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


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
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.

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