Taiwan IMO Selection Camp, 2022

Let the segments $TE$ and $TF$ cross $AC$ at $P$ and $Q$, respectively. Since $PE \parallel CD$ and $ED$ is tangent to the circumcircle of $ABCD$, we have
$$\angle EPA = \angle DCA = \angle EDA,$$
and so the points $A$, $P$, $D$, and $E$ lie on some circle $\alpha$. Similarly, the points $C$, $Q$, $D$, and $F$ lie on some circle $\gamma$.

We now want to prove that the line $DT$ is tangent to both $\alpha$ and $\gamma$ at $D$. Indeed, since $\angle FCD + \angle EAD = 180^\circ$, the circles $\alpha$ and $\gamma$ are tangent to each other at $D$. To prove that $T$ lies on their common tangent line at $D$ (i.e., on their radical axis), it suffices to check that $TP \cdot TE = TQ \cdot TF$, or that the quadrilateral $PEFQ$ is cyclic. This fact follows from
$$\angle QFE = \angle ADE = \angle APE.$$

Since $TD = TK$, we have $\angle TKD = \angle TDK$. Next, as $TD$ and $DE$ are tangent to $\alpha$ and $\Omega$, respectively, we obtain
$$\angle TKD = \angle TDK = \angle EAD = \angle BDE,$$
which implies $TK \parallel BD$.

Next, we prove that the five points $T$, $P$, $Q$, $D$, and $K$ lie on some circle $\tau$. Indeed, since $TD$ is tangent to the circle $\alpha$ we have
$$\angle EPD = \angle TDF = \angle TKD,$$
which means that the point $P$ lies on the circle $(TDK)$. Similarly, we have $Q \in (TDK)$.

Finally, we prove that $PK \parallel BC$. Indeed, using the circle $\tau$ and $\gamma$ we conclude that
$$\angle PKD = \angle PQD = \angle DFC,$$
which means that $PK \parallel BC$.

Triangles $TPK$ and $DCB$ have pairwise parallel sides, which implies the fact that $TD$, $PC$, and $KB$ are concurrent, as desired.