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Bosnia and Herzegovina JBMO TST 2014 Problem 2

Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2014

September 16, 2018
geometryRight AngledTriangle

Problem Statement

In triangle ABCABC, on line CACA it is given point DD such that CD=3⋅CACD = 3 \cdot CA (point AA is between points CC and DD), and on line BCBC it is given point EE (E≠BE \neq B) such that CE=BCCE=BC. If BD=AEBD=AE, prove that ∠BAC=90∘\angle BAC= 90^{\circ}
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