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Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of it's diagonal?
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Prove that the sum of the squares of the sides of a rhombus is equal t...
Given: ABCD is a Rhombus in which AC and BD are diagonals which intersect each other at O.
To Prove : AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof : We know that diagonals of rhombus bisect each other at right angle.
Therefore,
∠AOB=∠BOC = ∠COD = ∠DOA = 90DEG
Now, in right angle ∆AOB
AB2 = OA2 + OB2    ...(i)
[Using Pythagoras theorem]
In, ∆BOC right triangle we have
BC2 = OB2 + OC2    ...(ii)
[Using Pythagoras theorem]
In, ∆COD right triangle we have
CD2 = OC2 + OD2    ...(iii)
[Using Pythagoras theorem]
In, ∆AOD right triangle we have
AD2 = OD2 + OA2    ...(iv)
[Using Pythagoras theorem]
Adding (i), (ii), (iii) and (iv), we get

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Prove that the sum of the squares of the sides of a rhombus is equal t...
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Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of it's diagonal?
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