First note that $DE$ and $DF$ are orthogonal and $DE$ bisects $\angle BDA$, so $-1=(DA,DB;DE,DF)\overset{D}{=}(DA\cap EF, BC\cap EF,E,F)\overset{A}{=}(D,BC\cap EF; B,C)=(B,C;D,BC\cap EF)$. Now we have $$(B,C;D,BC\cap EF)=-1=(E,F;\infty_{EF},M)\overset{Y}{=}(E,F;X,T)\overset{A}{=}(B,C;D,BC\cap AT)$$so $BC\cap EF=BC\cap AT$.

From Ceva we have $AD,BF,CE$ concurrent.So if $S=EF\cap BC$ then $(B,C/D,S)=-1$ If $S'=AT\cap BC$ then $(B,C/D,S')=(E,F/X,T)=Y(E,F/X,M)=-1$ So $AT,EF,BC$ concurrent at $S$

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Let $S_1=AT \cap BC$ and $S_2=EF \cap BC$. We will show that $S_1=S_2$ $$\angle ETX =\angle EFX=\angle FEY=\angle FTY \implies ETFX \text{ is harmonic quadrilateral } \implies -1=(E,F;X,T) \overset{A}{=} (S_1,D;B,C) \qquad (*)$$ On the other hand by $\text{ Menelaus theorem}$ $$\frac{CF}{AF} \cdot \frac{AE}{EB} \cdot \frac{BS_2}{S_2C}=1 \iff \frac{CD}{AD} \cdot \frac{AD}{BD} \cdot \frac{BS_2}{S_2C}=1 \iff S_2C \cdot BD=CD \cdot BS_2 \iff (S_2,D ; B,C)=-1 \qquad (**)$$ By $(*)$ and $(**)$ we get $S_1=S_2$ which is desired result

Ibrahim_K wrote: Let $S_1=AT \cap BC$ and $S_2=EF \cap BC$. We will show that $S_1=S_2$ $$\angle ETX =\angle EFX=\angle FEY=\angle FTY \implies ETFX \text{ is harmonic quadrilateral } \implies -1=(E,F;X,T) \overset{A}{=} (S_1,D;B,C) \qquad (*)$$ On the other hand by $\text{ Menelaus theorem}$ $$\frac{CF}{AF} \cdot \frac{AE}{EB} \cdot \frac{BS_2}{S_2C}=1 \iff \frac{CD}{AD} \cdot \frac{AD}{BD} \cdot \frac{BS_2}{S_2C}=1 \iff S_2C \cdot BD=CD \cdot BS_2 \iff (S_2,D ; B,C)=-1 \qquad (**)$$ By $(*)$ and $(**)$ we get $S_1=S_2$ which is desired result Thanks for the solution Very nice solution

Suppose that $AX \cap EF=S$. Notice that $\angle TYX = \angle TMS = \angle TAX$, hence $ATSM$ is cyclic. Define $AT \cap EF =Q$, then we haveL $$ QA \cdot QT = QS \cdot QM = QE \cdot QF$$Ths implies that $(Q,S;E,F)=-1$. On the other hand, $(BC \cap EF, S; E,F) =-1$, this implies that $BC \cap EF =Q$, hence $AT, EF, BC$ are concurrent as needed.

Take $EF \cap BC$ at $X$ and $AX \cap (AEF)$ at $T'$.It is known that, $(X,D;B,C)=-1$ take pencil from $A$ to $(AEF)$ $\implies$ $ET'FX$ is harmonic.So, $TX$ is symmedian in $\triangle ETF$.Thus, $\angle ET'X=\angle MT'F=\angle XFE=\angle YEF$ it means $T-M-Y$.The $\angle XFE=\angle YEF$ gets from $XYFE$ isosceles trepazoid. We are done.

Using the Angle Bisector Theorem and Ceva's Theorem we get that $AD$, $BG$ and $CE$ concur, so $EF$ passes through the harmonic conjugate of $D$ wrt $\{B,C\}$. Hence it suffices to prove that $TEXF$ is harmonic, but this follows from$$\{T,X;E,F\}\underset{Y}{=}\{M,\infty ;E,F\}=-1.$$