$\boxed{\textbf{Case 1.}}$ $T$ lies on segment $EF $. $D=BC \cap EF $.Let the reflection of $T$ over $D$ be $T'$ and let $T'A \cap \Omega=Y' , T'A' \cap \Omega =X'$. $AA'$ is perpendicular to $EF$ is trivial by angle-chasing. Let the perpendicular foot be $S$. Also, $\measuredangle AX'A' =\measuredangle AY'A'$ , so $AX',A'Y',T'S$ are concurrent at a point and this point is the orthocenter of $\triangle AA'T'$(Let it be $T"$). Since $\measuredangle TX'T"=\measuredangle TY'T"$,we have $T" =T$ ,so $X=X',Y=Y',T'=W$. Because $BFSA',CESA'$ are cyclic ($\measuredangle FBA'=\measuredangle FSA' =\measuredangle ESA' =\measuredangle ECA'$) ,we have $AF \times AB=AE \times AC=AS \times AA' =AY \times AW=AT \times AX$, so $WYBF, WXCE$ cyclic. Then we have $\measuredangle WBA =\measuredangle WBF =\measuredangle WYF$ and $\measuredangle ACW =\measuredangle ECW =\measuredangle EXW =\measuredangle EYA$. Since $DT \times DW=DE \times DF$ , we have $(E,F;T,W)=-1$. And we have $\measuredangle TYW=\frac {\pi}{2}$ ,so $\measuredangle EYA=\measuredangle WYF$ ,done. $\boxed{\textbf{Case 2.}}$ $T$ doesn't lies on segment $EF $. This is the result of case1. just by changing $T$ and $W$. So $WYFB,WECY$ are cyclic. Then we have $\measuredangle WBA =\measuredangle WBF =\measuredangle WXF$ and $\measuredangle ACW =\measuredangle ECW =\measuredangle EXW $. By $(E,F;T,W)=-1$,we are done.

shanelin-sigma wrote: Triangle $ABC$ satisfying $AB<AC$ has circumcircle $\Omega$. $E, F$ lies on $AC, AB$, respectively, such that $BCEF$ is cyclic. $T$ lies on $EF$ such that $\odot(TEF)$ is tangent to $EF$ at $T$. $A'$ is the antipode of $A$ on $\Omega$. $TA', TA$ intersects $\Omega$ again at $X, Y$, respectively, and $EF$ intersects $\odot(TXY)$ again at $W$. Prove that $\measuredangle WBA=\measuredangle ACW$ Proposed by BlessingOfHeaven Typo:$(TEF)$->$(TBC)$