Question 147
Given a triangle $ABC$ with $\angle ACB \gt 90^\circ$, $\angle CBA \gt 45^\circ$. Points $P$ and $T$ belong to the sides $AC$ and $AB$, respectively, so that $PT = BC$ and $PT \perp BC$. Points $P_1$ and $T_1$ belong to the sides $AC$ and $AB$ so that $AP = CP_1$ and $AT = BT_1$. Show that $\angle CBA - \angle P_1T_1A = 45^\circ$.