Question 78
Let $ABC$ be an acute triangle with orthocentre $H$. The line through the point $A$ perpendicular to $AC$ and the line through the point $B$ perpendicular to $BC$ intersect at $D$. The circle with centre $C$ which contains $H$ intersects the circumcircle of the triangle $ABC$ at $E$ and $F$. Prove that $|DE| = |DF| = |AB|$. (Stipe Vidak)