China Mathematical Competition (Complementary Test)

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, $△ABD\sim △PCB$. 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 $△ABQ\sim △ACD$. 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. $\square$