2018 OIM Problems/Problem 2

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Problem

Let $ABC$ be a triangle such that $\angle BAC = 90^{\circ}$ and $BA = CA$. Let $M$ be the midpoint of $BC$. A point $D \ne A$ is chosen on the semicircle of diameter $BC$ that contains $A$. The circumcircle of the triangle $DAM$ intersects the lines $DB$ and $BC$ at the points $E$ and $F$, respectively. Show that $BE = CF$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

OIM Problems and Solutions