Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the lengths of the containing sides. In a triangle $ABC$, $I$ is the incentre, $H$ the orthocentre, and $O$ the circumcentre. Prove that $AI$ bisects $\angle OAH$. Hence, or otherwise, prove that if $O$, $H$ and $I$ are collinear the triangle must be isosceles.