PROBLEM
Let $\Gamma$ be the circumcircle of $∆ABC$. Let $D$ be a point on the side $BC$. The tangent to $\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\Gamma$ again at $F$. Suppose $B$, $D$, $F$, $E$ are concyclic. Prove that $AC, BF, DE$ are concurrent.
MY APPROACH
Let $X$ be some point on tangent line then, $$\angle XAB=\angle ACB=\angle ACD \quad(1)$$
Since $AB ||DE$, we cany say that $$\angle AED=\angle XAB \quad(2)$$
From $(1)$ and $(2)$, we can say that
$$\angle AED=\angle ACD$$
Therefore $ADCE$ is cyclic.
Consider the Circles of $BDEF$, Circles of $ADCE$ and Circumcircle of $∆ABC$. We see that $AC,BF$ and $DE$ are the radical axes of these circles which means they are concurrent at the radical Centre.
Hence, Proved!
But APMO-$2020$ haven't included my solution in their Official Answers. Is something wrong with my Proof?
DIAGRAM
