In an isosceles triangle $ABC$ with $AB = AC$,a circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ at $X$ and $Y$ respectively. Prove that $XY \parallel BC$.

(N/A) $1$. In $\triangle ABC$,we are given $AB = AC$. Therefore,$\angle ABC = \angle ACB$ (angles opposite to equal sides are equal).
$2$. Since $B, C, Y, X$ are concyclic points,$BCYX$ is a cyclic quadrilateral.
$3$. In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$. Thus,$\angle AXY + \angle AC B = 180^{\circ}$ is not directly useful,but rather the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
$4$. Specifically,for cyclic quadrilateral $BCYX$,the exterior angle $\angle AXY = \angle ACB$.
$5$. Since $\angle AXY = \angle ACB$ and we know $\angle ABC = \angle ACB$,it follows that $\angle AXY = \angle ABC$.
$6$. These are corresponding angles formed by the transversal $AB$ intersecting lines $XY$ and $BC$.
$7$. Since the corresponding angles are equal,$XY \parallel BC$.