As shown, points $X$ and $Y$ are on the extension of $BC$ in $\triangle{ABC}$ such that the order of these four points are $X$, $B$, $C$, and $Y$. Meanwhile, they satisfy the relation $BX\cdot AC = CY\cdot AB$. Let $O_1$ and $O_2$ be the circumcenters of $\triangle{ACX}$ and $\triangle{ABY}$, respectively. If $O_1O_2$ intersects $AB$ and $AC$ at $U$ and $V$, respectively, show that $\triangle{AUV}$ is isosceles.
Let $AZ$ bisects $\angle{BAC}$ and meet $BC$ at $Z$. Therefore $$\frac{BZ}{CZ}=\frac{AB}{AC}=\frac{BX}{CY}=\frac{BZ+BX}{CZ+CY}=\frac{XZ}{YZ}$$ or $$BZ\cdot ZY = CZ\cdot ZX$$ This implies that $Z$ lies on the two circle's radical axis. It follows that $AZ\perp BC$ and $AZ\perp UV$. Because $AZ$ also bisects $\angle{UAV}$, it must hold that $AU=AV$.
