Finding an angle of triangle with inner circle
Let the tangent point on $BC$ be $T$. So $\angle ITB =\angle ITC = 90^\circ$.
$$\begin{aligned} \angle BIC &= \angle BIT +\angle CIT\\ &=(90^\circ - \angle IBT) + (90^\circ - \angle ICT)\\ &=180^\circ - \frac12\angle ABC - \frac12\angle ACB\\ &= 180^\circ - \frac12(180^\circ- \angle BAC)\\ &= 90 + \frac12\angle BAC \end{aligned}$$
Since $\angle BIC = 135^\circ$, $\angle BAC = 90^\circ$.
Simply $$\angle BAC= \pi-2y-2\alpha$$