By angle chase or spiral sim, $SME \sim SBF$. Since $ME = AB$, we get $\frac{SE}{SF} = \frac{AB}{BF}$ and similarly $\frac{TE}{TF} = \frac{CE}{AC}$. We will prove $ST$ passes through $EF\cap BC = K$. By ratio lemma, this translates to $\frac{SE}{SF} \cdot \frac{TE}{TF} = \frac{KE}{KF}$, which can be checked to be true. (The first two ratios are known and last you can calculate by $\left(DE/DF\right)^2$

Solved with @GeoGuessr using...MMP. Here it goes... We will prove a more general version of the problem, namely that the statement holds always if $\frac{EM}{AB}=\frac{FN}{AC}$. Claim 1. It's enough to prove $SE, TF$ and $AD$ are concurrent. Proof: Assume we've proved this. Then let $AD\cap(DEF) = D'$ and $EF\cap BC=W$. Then $W$ is the polar of $W$ w.r.t the incircle, so if $DD'\cap EF=K$, $(F,E,K,W)=-1$ and after projecting through $ES\cap FT$ gives us $(S,T,DD'\cap ST, W) = -1$, so $W\in ST$. However, it is well-known that $(B,C,D,W)=-1$, so $BS, CT$ and $AD$ would be concurrent as desired. Now we go to the more interesting part of the problem, namely proving $FE, ES$ and $AD$ are concurrent: Let $ABC$ be a fixed triangle. We move point $M$ on the line $EF$. First we consider a composition of homothety centered at $E$ with coefficient $-\frac{AC}{AB}$ and a translation by the vector $\overrightarrow{EF}$. This projective transformation $f$ sends $M$ to $N$. We now consider a second projective transformation as a composition of the following: - An inversion at $B$ that preserves the incircle + a projection through $F$ onto the incircle sends $M$ to $S$ (since after the invesrion $(BMF)$ is sent to the line $FS$). - An invesrion centered at $W$ that preserves the incircle. This sends $S$ to a point $T'$, so that $S,T',W$ are collinear. - Invesrion centered at $C$ that preserves the incircle + a projection onto $EF$ sends $T'$ to a point $N'$. Let the proejctive maps that is a composition of the above be $g$, sending $M$ to $N'$. We want $N\equiv N'$, so we want $f\equiv g$ and to prove this it's enough to prove this for 3 different positions of $M$: - If $M$ is at an infinite point on $EF$, $S\equiv F$; $T,T'\equiv E$; $N\equiv N'\equiv M$. - IF $M\equiv E$; $S\equiv E$; $T\equiv T'\equiv N\equiv N'\equiv F$. - Finally, if $M$ is so that $S\equiv D$, we have the following trigbash: In that case $(BFD)$ and $(CED)$ both pass through $I$ and by angle-chase $M,I,C$ and $N',I,B$ are collinear and also $BMN'C$ is cyclic, so $\angle IFE=\angle IEF = \frac{\alpha}{2}, \angle IMN'=\frac{\beta}{2}, \angle IN'F = \frac{\gamma}{2}$. So by sine-law on $\triangle IME,\triangle IN'F,\triangle IMN'$ we have \[FN' = N'I\frac{\sin(90+\frac{\beta}{2})}{\sin\frac{\alpha}{2}} = EM\frac{\sin(90+\frac{\beta}{2})}{\sin\frac{\alpha}{2}}\frac{\sin(\frac{\beta}{2})}{\sin\frac{\gamma}{2}}\frac{\sin(\frac{\alpha}{2})}{\sin(90+\frac{\gamma}{2})} = \frac{\cos\frac{\beta}{2}\sin\frac{\beta}{2}}{\cos\frac{\gamma}{2}\sin\frac{\gamma}{2}} = \frac{\sin\beta}{\sin\gamma}EM\]so \[\frac{FN'}{EM} = \frac{\sin\beta}{\sin\gamma}=\frac{AC}{AB}\]so $N\equiv N'$. $\square$

Nice We will prove $\frac{\sin{BAD}}{\sin{CAD}}.\frac{\sin{CBS}}{\sin{ABS}}.\frac{\sin{ACT}}{\sin{BCT}}=1$. Note that $\frac{\sin{CBS}}{\sin{ABS}} = \frac{SD^2}{FS^2}$ and $\frac{\sin{ACT}}{\sin{BCT}} = \frac{ET^2}{TD^2}$ so we need to prove $\frac{BD}{CD}.\frac{AC}{AB}.\frac{SD^2}{FS^2}.\frac{ET^2}{TD^2}=1$. Note that $SME$ and $SBF$ are similar so $\frac{FS}{ES}=\frac{FB}{ME} = \frac{FB}{AB}$. similarly $TNF$ and $TCE$ are similar and $\frac{ET}{FT} = \frac{CE}{AC}$. Now we have $FS = \frac{FB.ES}{AB}$ and $ET = \frac{CE.FT}{AC}$ so $\frac{BD}{CD}.\frac{AC}{AB}.\frac{SD^2}{FS^2}.\frac{ET^2}{TD^2}=\frac{BD}{CD}.\frac{AC}{AB}.\frac{SD^2}{TD^2}.\frac{AB^2}{FB^2.ES^2}.\frac{FT^2.CE^2}{AC^2} = \frac{SD^2}{TD^2}.\frac{AB}{AC}.\frac{CE}{FS}.\frac{FT^2}{ES^2} = \frac{SD^2}{TD^2}.\frac{ET}{FT}.\frac{ES}{FS}.\frac{FT^2}{ES^2} = \frac{SD^2}{TD^2}.\frac{ET}{ES}.\frac{FT}{FS}$ so we need to prove $\frac{SD^2}{TD^2} = \frac{ES}{ET}.\frac{FS}{FT}$. Claim $: FE,ST,BC$ are concurrent. Proof $:$ Let $EF$ meet $BC$ at $P$. Note that $\frac{PF}{PE}.\frac{EC}{AC}.\frac{AB}{FB} = 1$ so $\frac{PF}{PE} = \frac{AC}{EC}.\frac{FB}{AB} = \frac{FT}{ET}.\frac{FS}{ES}$ which implies $P$ lies on $ST$. Now note that $\frac{SD^2}{TD^2} = \frac{PS}{PT} = \frac{SF}{FT}.\frac{SE}{ET}$ as wanted.

My first post Here is my solution using lenght bash and trig: Suppose $AD$, $BS$, $CT$ concur in a point $X$. Then, $AX\cap \omega$=$K$ $BX\cap \omega$=$X$ and $CX\cap \omega$= $T$. We will use the following lemma: Let the incircle of $\triangle ABC$ touch the sides $BC$, $CA$, $AB$ in $D$, $E$, $F$, respectively. Let $M$ be a point. Suppose $AM$, $BM$, $CM$ meet the incircle at $X$, $Y$, respectively $Z$. Then $DX$, $EY$, and $FZ$ concur. According to the lemma above, we are left to prove $FT$, $ES$, $KD$ concur. Since $FKETDS$ is a cyclic hexagon and $FT$, $ES$, $KD$ are its main diagonals, we should have: $$FK \cdot ET \cdot DS = FS \cdot KE \cdot DT (1)$$By Ptolemy theorem in $FKDS$, we get: $FD \cdot KS = FS\cdot KD + DS\cdot FK$ $(2)$ And by Ptolemy in $KETD$ we get: $KT \cdot ED= ET \cdot KD + KE \cdot DT$ $(3)$ By $(2), (3)$, after calculations, we get that $(1)$ is equivalent to: $ET\cdot FT\cdot KS = FS\cdot KT\cdot ED$ $(4)$ Apply Ptolemy theorem in $FSEK$ to get: $KS= \frac{FS\cdot KE+ FK\cdot SE }{FE}$ $(5)$ Also, Ptolemy in $KFTE$ gives: $KT=\frac{KE\cdot FT+ KF\cdot ET}{FE}$ $(6)$ Plugging $(5)$ and $(6)$ in $(4)$, we get that it suffices to prove: $$ ET\cdot FD\cdot FS\cdot KE + ET\cdot FD\cdot FK\cdot SE= FS\cdot ED\cdot KE\cdot FT + FS\cdot ED\cdot KF\cdot ET. $$ Using the sines law in the incircle to compute $FD$ and $DE$, we get: $$\cos(\frac{B}{2}) \cdot(ET\cdot FK \cdot SE+ ET\cdot FS\cdot KE)= \cos(\frac{C}{2})\cdot ( FS\cdot KE\cdot FT + FS\cdot KE\cdot FT)$$$(7)$ In $\triangle FAK$, applying the sines law gives: $FK=\frac{AF\sin(\angle BAD)}{\cos(\frac{B}{2})}$$(8)$ Similarily, in $\triangle AEC$: $EK=\frac{AE\sin(\angle DAC)}{\cos(\frac{C}{2})}$ $(9)$ Using $(8)$ and $(9), (7)$ becomes: $$ET\cdot SE\cdot \sin(\angle BAD) + ET\cdot FS\cdot \sin(\angle DAC)\cdot \frac{\cos(\frac{B}{2})}{\cos(\frac{C}{2})}=FS\cdot FT\cdot \sin(\angle DAC) + FS\cdot ET\cdot \sin(\angle BAD)\cdot \frac{\cos(\frac{C}{2})}{\cos(\frac{B}{2})} $$$(10)$ It's not difficult to prove that $\triangle MES$ and $\triangle BSF$ are similar, so $\frac{FS}{SE}$=$\frac{BF}{ME}$=$\frac{BD}{AB}$ $(11)$ Similarily, we have $\frac{FT}{ET}$=$\frac{AC}{CD}$ $(12)$. Multiplying $(11)$ and $(12)$$\Rightarrow$ $\frac{FS\cdot FT}{SE\cdot ET}$=$\frac{BD\cdot AC}{CD\cdot AB}$=$\frac{\sin(\angle BAD)}{\sin(\angle DAC)}$, last part is easy to prove by sines law. Using this in $(10)$ to get rid of the sides, we get: $$\sin(\angle DAC)\cdot \frac{\cos(\frac{B}{2})}{\cos(\frac{C}{2})}=\sin(\angle BAD)\cdot \frac{\cos(\frac{C}{2})}{\cos(\frac{B}{2})}$$which is equivalent to: $$\frac{BD\cdot AC}{DC\cdot AB}=\frac{\sin(\angle BAD)}{\sin(\angle DAC)}$$which is true, as desired.