Geometry Problem

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## Tuesday, January 25, 2011

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

Labels:
angle bisector,
circle,
cyclic quadrilateral,
measurement

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ReplyDeleteThis 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

By (i) Vertical Angle Bisector Theorem ,

ReplyDeleteand (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

In general

ReplyDeleteAH: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.

ReplyDeleteLikewise, 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

ReplyDeleteProblem #359