In $\triangle ABC$, let $E$ be a point on $AC$, $F$ be a point on $AB$, and let $D$ be the intersection of $CF$ and $BD$. We are given that $A,E,D,F$ are concyclic, let this circle be $\Omega$. Let $\omega$ be the circle with diameter $BC$. Also, let $\omega$ and $\Omega$ intersect at $X$ and $Y$. If $M$ is the midpoint of $BC$, prove that $MX$ and $MY$ are tangent to $\Omega$.
I am not sure how to prove this. It may be worth noting that it is easy to see the case where $CF$ and $BE$ are altitudes and $D$ is the orthocenter.
Also, applying Pascal's Theorem on cyclic hexagon $FAEEDF$ implies the intersection of the tangents to $\Omega$ at $E$ and $F$ intersect on $BC$. Not sure if this helps though.
I have an algorithmic solution, but it is very far from being elegant (its style is "I have to solve this tough problem of a math competition in a very limited amount of time. Unleash brutality").
Let $\widehat{FCB}=\theta$. Then $\widehat{CFA}=\widehat{B}+\theta$ and since $AFDE$ is cyclic, $\widehat{AEB}=\pi-(\widehat{B}+\theta)$ and $\widehat{CBE}=\pi-(\widehat{B}+\widehat{C}+\theta)=\widehat{A}-\theta$, also because $\widehat{BDC}=\pi-\widehat{A}$.
Now we may use trigonometry to compute the lengths of $AF$ and $AE$ in terms of $a,b,c,\theta$, the cosine theorem to compute the length of $EF$ and $EF=2R_{\Omega}\sin\widehat{A}$ to compute $R_\Omega$.
In a skew reference system with $AE$ and $AF$ being the directions of the axis, we may also easily compute $MO_\Omega$. The orthogonality of $\omega$ and $\Omega$ then follows from checking that: $$ MO_\Omega^2 = \frac{a^2}{4}+R_{\Omega}^2.$$
Interesting (and probably useful for an elementary solution) facts: