It suffices to proof that the incircle is the $A-excircle$ of $\triangle APQ$
Let $I'$ be the incircle of $\triangle APQ$, so obviously $A-I'-I$ are collinear.
Now, By alternate segment theorem, we have $\angle QIA=\angle QCI=\frac{\angle C}{2}$
similarly, $\angle PIA=\angle PBI=\frac{\angle B}{2}$
So, $\angle QIP=\angle QIA+\angle PIA=\frac{\angle C}{2}+\frac{\angle B}{2}=90^{\circ}-\frac{\angle A}{2}$
Therefore, $\angle QI'P+\angle QIP=(90^{\circ}+\frac{\angle A}{2})+(90^{\circ}-\frac{\angle A}{2})=180^{\circ}\Rightarrow Q,I',P,I$ are con-cyclic.
so we get that $\angle PQI=\angle PI'I=90^{\circ}-\frac{\angle AQP}{2}\Rightarrow \angle IQC=90^{\circ}-\frac{\angle AQP}{2}$
Hence $IQ$ is the angle-bisector of $\angle PQC$
Similarly, $IP$ is the angle-bisector of $\angle BPQ$
An thus $I$ is the $A-excenter$ of $\triangle APQ$, as the incircle of $ABC$ is tangent to the extension of $AP$ and $AQ$, we can say that the incircle of $ABC$ is indeed the excircle of $APQ$, and so it must be tangent to $PQ.\blacksquare$