Question 12
G3 (GBR) Let $ABC$ be an acute-angled triangle with $BC\gt CA$. Let $O$ be the circumcenter, $H$ its orthocenter, and $F$ the foot of its altitude $CH$. Let the perpendicular to $OF$ at $F$ meet the side $CA$ at $P$. Prove that $\angle FHP=\angle BAC$.
Possible second part: What happens if $|BC|\le|CA|$ (the triangle still being acute-angled)?
Possible second part: What happens if $|BC|\le|CA|$ (the triangle still being acute-angled)?