This problem is driving me crazy. It's from Andreescu's Mathematical Olympiad Challenges:
Let $AB$ be a chord in a circle and $P$ a point on the circle. Let $Q$ be the projection of $P$ onto $AB$ and $R$ and $S$ the projections of $P$ onto the tangents to the circle at $A$ and $B$. Prove that $PQ$ is the geometric mean of $PR$ and $PS$.
The solution given is as follows:
We will prove that the triangles $PRQ$ and $PQS$ are similar. This will imply $PR/PQ = PQ/PS$; hence $PQ^2 = PR\cdot PS$.
The quadrilaterals $PRAQ$ and $PQBS$ are cyclic, since each of them has two opposite right angles. In the first quadrilateral $\angle PRQ=\angle PAQ$ and in the second $\angle PQS = \angle PBS$. By inscribed angles, $\angle PAQ$ and $\angle PBS$ are equal. It follows that $\angle PRQ = \angle PQS$. A similar argument shows that $\angle PQR=\angle PSQ$. This implies that the triangles $PRQ$ and $PQS$ are similar, and the conclusion follows.
Now, just HOW by inscribed angles is $\angle PAQ=\angle PBS$? Those two do not subtend chords in the same circle, and I tried using angle chasing to find their values, but even if I consider the larger cyclic quadrilateral with vertices $P,R,S$ and the intersection of the tangents of $A$ and $B$, the two don't subtend any chords in common.
Can someone enlighten this tortured soul? I've been chasing angles for the past month now and it even haunts my dreams.