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R^2=OE^2+CD^2 [1- BC^2/(AB^2+AC^2) ], circle tangent to (ABC)

Source: Indonesia INAMO Shortlist 2008 G3

August 25, 2021
geometrytangent circles

Problem Statement

Given triangle ABCABC. A circle Γ\Gamma is tangent to the circumcircle of triangle ABCABC at AA and tangent to BCBC at DD. Let EE be the intersection of circle Γ\Gamma and ACAC. Prove that R2=OE2+CD2(1−BC2AB2+AC2)R^2=OE^2+CD^2\left(1- \frac{BC^2}{AB^2+AC^2}\right) where OO is the center of the circumcircle of triangle ABCABC, with radius RR.
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