Problem 1381

1. Definitions and Initial Setup:
- Let the circle be denoted by $\omega$ with center $O$.
- Points $P$, $B$, and $A$ lie on the circle $\omega$.
- Point $Q$ is inside the circle such that $\angle PAQ = 90^\circ$ and $PQ = BQ$.
- Extend $AQ$ to intersect the circle $\omega$ again at point $C$.
- Draw segments $\overline{BC}$, $\overline{PB}$, and the diameter $\overline{PC}$, noting that $O$ lies on $\overline{PC}$.
- Let $\overline{PB}$ intersect $\overline{AQ}$ at point $E$.
- Let $M$ be the midpoint of $\overline{PB}$ and the foot of the perpendicular from $O$ to $\overline{PB}$.

2. Collinearity and Parallel Lines:
- Since $M$ is the midpoint of $\overline{PB}$, $M$, $Q$, and $O$ are collinear.
- $\overline{MO}$ is the midline of $\triangle BPC$, so $\overline{MO} \parallel \overline{BC}$.
- This implies $\angle MQE = \angle BCA$.

3. Angle Chasing:
- First, note that $\angle MQB = \angle MQP$ because $PQ = BQ$ and $M$ is the midpoint of $\overline{PB}$.
- Therefore, $\angle MQB = \angle MQE + \angle PQA$.

4. Combining Angles:
- We need to find $\angle AQB - \angle PQA$.
- Using the previous results:
$$\angle AQB - \angle PQA = (\angle MQB + \angle MQE) - \angle PQA$$
$$= (\angle MQE + \angle PQA + \angle MQE) - \angle PQA$$
$$= 2\angle MQE$$
$$= 2\angle BCA$$
$$= \angle BOA$$
$$= \stackrel{\frown}{AB}$$

Thus, we have shown that $\angle AQB - \angle PQA = \stackrel{\frown}{AB}$.

$\blacksquare$