Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Let ABCD be a parallelogram. Let its diagonals AC and BD intersect at O. In △ABC, BO is the median. Diagonals of a parallelogram bisect each other ∴ By Apollonius theorem, AB²+BC²=2OB²+2OA² (1) In △ADC, DO is the median. Since diagonals bisect each other ∴ By Apollonius theorem, AD²+DC²=2OD²+2OC² (2) Adding (1) and (2) we get, AB²+BC²+AD²+DC²=2OB²+2OA²+2OD²+2OC² AB²+BC²+CD²+AD²=2OB²+2OA²+2OB²+2OA² AB²+BC²+CD²+AD²=4OB²+4OA² AB²+BC²+CD²+AD²=4(1/2×DB)²+4(1/2×CA)² [∵OB=DB/2 and OA=CA/2] AB²+BC²+CD²+AD²=4(1/4×DB²)+4(1/4×CA²) AB²+BC²+CD²+AD²=DB²+CA² [hence proved]