Problem

Source: 2014 Romania JBMO TST 5.3

Tags: geometry, equal angles, midpoints, diameter



Let $ABC$ be an acute triangle and let $O$ be its circumcentre. Now, let the diameter $PQ$ of circle $ABC$ intersects sides $AB$ and $AC$ in their interior at$ D$ and $E$, respectively. Now, let $F$ and $G$ be the midpoints of $CD$ and $BE$. Prove that $\angle FOG=\angle BAC$