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A simple one on cyclic quads

Source: India RMO 1992 Problem 4

October 15, 2005
geometrycircumcircletrigonometryparallelogramrectanglecyclic quadrilateralpower of a point

Problem Statement

ABCDABCD is a cyclic quadrilateral with AC⊥BDAC \perp BD; ACAC meets BDBD at EE. Prove that EA2+EB2+EC2+ED2=4R2 EA^2 + EB^2 + EC^2 + ED^2 = 4 R^2 where RR is the radius of the circumscribing circle.
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