Let $ABC$ be a triangle, and let $D$ be a point on side $AB$. Circle $\omega_1$ passes through $A$ and $D$ and is tangent to line $AC$ at $A$. Circle $\omega_2$ passes through $B$ and $D$ and is tangent to line $BC$ at $B$. Circles $\omega_1$ and $\omega_2$ meet at $D$ and $E$. Point $F$ is the reflection of $C$ across the perpendicular bisector of $AB$. Prove that points $D$, $E$, and $F$ are collinear.
Denote the circumcircle of $\triangle{ABC}$ as $\Gamma$. Note that $F\in\Gamma$: by reflection in the perpendicular bisector of $AB$: $\angle{BFA}=\angle{ACB}=\angle{BCA}$ hence $A,B,C,F$ are concyclic.
Now we can reparameterise the problem as follows: given $\triangle{ABC}$ with circumcircle $\Gamma$ and define $F$ as before. Now take a point $D$ on the line segment $AB$ and let $FD$ meet $\Gamma$ again at $E$. Prove that circle $ADE$ is tangent to $AC$ (and thus by symmetry circle $BDE$ is tangent to $BC$).
But this is an easy angle chase: $\angle{AED}=\angle{AEF}=\angle{ABF}=\angle{BAC}=\angle{DAC}$ - hence by converse of alternate segment theorem we are done.