Question 133
Let $ABC$ be a triangle with $AB \ne AC$, and let $H$ be its orthocenter. Let $T$ be a point on the arc $BC$ of the circumcircle of triangle $ABC$ that does not contain point $A$. Let $l$ be the line through $H$ parallel to $BC$, and let $l$ intersect lines $TB$ and $TC$ at points $P$ and $Q$, respectively. Let the circumcircles of triangles $PAB$ and $QAC$ intersect again at point $S$.
Suppose that $\angle PAQ = 2\angle BAC$. Prove the following.
(1)
The point $S$ lies on the circumcircle of triangle $BHC$.
(2)
$$
\angle PAH = \angle HAQ \text{ or } \angle PAB = \angle BAH.
$$
Suppose that $\angle PAQ = 2\angle BAC$. Prove the following.
(1)
The point $S$ lies on the circumcircle of triangle $BHC$.
(2)
$$
\angle PAH = \angle HAQ \text{ or } \angle PAB = \angle BAH.
$$