As shown in Fig. 1.2, we extend segment $DP$ to intercept with the circle at point $E$. Then $\angle CPE = \angle DPA = \angle BPA$. Since $P$ is the midpoint of $AC$, we get $\overline{AB} = \overline{CE}$, which means $\angle CDP = \angle BDA$. Furthermore, $\angle ABD = \angle PCD$. Therefore, $\triangle ABD \sim \triangle PCB$. Then $\frac{AB}{BD} = \frac{PC}{CD}$, i.e., $AB \cdot CD = PC \cdot BD$.

Then we have
$$AB \cdot CD = \frac{1}{2} AC \cdot BD = AC \cdot \left(\frac{1}{2} BD\right) = AC \cdot BQ,$$
or $\frac{AB}{AC} = \frac{BQ}{CD}$. Combining it with $\angle ABQ = \angle ACD$, we derive that $\triangle ABQ \sim \triangle ACD$. So $\angle QAB = \angle DAC$.

Extending segment $AQ$ to intercept with the circle at point $F$, we then have
$$\angle CAB = \angle QAB - \angle QAC = \angle DAC - \angle QAC = \angle DAF,$$
which means $\widehat{BC} = \widehat{DF}$. Furthermore, as $Q$ is the midpoint of $BD$, then $\angle CQB = \angle DQF$.

Since $\angle AQB = \angle DQF$, we then have $\angle AQB = \angle CQB$.

The proof is completed. $\square$