Chord tangent to two circles

Here is a question:

Let $Ω$ be a circle with a chord $AB$ which is not a diameter. Let $Γ_1$ be a circle on one side of $AB$ such that it is tangent to $AB$ at $C$ and internally tangent to $Ω$ at $D$. Likewise, let $Γ_2$ be a circle on the other side of $AB$ such that it is tangent to $AB$ at $E$ and internally tangent to $Ω$ at $F$. Suppose the line DC intersects $Ω$ at $X \neq D$ and the line $F E$ intersects $Ω$ at $Y \neq F$. Prove that $XY$ is a diameter of $Ω$.

At first I didn't see the condition that $X \neq D$ and $Y \neq F$, so I assumed both circles were tangent, which made this easy.

But when I try doing it without assuming them tangent, it gets difficult, and I have been unable to solve it. My main problem is that these circles could be arbitrary, so I know nothing more than the facts that their radii to AB will be parallel and that $\Omega$'s centre will be collinear with the centre of $\Gamma_1$ and $D$, and the centre of $\Gamma_2$ and $F$, separately.

This was asked in RMO 2017.

Let $O$ and $O_1$ be centers of $\Omega$ and $\Gamma_1$ respectively.
Thus, $$\measuredangle D=\measuredangle O_1CD=\measuredangle OXD,$$ which says that $OX||O_1C$ and since $O_1C\perp AB$, we obtain $$OX\perp AB,$$ which says that $X$ is a midpoint of the arc $AFB$.
Similarly, we get that $Y$ is a midpoint of the arc $ADB$, which says that $XY$ is a diameter of $\Omega$.