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

Construction:
Draw a parallelogram ABCD. Let diagonals AC and BD meet at P.

To prove : AC²+BD² = AB²+BC²+CD²+DA²
AB = CD [Opposite sides of parallelogram are equal]
BC = DA [Opposite sides of parallelogram are equal]
Since diagonals of a parallelogram bisect each other,
AP = ½ AC
BP = ½ BD
P is the midpoint of diagonals AC and BD.
InABC , BP is the median.
AB²+BC² = 2AP²+2BP² [Apollonius theorem]
AB²+BC² = 2[(1/2)AC]²+2[(1/2)BD]²
AB²+BC² =AC²/2+BD²/2
2(AB²+BC²) = AC²+BD²
2AB²+2BC² = AC²+BD²
AB²+ AB²+ BC²+BC² = AC²+BD²
AB²+ CD²+ BC²+DA² = AC²+BD²
AC²+BD² = AB²+BC²+CD²+DA²
Hence proved.