Let $ABC$ be an acute-angled triangle and let $D$, $E$, and $F$ be the midpoints of $BC$, $CA$, and $AB$ respectively. Construct a circle, centered at the orthocenter of triangle $ABC$, such that triangle $ABC$ lies in the interior of the circle. Extend $EF$ to intersect the circle at $P$, $FD$ to intersect the circle at $Q$ and $DE$ to intersect the circle at $R$. Show that $AP=BQ=CR$.