In a triangle $$ABC$$, if $$AD$$ is the median through $$A$$ intersecting the side $$BC$$ in $$D$$ then show that $$AB^2 + AC^2 = 2(AD^2 + BD^2)$$.

PLACEHOLDER ANSWER - this is a proof question. Let $$E$$ be the foot of the perpendicular from $$A$$ to $$BC$$ and $$DE = x$$, with $$BD = DC = m$$. Then $$AB^2 = AE^2 + (m + x)^2$$ and $$AC^2 = AE^2 + (m - x)^2$$ (or vice versa), so $$AB^2 + AC^2 = 2AE^2 + 2x^2 + 2m^2$$. Since $$AE^2 + x^2 = AD^2$$, this equals $$2(AD^2 + BD^2)$$ (Apollonius' theorem).