IGO 2022 I1

Given is a circle $\omega$ and a line $\ell$ tangent to $\omega$ at $Y$ . Point $X$ lies on $\ell$ to the left of $Y$ . The tangent to $\omega$ , perpendicular to $\ell$ meets $\ell$ at $A$ and touches $\omega$ at $D$ . Let $B$ a point on $\ell$ , to the right of $Y$ , such that $AX=BY$ . The tangent from $B$ to $\omega$ touches the circle at $C$ . Prove that $\angle XDA= \angle YDC$ .

Note: This is not the official wording (it was just a diagram without any description).