
1. Let $A B C$ be an acute angled triangle. The circle $\Gamma$ with $B C$ as diameter intersects $A B$ and $A C$ again at $P$ and $Q$, respectively. Determine $\angle B A C$ given that the orthocenter of triangle $A P Q$ lies on $\Gamma$.


Solution. Let $K$ denote the orthocenter of triangle $A P Q$. Since triangles $A B C$ and $A Q P$ are similar it follows that $K$ lies in the interior of triangle $A P Q$.

Note that $\angle K P A=\angle K Q A=90^{\circ}-\angle A$. Since $B P K Q$ is a cyclic quadrilateral it follows that $\angle B Q K=180^{\circ}-\angle B P K=90^{\circ}-\angle A$, while on the other hand $\angle B Q K=\angle B Q A-\angle K Q A=$ $\angle A$ since $B Q$ is perpendicular to $A C$. This shows that $90^{\circ}-\angle A=\angle A$, so $\angle A=45^{\circ}$.
