Let $ABC$ be an acute angled triangle. Let $M$ be the midpoint of $BC$, and let $BE$ and $CF$ be the altitudes of the triangle. Let $D \ne M$ be a point on the circumcircle of the triangle $EFM$ such that $DE = DF$. Prove that $AD \perp BC$.
Source:
Tags: geometry, 5th edition
Let $ABC$ be an acute angled triangle. Let $M$ be the midpoint of $BC$, and let $BE$ and $CF$ be the altitudes of the triangle. Let $D \ne M$ be a point on the circumcircle of the triangle $EFM$ such that $DE = DF$. Prove that $AD \perp BC$.