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Belarusian Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Belarus

The point $E$ lies on the altitude $BD$ of an acute triangle $ABC$ . It is given that $\angle AED = 50^\circ$ and the circumcircles of the triangles $ADE$ and $BEC$ tangent to each other at $E$ .
Find $\angle BCE$ .

Solution

$40^\circ$ .

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