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

Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,
$$AB^2 + BC^2 + CD^2 + AD^2 = AC^2 + BD^2$$
Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

In $$\triangle$$ AOB,
$$\angle$$ AOB = 90°
$$AB^2 = AO^2 + BO^2$$ …………………….. (i)
[By Pythagoras theorem]

Similarly,
$$AD^2 = AO^2 + DO^2$$ …………………….. (ii)
$$DC^2 = DO^2 + CO^2$$ …………………….. (iii)
$$BC^2 = CO^2 + BO^2$$ …………………….. (iv)

Adding equations (i) + (ii) + (iii) + (iv), we get,
$$AB^2 + AD^2 + DC^2 + BC^2 = 2(AO^2 + BO^2 + DO^2 + CO^2 )$$
= $$4AO^2 + 4BO^2$$
[Since, AO = CO and BO =DO]
= $$(2AO)^2 + (2BO)^2 = AC^2 + BD^2$$
$$AB^2 + AD^2 + DC^2 + BC^2 = AC^2 + BD^2$$
Hence, proved.