GEOMETRICAL FORISAIS. 127 



ing AC in E and e. The redangles BD, CE, and B^, Ce, are 

 each equal to the given redangle FP, PC. 



Let ED meet FK in L, join HL, HD. Becaufe by conflruc- 

 tion the angle HDP is equal to HKF ; therefore HDL is equal' 

 to HKL ; therefore the points H, K, D, L, are in a circle, and 

 the angle HLK is equal to HDK, that is HLF is equal to HDD, 

 now HFL is equal to HBD, alfo HF is equal to HB; therefore the 

 triangles HFL, HBD ar6 in all refperts equal, and FL is equal 

 to BD. Again, the triangles LFP, PCE are fimilar, therefore 

 FL is to FP as CP to CE, and the redlangle FL, CE is equal to 

 the redangle FP, PC, but FL is equal to BD, therefore the rec- 

 tangle BD, CE is equal to the redangle FP, PC, that is to the 

 given fpace. In the fame way it may be ihewn that the rec-- 

 tangle B(J, Cs is equal to FP, PC. 



PROP. XIIL PROBLEM, Fig. 18. PI. IV. 



Four ftraight lines DB, DF, CG, EG are given by pofition, it 

 is required todrawa hne to meet them in the points N,0,P,Q, 

 fo that the line NQjnay be divided at thefe points, fimilar- 

 ly to a given divided line n p q. 



Suppose the line NQ^drawn as required. Becaufe DB, DF,, 

 BF are three ftraight hues given by pofition, and that NQjs di- 

 vided by one of them at O into fegments, having to each other 

 a given ratid, if a circle be defcribed through the points B, D, F, 

 there is & given point E in the circumference, fuch, that the 

 points E, B, N, Q^are in a circle, (Prop. 5.) Again, becaufe 

 CB, CG, BG are three lines given by pofition, and NQ^is divi- 

 ded by one of them at P into fegments, having to each other a 

 griven ratio, if a cifcle be defcribed through B, C, G, there is a 

 given point A in the circumference, fuch, that A, N, B, Q^are 

 m a circle, (Prop, 5.) Thus it appears, tliat there are given 



three . 



