An acute triangle $ABC$ is given in which $AB <AC$. Point $E$ lies on the $AC$ side of the triangle, with $AB = AE$. The segment $AD$ is the diameter of the circumcircle of the triangle $ABC$, and point $S$ is the center of this arc $BC$ of this circle to which point $A$ does not belong. Point $F$ is symmetric of point $D$ wrt $S$. Prove that lines $F E$ and $AC$ are perpendicular.
The main claim is that $BEDF$ is an isosceles trapezoid, indeed note that $AS$ is the $A-$bisector on $\triangle ABC$ and since $AB=AE$ $\implies$ $BE \perp AS$ but also $FD \perp AS$ because $AD$ is diameter so $BE \| FD$. Now let's see that $\triangle ABF \cong \triangle AED$ for that note that $OS \|AF$ where $O$ is the circumcenter of $\triangle ABC$ this because $O$ is midpoint of $AD$ and $S$ is midpoint of $FD$ and since $S$ is also midpoint of $\hat {BC}$ se get $OS\perp BC$ hence $AF \perp BC$. It's well known that $\angle BAF =\angle EAD$, also notice that $AS$ is the perpendicular bisector of $FD$ so $AD=AF$ and since $AB=AE$ we have $\triangle ABF \cong \triangle AED$ which means $BF=ED$ and the claim is proved. Now we conclude seeing $\angle FEB=\angle DBE$ and $\angle BEA =\angle EBA$ so $\angle FEA= \angle FEB +\angle BEA=\angle DBE+\angle EBA=90^{\circ}$
