Consider a triangle $ABC,$ and let $D$ the foot of the bisector of angle $\angle{BAC},$ $M$ the midpoint of $BC,$ $D'$ the symmetric of $D$ with respect to $M.$ Consider the circle $\omega_1$ tangent externally to $BC$ in $D$ and internally to $\Gamma,$ the circumcircle of $ABC.$ Similarly consider the circle $\omega_2$ tangent externally to $BC$ in $D'$ and internally to $\Gamma.$ Let $E$ be the center of $\omega_1$ and let $E'$ be the center of $\omega_2.$ Prove that the following relation holds: $$(\overline{AB}-\overline{AC})^2=\overline{AE'}^2-\overline{AE}^2$$
All my attempts didn't give anything. I tried using coordinates but without any result; without coordinates, I know that the line between the center $O$ of $\Gamma$ and the center of $\omega_1$ intersect $\Gamma$ in the tangency point, and similarly for $\omega_2...$
Can anyone give me at least one idea on how to prove this fact?