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Geometry ProblemClick the figure below to see the complete problem 575.Zoom at: Geometry Problem 575
http://img40.imageshack.us/img40/5961/problem575.pngThis is a clarification to my previous solution.Let K is a point on CD such that FD=FK ( see picture)We have triangle EBC similar to tri. EDA and Tri. FDC similar to tri. FBA ( Case AA)Angle(EBC)=Angle(CKF) ( both angle supplement to angle(CDF) and tri. EBC similar to tri. FKC. ( case AA)Since FH is the angle bisector of angle (BFA) so HB/HA=FB/FA=FD/FC=FK/FCSince EG is the angle bisector of angle (AED) so GD/GA=ED/EA=EB/EC=FK/FCSo HB/HA=GD/GA and x=3.5Peter Tran
By (i) Vertical Angle Bisector Theorem , and (ii) Sine Rule:7:8=FB:FA =FD:FC (since FD.FA = FC.FB)=Sin DCF : Sin CDF = Sin BCE : Sin EBC (note EBC, CDF are supplementary)=EB:EC=ED:EA=x:4So x=3.5
In general AH:HB =AG:GD and so GH // BD
A slightly different solution for those interested. Join A to C & B to D. Triangles FDB and FAC are similar. So BD/AC = FD/FA and FD/FA = DH/HA by the angle bisector theorem.Likewise, triangles BDE and ACE are similar. So BD/AC = BE/AE = BG/GA again by the angle bisector theorem. Hence, DH/HA = BG/GA, BD//GH and x = 3.5 as b4.Vihaan
See also Problem #359