All three vertices of $\triangle ABC$ lie on the parabola $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The distance from $A$ to the line $BC$ is equal to the length of $BC$. What is the area of the triangle?

1. By symmetry $B=(-q,q^2)$ and $C=(q,q^2)$ for some $q>0$, so $BC=2q$, and the distance from $A=(0,0)$ to the line $y=q^2$ is $q^2$.
2. The condition says $q^2=2q$, so $q=2$: $BC=4$ and the height is $4$.
3. The area is $\frac12\cdot4\cdot4=\boxed{8}$.