$S$ is the $A-why$ point. Let $X$ be the reflection of $A$ with respect to the perpendicular bisector of $BC$ and let $S'$ be the intersects of $\omega$ with $XD$. Let also $M$ be the midpoint of $BC$, $W=\omega \cup AM$ and $L=BC \cup WS'$ we want to prove that $S'=S$ Then: $S(L,D/B,C)=(W,X/,B,C)=-1$ so $LM*LD=LB*LC=LS'*LW$ which means that $S',W,C,B$ are one a circle. So $<BDS'=<MWS'=<DKS'$ which means that the circle $DS'K$ is tangent to $BC$ but also circle $DSK$ is tangent to $BC$ so we are done

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Let $A'$ be the $A$-antipode. $E$, $D$, and $A'$ are clearly colinear. Now since $BC$, $KS$, $AE$ concur at $T$ there is an involution swapping $$(B,C),(K,S),(A,E)$$Projecting through $D$ from $\omega$ to $\omega$ we get an involution swapping $$(C,B),(A,X),(K,A')$$so $BC$, $AX$ and $KA'$ concur. It is well-known that $BC\parallel KA'$ so we conclude $AX\parallel BC$.

P2nisic wrote: $S$ is the $A-why$ point. Let $X$ be the reflection of $A$ with respect to the perpendicular bisector of $BC$ and let $S'$ be the intersects of $\omega$ with $XD$. Let also $M$ be the midpoint of $BC$, $W=\omega \cup AM$ and $L=BC \cup WS'$ we want to prove that $S'=S$ Then: $S(L,D/B,C)=(W,X/,B,C)=-1$ so $LM*LD=LB*LC=LS'*LW$ which means that $S',W,C,B$ are one a circle. So $<BDS'=<MWS'=<DKS'$ which means that the circle $DS'K$ is tangent to $BC$ but also circle $DSK$ is tangent to $BC$ so we are done Why-point?

Hasin_Ahmad wrote: Let $\Delta ABC$ be an acute triangle and $\omega$ be its circumcircle. Perpendicular from $A$ to $BC$ intersects $BC$ at $D$ and $\omega$ at $K$. Circle through $A$, $D$ and tangent to $BC$ at $D$ intersect $\omega$ at $E$. $AE$ intersects $BC$ at $T$. $TK$ intersects $\omega$ at $S$. Assume, $SD$ intersects $\omega$ at $X$. Prove that $X$ is the reflection of $A$ with respect to the perpendicular bisector of $BC$. By PoP, $TS \cdot TK = TE \cdot TA = TD^2$ thus $(BC)$ is tangent to $\Delta SKD$ so $\angle XAK=\angle XSK=\angle DSK=90$°