132 , GEOMETRICAL PORISMS, 



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

 is equal to NHK, the points N, C, H, K are 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 conftru(5lion of this problem follows readily from the 

 8th propofition, as follows: 



Drav/ AG, GK, fo as to form a triangle AGK, fimilar to the 

 given triangle d ef^ and having its angles upon the given lines 

 in the given order. 



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

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

 Through G, K, and B, the interfedlion of GC, KA, defcribe a 

 circle meeting the former in H. From the points H, P infledl- 

 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 conftru(5lion, 

 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 FBE or KHG,; 

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

 becaufe a circle pafiTes through H,G, D, E, the angle DHE is equal 

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



the 



