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perpendicular to bisector

Source: All-Russian Olympiad 1996, Grade 11, Second Day, Problem 6

April 19, 2013
geometrycircumcircleincenterparallelogramangle bisectorgeometry proposed

Problem Statement

In isosceles triangle ABCABC (AB=BCAB = BC) one draws the angle bisector CDCD. The perpendicular to CDCD through the center of the circumcircle of ABCABC intersects BCBC at EE. The parallel to CDCD through EE meets ABAB at FF. Show that BEBE = FDFD.
M. Sonkin
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