Bump....

Let $N=FE \cap BC$ and $W=SX\cap BC,AN\cap \odot(ABC)=Y\neq A$ Equivalently i have to show that $WX\cdot WM=WT\cdot WM$. For this it's enough to show that $(T,N;B,C)=-1$ because then $WT\cdot WM=WB\cdot WC= WX\cdot WM$ Note that $(N,D;B,C)=-1$ so $A(N,D;B,C)=-1$ and thus $YBXC$ is harmonic quadrilateral, so $S(Y,X;B,C)=-1$ and now the only thing remains to prove is that $S,Y,T$ are collinear. For this it is enought to show that $\dfrac{YB}{YC}\cdot \dfrac{SB}{SC}=\dfrac{TB}{TC}\,\,\,(*)$ $S$ is the miquel point of $BCEF.AN$ so $\Delta SBF\sim \Delta SCE$ and so $\dfrac{SB}{SC}=\dfrac{BF}{EC}$ Also $\dfrac{NB}{SC}=\dfrac{YB}{YC}\cdot \dfrac{AB}{AC}\Leftrightarrow \dfrac{YB}{YC}=\dfrac{NB}{NC}\cdot \dfrac{AC}{AB}$ and $(N,D;B,C)=-1\Rightarrow \dfrac{NB}{NC}=\dfrac{DB}{DC}$. As a result $\dfrac{YB}{YC}\cdot \dfrac{SB}{SC}=\dfrac{DB}{DC}\cdot \dfrac{AC}{AB}\cdot \dfrac{BF}{EC}$ $TP\parallel NF\Rightarrow \dfrac{TN}{TC}=\dfrac{PF}{PC}\,\,(1)$ $TP\parallel NE\Rightarrow \dfrac{TN}{TB}=\dfrac{PE}{PB}\,\,(2)$ $(1)/(2)\Rightarrow \dfrac{TB}{TC}=\dfrac{FP}{PC}\cdot\dfrac{BP}{PE}=\dfrac{AF}{AC}\cdot \dfrac{\sin\angle BAD}{\sin\angle DAC}\cdot \dfrac{AB}{AE}\cdot \dfrac{\sin\angle BAD}{\sin\angle DAC}=\dfrac{BD}{DC}\cdot \dfrac{AF}{AE}\cdot \dfrac{\sin\angle BAD}{\sin\angle DAC}$ Now (*) becomes $\dfrac{BD}{DC}\cdot \dfrac{AF}{AE}\cdot \dfrac{\sin\angle BAD}{\sin\angle DAC}=\dfrac{DB}{DC}\cdot \dfrac{AC}{AB}\cdot \dfrac{BF}{EC}\Leftrightarrow $ $ \dfrac{AF}{AE}\cdot \dfrac{EC}{BF}=\dfrac{AC}{AB}\cdot\dfrac{\sin\angle DAC}{\sin\angle BAD}\Leftrightarrow \dfrac{AF}{AE}\cdot \dfrac{EC}{BF}\cdot \dfrac {BD}{DC}=1$ which is true (trig.Ceva)

Here is my way during the camp, I think it is intriguing that to use $-(E,F;Y,\infty_{EF})$ to replace $(E,F;Y,N)$ and can avoid any bash.

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Let $AX$ and $SX$ meet $BC$ at $R$ and $K$ and let $M$ be midpoint of $BC$. Let $EF$ meet $BC$ at $Q$. Note that in fact we need to prove $MK.KT = BK.KC$ and since $M$ is midpoint of $BC$ we need to prove $(BC;TK) = -1$. Claim $: (BC;RQ) = -1$. Proof $:$ Note that $\frac{BR}{RC} = \frac{\sin{BAR}}{\sin{RAC}} . \frac{AB}{AC} = \frac{PF}{PE} . \frac{BE}{CF}$. Also Note that By Menelaus in $PBC$ and with line $FE$ we have that $\frac{BQ}{CQ} = \frac{BE}{PE}.\frac{PF}{CF}$ so $\frac{BR}{RC} = \frac{BQ}{CQ}$. Let $AQ$ meet $ABC$ at $Y$. Since $(BC;RQ) = -1$ then $YCXB$ is Harmonic so instead of proving $(BC;TK) = -1$ we can prove $S,Y,T$ are collinear or instead $\frac{TC}{TB} = \frac{SC}{SB} . \frac{YC}{YB}$. $\frac{SC}{SB} . \frac{YC}{YB} = \frac{EC}{FB} . \frac{QC}{QB} . \frac{AB}{AC} = \frac{EC}{FB} . \frac{AB}{AC} . \frac{RC}{RB}$ Also Note that $\frac{TC}{TB} = \frac{\frac{TQ}{TB}}{\frac{TQ}{TC}} = \frac{PC}{FP} . \frac{PE}{BP} = \frac{RC}{BR} . \frac{AE}{AF} . \frac{\sin{RAC}}{\sin{BAR}}$ so we need to prove $\frac{EC}{FB} . \frac{AB}{AC} = \frac{AE}{AF} . \frac{\sin{RAC}}{\sin{BAR}}$ or $\frac{RC}{RB} = \frac{AC}{AB} . \frac{\sin{RAC}}{\sin{BAR}}$ which is true. so $S,Y,T$ are collinear as wanted and we're Done.