Let $P$ be a point such that after inversion wrt $P$ $\omega_1^*$ and $\omega_2^*$ are concentric, which is well known to exist (let $O$ be the new common center). Now, the angle condition gets simply translated to $\angle A^*PC^*=\angle B^*PD^*$, where $(A^*B^*C^*D^*P)$ is cyclic. But this cyclicity + the two concentric circles implies by symmetry that the arc $A^*C^*$ is equal to the arc $B^*D^*$, so we are done.

Let $P$ be a point such that $\omega_1,\omega_2$, and the point circle $P$ are coaxial. Let the radical axis of $\omega_1,\omega_2$ meet $\ell$ at $Q$. Then the circle centered at $Q$ with radius $\sqrt{QA\cdot QB}=\sqrt{QC\cdot QD}=QP$ is an Appolonian circle with respect to $A,B$ and $C,D$, so the angle bisectors of $\angle APB$ and $\angle CPD$ meet at the same point on $\ell$, or are the same, as desired.