Let $\vartriangle ABC$ be isosceles with $AB = AC$. Let $\omega$ be its circumscribed circle and $O$ its circumcenter. Let $D$ be the second intersection of $CO$ with $\omega$. Take a point $E$ in $AB$ such that $DE \parallel AC$ and suppose that $AE: BE = 2: 1$. Show that $\vartriangle ABC$ is equilateral.