In the triangle $ABC$ it is known that $\angle ACB> 90 {} ^ \circ$, $\angle CBA> 45 {} ^ \circ$. On the sides $AC$ and $AB$, respectively, there are points $P$ and $T$ such that $ABC$ and $PT = BC$. The points ${{P} _ {1}}$ and ${{T} _ {1}}$ on the sides $AC$ and $AB$ are such that $AP = C {{P} _ {1}}$ and $AT = B {{T} _ {1}}$. Prove that $\angle CBA- \angle {{P} _ {1}} {{T} _ {1}} A = 45 {} ^ \circ$. (Anton Trygub)