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
You have a beautiful, neat proof for this problem, and it is absolutely correct, in fact it looks much more neat than the official solutions. There is nothing wrong with your solution.
But when exams are created, the creators do not know all the possible solutions to a problem, and they don't list all of them. However, as we try to solve them we can discover many new proofs which were not originally given as official solutions.
This is your own solution to the problem, and it doesn't have to be among the official solutions. Hope this helps :)