Friday, August 19, 2016

Geometry Problem 1249: Cyclic Quadrilateral, Circle, Triangle, Circumcircle, Angle Bisector, Parallel Lines, Area

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click the figure below to view more details of problem 1249.


Geometry Problem 1249: Cyclic Quadrilateral, Circle, Triangle, Circumcircle, Angle Bisector, Parallel Lines, Area

1 comment:

  1. Problem 1249
    Is OO1 perpendicular bisector in AC and EY ,thenCE=AY. But <YAX=<YAF=<AED=<BEC and
    <BCE=<BDA=<EDA=<EDX=<AYX (XYED=cyclic ). So triangleAYX=triangleBCE ie AX=BE. The distances of point E from the AX and AF are equal (AE is bisector of <DAF) .Therefore area triangle AEB=area triangleAEX.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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