



OM aid* of AD. or on 



PROPOSITION XLL THBOBBM. 



If a para'l,l.-,r:,m (A B C D) and a triangle (EBC)6e 

 upon the tame ba*e and betweett the tame parallel*, the 

 parallelogram ihall be double the triangle. 



Draw A C, then triangle ABC 

 = triangle K It C, because they 

 are on the same base and are be- 

 tween the same parallels. * But 

 Pr. 97. ABC 1) is double the 

 + IT. M. triangle A H C,t . ' 

 A B C 1) is alto double the tri- 

 angle E B C, .-. if a paral- 

 lelogram, &c. Q. E. D. 



Sun. The learner will perceive that this thorem i unnreeasarily 

 restricted to the case in which the baso* oovieidt; he may Rfiicr.il i-r 

 It himelf, proving that if a parallelogram and a triangle be upon 



lelogram FECG 



PROPOSITION XLII PROBLEM. 

 To detcribe a parallelogram that shall be equal to a given 

 triangle (A B C), and hate one of its angles equal to a 

 given angle (D). 



Pr. 10. Bisect B C in E, * and make the angle CEF 

 t Pr. M. . D. t Also through A draw A F G parallel 



Pr si. to B C, and through C, C G parallel to E F :* 

 then F E C G is the parallelogram required. For draw 



A E : the triangle A B E = 

 G tPr. 38. triangle A EC, t 

 since they are on equal 

 bases and between the same 

 parallels, . . triangle A H C 

 w double the triangle 

 A EC. But FECG is like- 

 wise double the triangle 

 Pr. 41. AEC,*.'.paraf- 

 = triangle ABC, and the angle CEF 

 . . a parallelogram has been described 

 as required. 



PROPOSITION XLIII. THEOREM. 

 The complement* (B K, K D) of the parallelogram* (E H, 

 G F) which are about the diagonal (A C) of a parallelo- 

 gram art equal. 



The parallelogram E H. O I), through which the diagonal A C panel- 

 are paid to be ahttui thf diagonal, and the remaining parallelograms 

 H K. K I). Mch make up the whole ngure, are called the comple- 

 ment* of the former. 

 Because B D is a parallelogram, and A C its diagonal, 



IT. S4. .-. triangle A B C = triangle ADC.* 

 because E H U a parallelogram, . . triangle A E K = 



+ Pr. 4. triangle A H K ; 

 and, for a like reason, triangle 

 K G C = triangle K F C. . . 

 AEK+KGC-AHK+ 

 K F C. But the whole ABC 

 _ the whole ADC, . . the 

 remainders that is, the com- 



B 6 C ptrmrnf* BK, K.D are equal, 



.-. the complement*, <tc. Q. E. D. 



PROPOSITION XLIV. PROBLEM. 

 To a givtn finite straight line (A B) to ajrpltj a parallelo- 

 gram which shall be equal to a gin-it ', ,:!', (C), and 

 hate one of U* angle* equal to a girtn anile (D). 

 Make the parallelogram B F triangle C, and having 



11 



. pr. 41. the angle E B O - D, no that B E may be in 



the same straight line with A B ; ami prolong F G to H. 



tPr.il. Through A draw A H, parallel to B(i orKK.t 



meeting the prolongation of Vi! in II, ainl draw II It. 



Then because H F falls on the parallel* A 11, E F, the 



angles A II !'. 



t E 5 H FE are together 



_ two right 

 Pr . angles,* 



.-. r. HF + HFE 



are lea than two 

 riulit angles, .'. 

 H 15, F E, if pro- 

 ved, must 



meett Let them moot in K ; and draw K L 

 parall.-l to K A. or F II ; and prolong II A, 



t AX. n. 



Pr. 51. 



LB~C. 



A B M,t and 

 , .'. to the 

 allied = 0, 

 Which was to be 



Uel to Cj n. or r n ; " i"" " " "i 



G B~~ to L.'M. Then LF U a parallelogram, of which 

 the diagonal is H K ; and 1. M. !' '' aro *nt 



of the parallelograms A G, ME, about the diagonal, 

 tcJu .-.LB-BF;t but B F = C ' 

 t Pr. is. And because the angle G B E 

 con.t. G B E - D,* . . A B M = v 



straight line A B the paralltlnyram L H is 

 and ' having an angle A B M = D. \M 

 done 



PROPOSITION XLV. PROBLEM. 



To describe a parallelogram equal to a yiren rectilineal 

 figure, and h-iriny an 'angle equal to a given angle (E). 



First, let the figure be a /nr-i<M one, A B C D. 

 Draw D B, and describe the parallelogram F H = the 



. Pr 42 triangle A D B, and having the angle K = b ; 

 and to G H apply the parallelogram G M = the triangle 



+ Pr 44 DEC, having the angle G H M - E.f Me 

 figure F M shall be the parallelngram required. 



const Because the angles K, G H M are each = E,* 

 .-. K - O H M. 



Add to each of these A 

 the angle K H G, 

 .-. K +KHG = 

 G H M + K H G ; 

 but K + K H G = 

 two right angles,t 

 + Pr.29. .'.OHM 

 + K H G = two 

 right angles, and consequently K H. H M are in the 



Pr 14 same straight line.* And because H G meets 

 the parallels K M, FG, the angle M H G = H G F ; t 



+ Pr 29. add to each of these the angle H G L, . . 

 M H G -f-HGL=HGF + HGL; but the former are 



Pr. 29 = two right angles ;* .-.HGF+HGL = 

 two right angles, . . F G, G L are in the same straight 



1 Pr. 14. line.t And because K F, ML are each 



Pr. so. parallel to H G, . . K F is parallel to M L,* 

 and K M, F L are by construction parallel ; . . F M is a 



+ Pr. M. parallelogram.? And because the triangle 

 Def. A B D - F H, and the triangle D E C - 



Const. G M,* .' . the tchole A C = the paMtUegram 

 FM. Again, let the figtire be fire-sided, then, having 

 drawn D C, there will be, besides the four-sided figure 

 A C, a triangle upon DC. A parallelogram, equal to 

 this triangle, may be applied to LM, just as G M 

 DEC was applied to G H ; and thus, however numerous 

 be the sides, a parallelogram may be described equal to 

 the given rectilineal figure, and having the angle K. the 

 giren angle E. 



COR. From this it is manifest how, to a given straight 

 line, to apply a parallelogram which shall have an angle 

 equal to a given rectilineal angle, and shall be equal to a 

 given rectilineal figure. 



PROPOSITION XL VI. PROBLEM. 



To describe a square vpnn a given finite straight lint 

 (AB). 



Pr. 11. Dnw A C at right angles to A B.* and make 



IT. 3. AD AB.t Tnrough U draw I) K parallel to 



Pr. II. A B, and through B dra it B E [lanillel to A D,* 



