GEOMElRlCJL FORISMS. 115 



are in a circle, therefore (Lemma.) the points H,!], E, F are alfo 

 in a circle. The angle HDE is equal to HBC, that is, to HAK, 

 and fince FIEF is equal to f-lCF, therefore HKD is equal to 

 HCB, that is, to HGB or HKA; hence the triangles HDE, HAK 

 are fimilar, and fince HFE is equal to HCK, the triangles HEF, 

 HKG are alfo fimilar; therefore DE is to EF as AK to KG, that 

 is, as de to ef. 



Cor. I. The lines DH, EH, FH contain given angles, and 

 have to each other the given ratios of AH, KH, CH. 



Cor. 2, The line DF cuts off fegments DA, EK, FC from 

 the given lines, adjacent to given points in them, and having to 

 each other the given ratios of HA, HK, HC. For the angles 

 HDB, EIEK, HFC are equal among themfelves, and fince BOH 

 or BGH, that is, AKH, is the fupplement of each of the angles 

 HCF, HAD, HKE, the angles HAD, HKE, HCF are equal a- 

 mong themfelves, therefore the triangles HAD, HKE, HCF 

 are fimilar, and AD, KE, CF are proportional to the given lines 

 AH, KH, CH:. 



PROP. VI. PORISM, Fig. II. PI. ir. 



Let AB, AC, BE, DE be four ftraight lines given by pofi- 

 tion ; a point P may be found, fuch, that if any circle be 

 defcribed through it and B, any of tlie fix interfedlions of 

 the given lines, to meet the lines through whofe interfe(?tion 

 it pafles in G and L, and if GL be joined, meeting the re- 

 maining lines in H and K, the fegments GH, HK, KL have 

 given ratios to one another, which ratios are to be found. 



Because, by hypothefis, the points P, A, G, H are in a circle, 

 and alfo the points P, F, H, K, it will appear, as in the analyfis 

 of laft propofition, that P is in a circle defcribed about the tri- 

 angle ADF ; in the fame way it will be found, that P muft be 

 in circles defcribed about each of the triangles ABC, DBE, 



O 2 FCE. 



