Let $\omega$ be the circumscribed circle of triangle $ABC$, and let $\omega'$ 'be the circle tangent to the side $BC$ and the extensions of the sides $AB$ and $AC$. The common tangents to the circles $\omega$ and $\omega'$ intersect the line $BC$ at points $D$ and $E$. Prove that $\angle BAD = \angle CAE$.