In triangle $ABC,\angle B > \angle C, T$ is the midpoint of arc $BAC$ of the circumcicle of $ABC$, and $I$ is the incentre of $ABC$. Let $E$ be point such that $\angle AEI = 90^0$ and $AE$ is parallel to $BC$. If $TE$ intersects the circumcircle of $ABC$ at $P(\ne T)$ and $\angle B = \angle IPB$, determine $\angle A$.