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.
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Friday, August 19, 2016
Geometry Problem 1249: Cyclic Quadrilateral, Circle, Triangle, Circumcircle, Angle Bisector, Parallel Lines, Area
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Problem 1249
ReplyDeleteIs 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