Solution:
Let $M$ be the midpoint of $AB$ and $X$ the midpoint of $MB$. Construct the circle center $B$, radius $BC/2$ and the circle diameter $AX$. If they do not intersect (so $BC < AB/2$ or $BC > AB$) then the construction is not possible. If they intersect at $N$, then take $C$ so that $N$ is the midpoint of $BC$. Let $CM$ meet $AN$ at $O$. Then $AO/AN = AM/AX = 2/3$, so the triangles $AOM$ and $ANX$ are similar. Hence $\angle AOM = \angle ANX = 90^{\circ}$.