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, $AP \cdot AE = AB^2$.

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 $DQ \cdot DE = DF^2$. Therefore,
$$\frac{AB^2}{DF^2} = \frac{AP \cdot AE}{DQ \cdot DE} = \frac{AE^2}{DE^2},$$
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
$$\triangle PYD \sim \triangle PAY.$$
This implies $PA \cdot PD = PY^2$ and $\frac{PA}{PD} = \left(\frac{YA}{YD}\right)^2$.

Let $\omega$ be the circumcircle of $\triangle ABC$ with center $O$, and $\Omega$ be the circumcircle of $\triangle DEF$ 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 $PE^2 = PA \cdot PD$. In fact, let $E'$ be the image of $E$ under the homothety centered at $P$ with ratio $\frac{PA}{PD}$. Then $AE \parallel DE'$, implying $\angle PEA = \angle PE'D = \angle PDE$. Thus, $PE^2 = PA \cdot PD$.

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