ijo GEOMETRIC JL P R I S M S. 



EPG, the ratio of EP to PH may be the fame with the given 

 ratio of a. to /3. (Loci Plani, Prop. 6. Lib. i.) Find alfo a 

 ftralght line LM given by pofition, fuch, that PF drawn to any 

 point in AB, may be bife(5led by it in K. Through L, M, C, 

 the interfccflions of the given lines LM, AC, LC, defcribe a cir- 

 cle. Draw CO parallel to LM, meeting the circle in O ; bifedl 

 ML in Q^; join OQ^ meeting the circle in N ; join NM, and 

 infledt NG, PG to AC, fo that the angle NGP may be equal to 

 NML ; draw PE, fo that tlie angle EPG may be fuch as is re- 

 quired. 



Let GP meet CL in H, and AB in F, alfo LM in K ; join NH, 

 NK, NL. Since NGP is equal to NML, the points N, K, G, M 

 are in a circle, and the angle NKH is equal to NMG or NMC, that 

 is to NLH ; therefore the points N, K, L, H are in a circle, and 

 the angle NHK is equal to NLQ^; now NKH is equal to NMG 

 or NOC, that is (OC being parallel to ML) to NQL ; therefore 

 the triangles NKH, NQL are fimilar. In like manner it ap- 

 pears, that NKG, KQAl are fimilar ; therefore ML and GH 

 are fimilarly divided at Q^ and K, but ML is bifecfled at Q^; 

 therefore GH is bifeiflcd at K ; now PF is alfo bifedled at K ; 

 therefore OF is equal to PH, and EP is to FG as EP to PH, 

 that is, by conftrudtion, as « to /3. 



PROP. XV. PROBLEM, Fig. 20. PL FV. 



Three ftraight lines AB, AC, BC are given by pofition, and 

 three points D, E, F are given in thefe lines. It is required 

 to draw a ftraight line GHK to meet them, fo that DG-, 

 EH, FK may have to each otlier the given ratios that P, Q^ 

 R have among tliemfelves. 



Suppofe 



