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Geometry Difficulty 6.1 National Olympiad Prove it Taiwan

Let $ABC$ be a triangle with circumcircle $\Gamma$ , and points $E$ and $F$ be chosen from sides $CA$ , $AB$ , respectively. Let the circumcircle of triangle $AEF$ and $\Gamma$ intersect again at point $X$ . Let the circumcircles of triangle $ABE$ and $ACF$ intersect again at point $Y$ . Line $AK$ intersect with $\Gamma$ again at point $M$ other than $A$ , and $N$ be the reflection point of $M$ with respect to line $BC$ . Line $XN$ intersect with $\Gamma$ again at point $S$ other than $X$ .
Prove that $SM$ is parallel to $BC$ .

Solution

Take a point $S'$ on $\odot(ABC)$ such that $S'M$ is parallel to $BC$ ; then $NBS'C$ is a parallelogram, hence $S = S'$ if and only if $XS'$ bisects $\overline{BC}$ . Note that
$\angle KBF = \angle KFC, \quad \angle BFK = \angle ECK \Rightarrow \triangle KBF \sim \triangle KEC,$
$\angle XBF = \angle XCE, \quad \angle BFX = \angle CEX \Rightarrow \triangle XBF \sim \triangle XCE.$
Therefore
$\frac{\overline{BS'}}{S'C} = \frac{\overline{BM}}{\overline{MC}} = \frac{\sin \angle BAK}{\sin \angle KAC} = \frac{\overline{BK}}{\overline{KE}} = \frac{\overline{BF}}{\overline{EC}} = \frac{\overline{BX}}{\overline{XC}}.$
We have
$\frac{[\triangle XBS']}{[\triangle XS'C]} = \frac{\frac{1}{2} \cdot \overline{BX} \cdot \overline{BS'} \cdot \sin \angle XBS'}{\frac{1}{2} \cdot \overline{CX} \cdot \overline{CS'} \cdot \sin \angle XCS'} = 1,$
that is, $XS'$ bisects $\overline{BC}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.