Let $DM \cap \Omega = \{R\}$ . Then the whole problem boils down to proving the following claim : Claim: $A , X$ and $R$ are collinear. The proof is just a trig bash, using only the sine rule and the trigonometric version of Ceva’s theorem : $$\prod \frac{\sin{\angle BAR}}{\sin{\angle CAR}} = \prod \frac{\frac{FR \sin{\angle AFR}}{AR}} {\frac{ER \sin{\angle AFR}}{AR}} = \prod \frac{FR}{ER}\cdot \frac{\sin{\angle FDR}}{\sin{\angle EDR}} = \prod \frac{FR^2}{ER^2} = \frac{FR^2}{ER^2} \cdot \frac{DP^2}{FP^2} \cdot \frac{QE^2}{QD^2} = 1$$since $FR = QD , PD = ER$ and $PF = QE$ from the isosceles trapezoids inscribed in $\Omega$ With this claim proved we can reformulate the problem like this : Let $\bigtriangleup DEF$ and $\Omega$ its circumcenter . The median $DM$ intersects $\Omega$ again at $R$ . If $A$ is the intersecton of the tangents at $E$ and $F$ the prove that $\angle RAM = \angle DAM$ Let $DR \cap \Omega =\{G\}$ . Now if we reflect every point about the perpendicular bisector of $BC$ is easy to see that $R$ goes to $G$ since arcs $EG$ and $FR$ are equal because $DG$ is symmedian and $DM$ is median . Then the line $AR$ goes to line $AD$ . Since the refelction is about $AM$, the conclusion follows.

Diagram: https://youtu.be/xNqNUgSpo5k

Attachments:

Let $ T $ be the second intersection of $ AD $ with $ \Omega. $ Note that $ AD $ is the D-symmedian of $ \triangle DEF, $ so $$ \measuredangle BDP = \measuredangle EDM = \measuredangle TDF = \measuredangle TFB. $$i.e. $ BT, BX $ are isogonal conjugate WRT $ \angle B. $ Similarly, we can prove that $ CT, CX $ are isogonal conjugate WRT $ \angle C, $ so $ T, X $ are isogonal conjugate WRT $ \triangle ABC. $ $ \qquad \blacksquare $

Let $DM \cap \Omega=\{D,R\}.$ As in #2, it suffices to show $A,X,R$ are collinear. We prove a more general result: General 2020 Taiwan TST Round 3 wrote: Let $ABC$ be a triangle with excircle $\gamma$ and tangent to $BC,CA,AB$ in $D,E,F$ respectively. Let $K,L,M \in \gamma$ such that $DK,EL,FM$ are concurrent at some point $T.$ Then $AK, BL, CM$ are concurrent. (In the problem $T$ is a point at infinity and $DK$ bisects $EF).$ I believe there's a simpler proof, and will add it in when I find it. For now define $EF \cap KK=K'$ and $L',M'$ similarly. The projective dual of this result is to show $K',L',M'$ are collinear. Apply Pascal: on \begin{align*} &KKXYZM \implies \{K', T, XY \cap MK\} \text{ collinear} \\ &LLYZXK \implies \{L',T, YZ \cap KL\} \text{ collinear} \\ &MMZXYL \implies \{M', T, ZX \cap LM\} \text{ collinear.} \end{align*}By Desargues theorem, $\{XY \cap MK, YZ \cap KL, ZX \cap LM\}$ are collinear. Thus, $K',L',M'$ are collinear. 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A homography or moving points can be also be used to trivialize this generalization.

@above Ig your Generazilation is the Extraversion of the Steinbart Rabonitwz Theorem. Anyway here is another Generalization. Jerabek's Theorem: Let $\Delta UVW$ and $\Delta XYZ$ be the circumcevian triangles of $P$ and $Q$ with respect to $\triangle ABC$. Then the triangle formed by the lines $\{XU, YV, ZW\}$ is perspective with $\Delta ABC$. $(\bigstar)$ Let $UX\cap BC=A^*$; $VY\cap AC=B^*$ ; $ZW\cap AB=C^*$. Fix Point $Q$ ,$\Delta ABC$ and a line $\ell$ passing through $A$. Now Animate $\{P\}$ on $\ell$. Now $P\mapsto V\mapsto B^*$ and $P\mapsto W\mapsto C^*$ are homographies. Let $A^*B^*\cap AB=C'$. Then $P\mapsto C^*$ is a Homography. Hence, $C^*,C'$ coincide on three Input values of $\{P\}$ on $\ell$. Now considering the cases when $P\equiv A,U,\ell\cap BC$ we are done. $\blacksquare$ Now the above generalization mentioned by Wizard_32 is the Extraversion of $(\bigstar)$ when $P=Q$. $\blacksquare$