Let $t$ be the radical axis of the two circles. Notice that the midpoints $M$ of $CD$ is on this line. Let the angle be $\angle AXF$, where $AX$ touches $\omega, \Omega$ at $Y, Z'$ and $FX$ touches $\Omega, \omega$ at $Z, Y'.$ Let $A_1$ be the point on $FX$ so that $A_1Z = AY$ and $A_1Y = AZ'$, and let $A'$ be the reflection of $A$ over $M.$ Notice that the powers of $A', A_1$ with respect to $\omega$ are both equal to the power of $A$ with respect to $\Omega.$ Also, the powers of $A', A_1$ with respect to $\Omega$ are both equal to the power of $A$ with respect to $\omega.$ Hence, if we let $O_1, O_2$ be the centers of $\omega, \Omega$ respectively, we have that $A'O_1 = A_1O_1$ and $A'O_2 = A_1O_2.$ This means that either $A' = A_1$ (in which case we're done as $A' \in FX \Rightarrow A' = F$) or $A'$ is the reflection of $A_1$ over $O_1O_2.$ In the latter case, that would imply $A' \in AX \Rightarrow A' = CD \cap AX = A$. However $A' = A$ is clearly absurd, and so the former case holds and we're done. $\square$