$\textbf{Problem.}$ $ABCD$ is a cyclic quadrilateral and the tangent to the circle at $D$ intersect $AB$ and $BC$ at $E$ and $F$ respectively. $T$ is a point inside $\triangle ABC$ such that $TE$ is parallel to $CD$ and $TF$ is parallel to $AD$. Say $TE$ and $TF$ intersect $AC$ at $M$ and $N$ respectively, it is given that $AEDM, MTND, DNCF$ are concyclic. $D$ is tangent to both $AEDM$ and $DNCF$. Let $AC$ intersect $TD$ at $L$. Let $BL$ intersect the circumcircle of $MTND$ at $X$. Prove that $NX$ is parallel to $AB$.
$\textbf{Attempt:}$ The first thing that came into my mind after drawing the diagram was to somehow use Reim's theorem. If we can prove that $A,D,X$ are collinear then we will be done. Some approaches that can be taken are as follows:
- Proving $\angle CDX= \angle CAD$ as then bt Alternate Segment Theorem, it would be proven that $X$ lie on $EF$.
- Proving $\angle TDX + \angle TDE=180°$.
- Proving $\angle MDE+ \angle MDX= 180°$.
- Proving $\angle BAN = \angle ANX$.
- Proving $\angle MDE = \angle MNX$. Proving any one of the above claims will solve the problem, however there can be other ways to solve it as well.
I have attached a diagram.