Parallel Lines Associated with a Mixtilinear Incircle

Let $I$, $O$ and $\tau$ be the incenter, circumcenter and circumcircle of triangle $ABC$. The circle $\omega$ is tangent to lines $AB$ and $BC$, and touches internally $\tau$ at $T$. The tangents to $\tau$ at $T$ and $B$ intersect at $P$. Prove that $IP\parallel{AC}$.

Proof. It suffices to show that $\angle{\frac{ACB}{2}}=\angle{CIP}$. We know from lemma 2 in a previous problem that $\triangle{BPI}$ is isosceles with $\angle{IBP}=\angle{BIP}=\frac{\angle{ABC}}{2}+\angle{BAC}$. We have

$$\angle{CIP}=\angle{BIC}-\angle{BIP}=180^\circ-\frac{\angle{ABC}}{2}-\frac{\angle{ACB}}{2}-\frac{\angle{ABC}}{2}-\angle{BAC}.$$

But $180^\circ-\angle{ABC}-\angle{BAC}=\angle{ACB}$. Hence, $\angle{CIP}=\frac{\angle{ACB}}{2}$.

$\square$