GEOMETRICAL POR/SMS. 131 



Suppose that the line is drawn as required. Becaiife the ra- 

 tio of DG to EH is given, there is given (prop, i.) a point M in 

 the circumference of a circle paffing through A, D, E, fuch, that 

 the points A, M, G, H are in a circle. If this point be found, 

 and MG, MH, MD, ME joined, the angle GMH is equal to 

 GAH or to DME. Alfo if MA, DE be joined, the angle MHG 

 is equal to MAG or to MED. Therefore the triangle MHG is ii- 

 milar to thegiven triangle MED, and the angle MHG is given. 



In like manner, becaufe the ratio of EH to FK is given, there 

 is given a point N in the circumference of a circle paffing through 

 E, C, F, fuch, that N, C, H, K are in a circle. If NH, NK, NE, 

 NF, NC, EF be joined, it may be proved, in the fame way, that 

 the triangle NHK is fimilar to NEF, hence the angle NHK is 

 given. Now, the angles MHG, NHK being each proved to be 

 given, the angle MHN is given, and the points M, N being 

 alfo given, the point H is in the circumference of a given circle ; 

 but it is alfo in a ftraight line given by pofition ; therefore the 

 point H is given, and the angles MHG, NHK being given, the 

 line GK is given by pofition, which was to be found. 



Construction. Through the points A, D, E defcribe a cir- 

 cle, and infledl DM, EM to the circumference, fo that DM may 

 be to EM as P to Q^ Defcribe alfo a circle through C, E, F, 

 and infled: EN, FN to the circumference, fo that EN may be to 

 FN as Q^to R. Join DE, EF, and infled MH, NH to the 

 ftraight line AE, fo that the angle MHN may be the fupplement 

 of the fum of MED and NEF ; draw HG, fo that the angle 

 MHG may be equal to MED ; then NHK is equal to NEF. 



Join MG, MA. Becaufe the angle MHG is equal to MED 

 or to MAG, the points M, A, H, G are in a circle j hence the 

 angle MHE is equal to MGD ; now MEH is equal to MDG ; 

 for MEA is equal to MDA; therefore the triangles MEH, MDG 

 are fmiilar, and DG is to EH as DM to ME, that is as P to Q^ 



0^2 In 



