132 , GEOMETRIC JL PORISMS. 



In like manner it may be proved, that becaiife the angle NEF 

 is equal to NHK, the points N, C, H, K ai-e in a circle, and 

 hence that the triangle NEH is fimilar to NFK ; hence EH is 

 to FK as EN to FN, that is as Q^ to R. Therefore GHK is 

 drawn as required. 



PROP. XVI. PROBLEM, Fig. 20. PI. IV. 



It is required to defcribe a triangle DEF fimilar to a given 

 triangle def, having one of its fides EF pafling through 

 P a given point, and having its angles in a given order up- 

 on three ftraight lines AB, AC, BC given by pofition. 



The conflrudlion of this problem follows readily from the 

 8th proportion, as follows : 



Draw AG, GK, fo as to' form a triangle AG K, fimilar to the 

 given triangle def, and having its angles' upon the given lines 

 in the given order. 



Through A, G, any two of its angles, and C, the interfecflion 

 of the lines upon which they are placed, defcribe a circle. 

 Through G, K, and B, the interfe«flion of GC, KA,^ defcribe a 

 circle meeting the former in H. From the points H, P inflec5l 

 HE, PE to CB, fo that the angle HEP may be equal to HGK 'y 

 let PE meet AB in F. Through H, C, E defdribe a circle to 

 meet CA in D ; join DE, DF, and the triailgle DEF fhall be fi- 

 milar to AGK or to d e f. 



Join HD, HF, HA, HK, HB, HC. Becaufe, by conftrudlon, 

 the angle HEF is equal to HGK or to HBK., the points H, B, E, 

 F are in a circle, and the angle FHE is equal to FEE or KHG, 

 therefore the triangles EHF, GHK are fimilar. In like manner, 

 becaufe a circle pafTes through H,C, D, E, the angle DHE is equal 

 to DCE or AHG, and HDE is equal to HCE or HAG,, therefore 



the 



