Clearly $X$ is the A-humpty point of $\triangle ABC$ so using its config we get $TDXA$ cyclic, now since $ME,MF$ are tangent to $(AEF)$ by PoP $$MF^2=MA \cdot MX=MD \cdot MT \implies MF \; \text{tangent to} (TFD)$$Which means that $(TFD)$ is tangent to $(AEF)$ at $F$ thus we are done

Guten Tag. From Euler theorem we have EFDM cyclic. MT is radical for EFDM and XHDM. MH diametr then angle (AXH) is right angle. We have EAFHX cyclic. XT is radical of the AEXHF and MXHD As the radicals of three circles intersect at one point, then radical of AFHXE and EMDF is ET . We have points E F T collinear. From angle chasing FM tangent to the circle (AEF). MF^2=MX*MA=MD*MT then MF is tangent to the circle (TDF).

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We have $\angle FMC=2\angle B,\angle FED=2\angle FCB=180^{\circ}- 2\angle B$ So $X,H,M,D$ is cyclic Consider $F,X,H,E,\ \ \ \ $$X,H,M,D$is cyclic By root axis theorem,we can know $F,E,T$ is collinear Since $\angle A = \angle FDB= \angle DFT+ \angle FTD$ Hence the circumcircles of $\triangle TFD$ and $\triangle AEF$ are tangent.

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Clearly $X$ is the $A$ humpty point so $T$ is just $EF \cap BC$. Since $MF$ is tangent to $(AEF)$, it remains to show it is tangent to $(TFD)$ as well but since $(TD;BC)=-1$, we have that $D$ inverts to $T$ under inversion about $(BC)$ so $MD.MT=MB^2=MF^2$ which finishes.