Tuesday, January 25, 2011

Problem 575: Cyclic quadrilateral, Angle bisector, Measurement

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

Problem 575: Cyclic quadrilateral, Angle bisector, Measurement.
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Posted by Antonio Gutierrez at 7:54 PM
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5 comments:

  1. Peter TranJanuary 25, 2011 at 10:50 PM

    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

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  2. PravinJanuary 28, 2011 at 10:11 PM

    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

    ReplyDelete
  3. PravinJanuary 29, 2011 at 12:22 AM

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

    ReplyDelete
  4. VihaanJanuary 30, 2011 at 5:14 PM

    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

    ReplyDelete
  5. Antonio GutierrezJanuary 30, 2011 at 9:46 PM

    See also
    Problem #359

    ReplyDelete

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