作 $AB$ 的中垂線 $OM$。故 $\triangle XES \sim \triangle OMS$，於是 $SX = XE$。

Draw the perpendicular bisector $OM$ of $AB$. Thus $\triangle XES \sim \triangle OMS$, so $SX = XE$.

畫以 $XE$ 為半徑的圓 $X$。圓 $X$ 與弦 $AB$ 及直線 $CS$ 均相切。又作 $\triangle PQR$ 的外接圓，以及直線 $MA$ 與 $MC$，如圖所示。

Draw circle $X$ with radius $XE$. Circle $X$ is tangent to both the chord $AB$ and the line $CS$. Also draw the circumcircle of $\triangle PQR$, together with the lines $MA$ and $MC$, as shown in the figure.

因為 $\triangle AMR \sim \triangle PMA$，所以有
$$MR \cdot MP = MA^2 = ME \cdot MS.$$
又由圓幂定理知 $CQ \cdot CP = CS^2$。故 $M$, $C$ 兩點皆位於圓 $Z$ 與圓 $X$ 的根軸上，得 $ZX \perp MC$。同理可知 $ZY \perp MC$。所以 $X$, $Y$, $Z$ 三點共線，得證。

Since $\triangle AMR \sim \triangle PMA$, we have
$$MR \cdot MP = MA^2 = ME \cdot MS.$$
Also, by the power of a point theorem, $CQ \cdot CP = CS^2$. Hence both $M$ and $C$ lie on the radical axis of circle $Z$ and circle $X$, giving $ZX \perp MC$. By the same reasoning, $ZY \perp MC$. Therefore $X$, $Y$, $Z$ are collinear, as required.