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.

## Friday, August 19, 2016

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

Labels:
angle bisector,
area,
circle,
circumcircle,
cyclic quadrilateral,
inscribed,
parallel,
triangle

Subscribe to:
Post Comments (Atom)

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