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Let $ABC$ be an acute triangle with $AB < AC$. Denote by $D$ the foot of the perpendicular from $A$ to $BC$. Let $E$ be the point such that $ABEC$ is a parallelogram. Let $M$ be a point inside triangle $ABC$ such that $MB = MC$. Let $F$ be the reflection of point $D$ across the tangent to the circumcircle of triangle $ADM$ at point $M$. Prove that $AF = DE$.