Let $T$ be the point on $DE$ such that $AT\parallel BC$. So $\frac{DT}{DE}=\frac{AC}{CE}$, $DT=b\cdot\frac{DE}{CE}=2b\sin\frac{C}2$. We also have $DF=2BD\sin\frac{B}2=(a-b+c)\sin\frac{B}2$. Thus \begin{align*}\frac{DF}{DT}&=\frac{(a-b+c)\sin\frac{B}2}{2b\sin\frac{C}2}\\&=\frac{(\sin A-\sin B+\sin C)\sin\frac{B}2}{2\sin B\sin\frac{C}2}\\&=\frac{\sin(B+C)-2\cos\frac{B+C}2\sin\frac{B-C}2}{4\cos\frac{B}2\sin\frac{C}2}\\&=\frac{2\sin\frac{B+C}2\cos\frac{B+C}2-2\cos\frac{B+C}2\sin\frac{B-C}2}{2\sin\frac{B+C}2-2\sin\frac{B-C}2}\\&=\cos\frac{B+C}2\\&=\cos\angle FDT.\end{align*} Hence $\angle DFT=90^{\circ}$, which gives $T,K,F$ are collinear and $T=S$. So $AS\parallel BC$.

Dear Mathlinkers, the result can be obtains by using three times the Pascal's in a degenerated form... Sincerely Jean-Louis

It's known that $DK$, $EF$ and $A$ median intersect in one point. Using this we can easily prove that $AS$ is polar of this point with respect to inscribed circle, which means that $AS\perp DK$ which we wanted to prove.

Dear J-L, What about not using Pascal or Stanley Rabinowitz? A new challenge ... M.T.

Martin N. wrote: The incircle of the triangle $ABC$ touches the sides $BC$, $CA$, and $AB$ in the points $D$, $E$ and $F$, respectively. Let $K$ be the point symmetric to $D$ with respect to the incenter. The lines $DE$ and $FK$ intersect at $S$. Prove that $AS$ is parallel to $BC$. (4th Middle European Mathematical Olympiad, Team Competition, Problem 5)

Dear Armpist (M.) and Mathlinkers, your idea seems to be very nice, but I don't see the way you have in mind. Sincerely Jean-Louis

armpist wrote: Dear J-L, What about not using Pascal or Stanley Rabinowitz? A new challenge ... M.T.

Let $R \equiv DF \cap KE$. Apply Pascal theorem to degenerate hexagon $KFFDEE$ to obtain that $A,S$ and $R$ are collinear. Since $KD$ is a diameter of the incircle we have $\angle KFD = \angle KED = 90$ and therefore $K$ is the orthocenter of triangle $DRS$ which implies $DK \perp AS => AS \parallel BC$

by angle chasing can be seen easily that FAE=2(FSE) and A lies on perpendicular bisector of FE imply A is circumcenter of FES so AS=AE , angle AES=DEC and DEC is isosceles imply AES similar to CED so the results.

Solution: Join $AI$ and $IF$. Clearly, $K$ lies on the incircle. We have, $\angle AFI$ = $90$ and $\angle AIF$ = $\angle FDS$. So, $\Delta AFI$ ~ $\Delta SFD$. Thus, $\angle FAI$ = $\angle FSD$, or, $\angle FSD$ = $\frac{\angle A}{2}$. Also, $AF$ = $AE$. So, $A$ is the circumcenter of $\Delta FES$. This implies that $AE$ = $AS$. Therefore, $\angle ASE$ = $\angle AES$ = $\angle CED$ = $\angle CDE$. Hence, $AS$ is parallel to $BC$.

How has noone beat me to this?

Pascal theorem: $DEEKFF$

Let $T$ be the point of intersection of the lines $EK$ and $DF$. By Pascal on $DEEKFF$, we have that $T,A,S$ are colinear points. Since we know that $\angle KFD = \angle KED = 90$, we have that $I \in KD$. Let $G$ be the intersection points of $EF$ and $KD$. By Brokard's we have that $ IG \perp TS$, and since $IG \perp BC$, we have that $AS \parallel BC$.

denote: $KE \cap DF=T$, second intersection of AD with incircle $ABC$ as $Q$ let's prove that the points $T,A,S$ lie on the same line. Let it not be so. By the Brocard's theorem to the inscribed quadrilateral $DEKF$, we know that $DK \bot TS$, and by virtue of the fact that $DK$ is a diameter of the circumcircle of the triangle DEF, $DK \bot BC$, so $TS \parallel BC$. and we know that $SF \bot TD$, $TE \bot SD$, so $K$ is the orthocenter of the triangle $TDS$ denote $TS \cap BC=P_{\infty}$, $TS \cap AD=R$ note that the $DEQF$ is a harmonic quadrilateral. By taking perspectivity in point $D$, $(F,E;Q,D)=(T,S;R,P_{\infty})=-1$, so $TR=SR$, so $TR=SR=FR=ER$ note that $\angle STE=\angle SFE$, because of $AC$ is a tangent to circumcircle of $DEF$, $\angle SFE=\angle TEA$, and because of $RT=RE$, $\angle RTE=\angle RET$. so $\angle REA=\angle RET- \angle TEA=0$, contradiction, so $T,A,S$ lie on the same line. And so because of $TS \parallel BC$, $AS \parallel BC$