## Tuesday, January 25, 2011

### Problem 575: Cyclic quadrilateral, Angle bisector, Measurement

Geometry Problem
Click the figure below to see the complete problem 575.

Zoom at: Geometry Problem 575

1. http://img40.imageshack.us/img40/5961/problem575.png
This 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/FC
Since EG is the angle bisector of angle (AED) so GD/GA=ED/EA=EB/EC=FK/FC
So HB/HA=GD/GA and x=3.5

Peter Tran

2. 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:4
So x=3.5

3. In general
AH:HB =AG:GD
and so
GH // BD

4. 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