IR-TST 2015 Problem 2

$I_b$ is the $B$ -excenter of the triangle $ABC$ and $\omega$ is the circumcircle of this triangle. $M$ is the middle of arc $BC$ of $\omega$ which doesn't contain $A$ . $MI_b$ meets $\omega$ at $T\not =M$ . Prove that $$TB\cdot TC=TI_b^2.$$