SELECTION EXAMINATION

Let $X$ be the second intersection point of $\Gamma$ with $I{A}^{\prime }$, where $A^{\prime }$ is the antipodal of $A$ with respect to $\Gamma$. We will prove that the points $X,S,I$ are collinear.

We have $\widehat{IXA}=90^{\circ }$, so $X$ belongs to the circle of diameter $AI$. The same holds for the points $E$ and $F$, thus $X$ is on the circumcircle of $AEIF$. Therefore $\widehat{AFX}=\widehat{AEX}$. However, $\widehat{ACX}=\widehat{ABX}$, and as a result the triangles $BFX,CEX$ are similar. It follows that, $$\frac{XF}{XE}=\frac{BF}{CE}=\frac{BD}{CD}(1).$$ Since the intersection of $EF$ with $BC$ is the isogonal conjugate of $B$, and $\widehat{SDF}=90^{\circ }$, we get that $SD$ is the bisector of $\widehat{BSC}$, thus $$\frac{BS}{CS}=\frac{BD}{CD}(2).$$ Finally, since $\widehat{AFE}=\widehat{AEF}$, and $\widehat{FSB}=\widehat{CSE}$, the triangles $BFS,CES$ are similar, so $$\frac{BS}{CS}=\frac{SF}{SE}(3).$$ From (1), (2), (3) we get $\frac{XF}{XE}=\frac{BD}{CD}=\frac{BS}{CS}=\frac{SF}{SE}$, which gives us that $XS$ is the bisector of $\widehat{EXF}$. Since $I$ is the midpoint of the arc $EF$ at ($AEIF$), we have that $XI$ is the bisector of $\widehat{EXF}$, so $X,S,I$ are collinear

fig. 5 fig. 6

Let $M,N$ the midpoints of $FD,DE$, then $\widehat{DFE}=\widehat{DEC}$, and the right-angled triangles DFS, DCN are similar, so: $$\frac{FS}{DF}=\frac{EN}{DC}(4)$$ Similarly, from the similar triangles DES, BDM we get: $$\frac{SE}{ED}=\frac{MD}{BD}(5)$$ The relations (4) and (5) give $$\frac{FS}{SE}=\frac{BD}{CD}(6)$$ The last one with the help of (1) gives that XS is the bisector of $\widehat{EXF}$. Since I is the midpoint of the arc EF at (AEIF), we have that XI is the bisector of $\widehat{EXF}$, so X, S, I are collinear.