The circumcircle of triangle ABC has centre $O$. $P$ is the midpoint of $\widehat{BAC}$ and $QP$ is the diameter. Let $I$ be the incentre of $\triangle ABC$ and let $D$ be the intersection of $PI$ and $BC$. The circumcircle of $\triangle AID$ and the extension of $PA$ meet at $F$. The point $E$ lies on the line segment $PD$ such that $DE=DQ$. Let $R,r$ be the radius of the inscribed circle and circumcircle of $\triangle ABC$, respectively.
Show that if $\angle AEF=\angle APE$, then $\sin^2\angle BAC=\dfrac{2r}{R}$.
China TST 2013 Day 5 Problem 2
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$