I30 GEOME'TRICAL PORISMS. 



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

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

 ftraight hne LM given by pofition, fuch, that PF drawn to any 

 point in AB, may be bifed:ed by it in K. Through L, M, C, 

 the interfccftions of the given Hnes 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 

 infle^l NG, PG to AC, fo that the angle NGP may be equal to 

 NML ; draw PE, fo that the 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, NQ^l are fimilar ; therefore ML and GH 

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

 therefore GH is bifecfled 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 conftrudlion, as a, to /3. 



PROP. XV. PROBLEM, Fig. 20. PI. IV. 



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 themfelves, 



■ Suppofe 



