GKOME'TRIGAL FORI SMS. 



II 



are in a circle, therefore (Lemma.) the points H,*C, E, F are'alfo 

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

 and fmce HEF is equal to HCF, therefore HED is equal to' 

 HCB, that is, to E]GB or HKA; hence the triangles HDE, HAK 

 are fimilar, and fmce HFE is equal to Hl:K, the triangles' HEF, 

 HKC are alfo fimilar; therefore DE is to EF as AK to KC, that 

 is, as de to ef. 



Cor. I. The Hues 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, F:K, FC from 

 the given lines, adjacent to given points in them, and havinr^ to 

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

 HDB, EIEK, HFC are equal among themfelves, and fmce BCH 

 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 giveA lines 

 AH, KH, CH. 



PROP. VI. PORISM, Fig. II. PLir, 



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 

 defcnbed through it and B, any of the fix interfedlions of 

 the given lines, to meet the lines through whofe interfedion 

 It pafTes in G and L, and if GL be joined, meeting the re- 

 , maming 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 

 ot lafl 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 

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



O 2 FCE. 



