Question

prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides

Answer 1

using a diagonal per .
1) the diagonal of a square are a congruent.
2) the all sides of a square are a congruent
3)the diagonals of a parallelogram are bisect each other.

Answer 2

Given : ABCD is a parallelogram in which AC and BD are diagonals which intersect each other at O.

To Prove : AB² + BC² + CD² + DA² = AC² + BD² .

Concept used :-

• Diagonals of a parallelogram bisect each other.
• Apollonius theorem says that, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side. { Refer to second diagram of image. }

Solution :-

Since, diagonals of parallelogram bisect each other.

Therefore, BO and DO are medians of ∆ABC and ∆ADC, respectively.

Now, in ∆ABC we have BO is a median . Using apollonius theorem we get,

→ AB² + BC² = 2BO² + (1/2)AC² --------- Eqn.(1)

similarly, in ∆ADC we have DO is a median . Using apollonius theorem we get,

→ AD² + DC² = 2DO² + (1/2)AC² ----------- Eqn.(2)

Adding Eqn.(1) & Eqn.(2) , we get,

→ AB² + BC² + AD² + DC² = 2BO² + 2DO² + (1/2)AC² + (1/2)AC²

→ AB² + BC² + AD² + DC² = 2(BO² + DO²) + (1/2)(AC² + AC²)

→ AB² + BC² + AD² + DC² = 2(BO² + DO²) + (1/2)(2AC²)

→ AB² + BC² + AD² + DC² = 2(BO² + DO²) + AC²

now, we know that,

• BO = DO = (1/2)BD

Putting value of BO and DO and terms of BD in RHS, we get,

→ AB² + BC² + AD² + DC² = 2[{(1/2)BD}² + {(1/2)BD}²] + AC²

→ AB² + BC² + AD² + DC² = 2[(1/4)BD² + (1/4)BD²] + AC²

→ AB² + BC² + AD² + DC² = 2 * (1/4)[BD² + BD²] + AC²

→ AB² + BC² + AD² + DC² = (1/2)(2BD²) + AC²

→ AB² + BC² + AD² + DC² = BD² + AC² . (Proved.)

Hence, we can conclude that, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

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