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<DOE=90^o if BD=AD, CB=CE, right triangle (Kyiv City Olympiad 2004 7.3)

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June 27, 2021
geometryright triangleright angle

Problem Statement

Given a right triangle ABCABC (∠A<45o\angle A <45^o,∠C=90o \angle C = 90^o), on the sides ACAC and ABAB which are selected points D,ED,E respectively, such that BD=ADBD = AD and CB=CECB = CE. Let the segments BDBD and CECE intersect at the point OO. Prove that ∠DOE=90o\angle DOE = 90^o.
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