Answer: Apply Apollonius theorem to the two triangles formed by one diagonal, using the fact that the diagonals bisect each other.
Let □ABCD be a parallelogram whose diagonals AC and BD intersect at O. The diagonals bisect each other, so AO = OC, BO = OD, and \(AC^2 = 4AO^2\), \(BD^2 = 4BO^2\).
In \(\triangle ABD\), AO is a median to BD. By Apollonius theorem,
\[AB^2 + AD^2 = 2AO^2 + 2BO^2\quad \text{(I)}\]In \(\triangle CBD\), CO is a median to BD:
\[CB^2 + CD^2 = 2CO^2 + 2BO^2\quad \text{(II)}\]Adding (I) and (II) and using CO = AO:
\[AB^2 + BC^2 + CD^2 + DA^2 = 4AO^2 + 4BO^2\] \[= AC^2 + BD^2\]Hence the sum of the squares of the sides equals the sum of the squares of the diagonals.