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Question:

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

Solution:

Given:ABCD is a rhombus.Hence,AB=BC=CD=ADAnd, AC perpendicular to BDDO=1/2DBandAO=1/2ACTo Prove:AB^2+BC^2+CD^2+AD^2=AC^2+BD^2Proof:In △AOD, By Pythagoras Theorem,AD^2=AO^2+OD^2 (1)Similarly,DC^2=DO^2+OC^2 (2)BC^2=OB^2+OC^2 (3)AB^2=AO^2+OB^2 (4)Adding 1, 2, 3, 4 we get,AB^2+BC^2+CD^2+AD^2=2AO^2+2BO^2+2CO^2+2DO^2(5)Since,DO=OB=1/2DBandAO=OC=1/2AC(6)From 5 and 6,AB^2+BC^2+CD^2+AD^2=AC^2/2+BD^2/2+AC^2/2+BD^2/2∴AB^2+BC^2+CD^2+AD^2=AC^2+BD^2Hence Proved.