In the triangle $ABC$ the angle bisector $AD$ is drawn, $E$ is the point of tangency of the inscribed circle to the side $BC$, $I$ is the center of the inscribed circle $\Delta ABC$. The point ${{A} _ {1}}$ on the circumscribed circle $\Delta ABC$ is such that $A {{A} _ {1}} || BC$. Denote by $T$ - the second point of intersection of the line $E {{A} _ {1}}$ and the circumscribed circle $\Delta AED$. Prove that $IT = IA$.