of the Trapezium. 



123 



Through either angle as 



Prop. VIII. 



Fig. 7- 



C of any quadrilateral figure 

 ABCD draw a line FG li- 

 mited by the other sides AB, 

 AD in F and G. Join GB, 

 FD cutting the sides BC, 

 CD in O and P, and let 

 AO, AP cut BG, FD in 

 Qand R. Then FQ and 

 GRwill pass through E, 

 the intersection of the dia- 

 gonals AC, BD. 



Bern. — The triangles FCA, PBD being constituted in the 

 trapezium FPDA according to the conditions of Pr. 6. show 

 that E, R, G, are three points in the same straight line. Also, 

 the triangles GCA, BDO in the trapezium GOBA, show in 

 like manner that E, Q, F, are points in the same straight line. 



Q. E. D. 



Prop. IX 



Let EFDG be a trapezium, whose opposite pairs of sides 

 EF, GD and EG, FD*, meet in A and C. From any point 

 B in the line AC, draw lines to cut the adjacent pairs of sides 

 of the trapezium, viz. BIH, to cut DG in I and EG in H, 

 and BKL to cut FD in K and FE in L. Then LI, KH will 

 always intersect in the diagonal FG. 



Dem. — 1. The transversal AC cutting the triangles KEL 

 and IGH give 



BK : BL : : AF . CK : AL . FC, (Bland's Geom. Prob. 

 Pr. 47. Sect. 4.) and 



BI : BH : : CG . AK : CH . AG. Hence, compounding 



* It is to be taken for granted, that all lines given by position may be 

 prolonged when necessary, as well as that lines joining given points, are 

 also drawn. 



Q2 BK.BI 





