XXIX Rioplatense Mathematical Olympiad

Let's assume the incircle is tangent to $AB$ at $F$, then we have that $\angle FAD=\angle IAQ$ and $\angle AFD=\frac{180^{\circ }-\angle B}{2}=\angle AIQ$ where the first equality is from the statement and the other are well-known identities. It follows that $△AFD$ and $△AIQ$ are similar to each other and hence $△ADQ$ and $△AFI$ are also similar to each other. In particular $\angle ADQ=\angle AFI=90^{\circ }$ and moreover, by a similar argument, we get the analogous equality $\angle ADP=90^{\circ }$. Finally, we observe that $\angle PAD=\angle CAI=\angle BAI=\angle QAD$ and hence $AD$ is the angle bisector and also the altitude that corresponds to vertex $A$ in triangle $△PAQ$. Thus $AP=AQ$.