Let $A B C$ be a triangle, $I$ its incenter and $D$ a point on $B C$. Consider the circle that is tangent to the circumcircle of $A B C$ but is also tangent to $D C$, $D A$ at $E, F$ respectively. Then $E, F$ and $I$ are collinear.
Proof. There is a "computational" proof using Casey's theorem and transversal theorem. You can try to work that out yourself. Here, we show a clever but difficult synthetic proof (communicated to me via Oleg Golberg).
Denote $\Omega$ the circumcircle of $A B C$ and $\Gamma$ the circle tangent tangent to the circumcircle of $A B C$ and lines $D C, D A$. Let $\Omega$ and $\Gamma$ touch at $K$. Let $M$ be the midpoint of arc $\overparen{B C}$ on $\Omega$ not containing $K$. Then $K, E, M$ are collinear (think: dilation with center $K$ carrying $\Gamma$ to $\Omega$ ). Also, $A, I, M$ are collinear, and $M I=M C$.
Let line $E I$ meet $\Gamma$ again at $F^{\prime}$. It suffices to show that $A F^{\prime}$ is tangent to $\Gamma$.
Note that $\angle K F^{\prime} E$ is subtended by $\overparen{K E}$ in $\Gamma$ and $\angle K A M$ is subtended by $\overparen{K M}$ in $\Omega$. Since $\overparen{K E}$ and $\overparen{K M}$ are homothetic with center $K$, we have $\angle K F^{\prime} E=\angle K A M$, implying that $A, K, I^{\prime}, F^{\prime}$ are concyclic.
We have $\angle B C M=\angle C B M=\angle C K M$. So $\triangle M C E \sim \triangle M K C$. Hence $M C^{2}=M E \cdot M K$. Since $M C=M I$, we have $M I^{2}=M E \cdot M K$, implying that $\triangle M I E \sim \triangle M K I$. Therefore, $\angle K E I=\angle A I K=\angle A F^{\prime} K$ (since $A, K, I^{\prime}, F^{\prime}$ are concyclic). Therefore, $A F^{\prime}$ is tangent to $\Omega$ and the proof is complete.
The transversal theorem is a criterion for collinearity. It states that if $A, B, C$ are three collinear points, and $P$ is a point not on the line $A B C$, and $A^{\prime}, B^{\prime}, C^{\prime}$ are arbitrary points on lines $P A, P B, P C$ respectively, then $A^{\prime}, B^{\prime}, C^{\prime}$ are collinear if and only if
$$
B C \cdot \frac{A P}{A^{\prime} P}+C A \cdot \frac{B P}{B^{\prime} P}+A B \cdot \frac{C P}{C^{\prime} P}=0,
$$
where the lengths are directed. In my opinion, it's much easier to remember the proof than to memorize this huge formula. The simplest derivation is based on relationships between the areas of $[P A B],\left[P A^{\prime} B^{\prime}\right]$, etc.
