Question 79
In acute-angled triangle $\triangle ABC$, altitudes $BE, CF$ meet at $H$. A perpendicular line is drawn from $H$ to $EF$ and intersects arc $BC$ of the circumcircle of $\triangle ABC$ (the one that doesn't contain $A$) at $K$. If $AK, BC$ meet at $P$, prove that $PK = PH$.