The 65th IMO China National Team Selection Test

Proof 1. Let $AE$ and $DE$ intersect line $BC$ at points $P$ and $Q$, respectively. Since $AB=AC$ and $A$, $B$, $E$, $C$ are concyclic, we have $\angle ABP=\angle ACB=\angle AEB$. Hence, $AP\cdot AE=A{B}^2$.

Note that $Q$ might be on the extension of $BC$, but $F$ can only be on the extension of $BQ$. Otherwise, if $F$ is on the ray $QB$, combining with $F$ being on the extension of $BC$ would lead to $\angle DFE>\angle DCE>\angle FCE>90^{\circ }$, while clearly $\angle ADE<90^{\circ }$, which contradicts $\angle ADE=\angle DFE$.

Since $AD\parallel BC$, we have $\angle DQF=\angle ADE=\angle DFE$, thus $DQ\cdot DE=D{F}^2$. Therefore, $$\frac{A{B}^2}{D{F}^2}=\frac{AP\cdot AE}{DQ\cdot DE}=\frac{A{E}^2}{D{E}^2},$$ implying $\frac{AB}{DF}=\frac{AE}{DE}$. Hence $\frac{AX}{DX}=\frac{AB}{DF}=\frac{AE}{DE}$. Similarly, $\frac{AY}{DY}=\frac{AC}{DF}=\frac{AE}{DE}$.

Take a point $T$ on segment $AD$ such that $\frac{AT}{TD}=\frac{AE}{DE}$. Then $X$, $Y$, $E$, and $T$ are concyclic (Apollonian circle). Notice that $XT$ and $YT$ bisect $\angle AXD$ and $\angle ATD$, respectively. Therefore, $$\angle XEY=\angle XTY=\angle TXD-\angle TYD=\frac{1}{2}\angle AXD-\frac{1}{2}\angle AYD=\frac{1}{2}\angle XAY=\frac{1}{2}\angle BAC$$ is a fixed value. $\square$



Proof 2. Since $AB=AC$ and $AD\parallel BC$, $AD$ bisects the external angle $\angle XAY$. Let $P$ be the intersection of the perpendicular bisector of $XY$ and $AD$. It is known that $A$, $X$, $Y$, and $P$ are concyclic. Hence, $\angle PYD=\angle PXY=\angle PAY$, thus $$△PYD\sim △PAY.$$ This implies $PA\cdot PD=P{Y}^2$ and $\frac{PA}{PD}={\left(\frac{YA}{YD}\right)}^2$.

Let $\omega$ be the circumcircle of $△ABC$ with center $O$, and $\Omega$ be the circumcircle of $△DEF$ with center $Q$. Clearly, $PA$ is tangent to $\omega$ at point $A$, and since $\angle DFE=\angle ADE$, $PA$ is also tangent to $\Omega$ at point $D$. Since $AD\parallel CF$, we have $$\frac{AO}{DQ}=\frac{\frac{AC}{\sin \angle ABC}}{\frac{DF}{\sin \angle DEF}}=\frac{AC}{DF}\cdot \frac{\sin \angle YDA}{\sin \angle DAY}={\left(\frac{YA}{YD}\right)}^2=\frac{PA}{PD}.$$ Thus, $P$ is the external homothety center of circles $\omega$ and $\Omega$.

Since $E$ is the intersection of $\omega$ and $\Omega$, it is known that $P{E}^2=PA\cdot PD$. In fact, let $E^{\prime }$ be the image of $E$ under the homothety centered at $P$ with ratio $\frac{PA}{PD}$. Then $AE\parallel D{E}^{\prime }$, implying $\angle PEA=\angle P{E}^{\prime }D=\angle PDE$. Thus, $P{E}^2=PA\cdot PD$.

Therefore, $PE=\sqrt{PA\cdot PD}=PY=PX$. Hence, $P$ is the circumcenter of $△EXY$. Therefore, $\angle XEY=\frac{1}{2}\angle XPY=\frac{1}{2}\angle XAY=\frac{1}{2}\angle BAC$ is a fixed value. $\square$