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

To do:

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

Solution:

We know that,

The diagonals of a parallelogram bisect each other.

Let $ABCD$ be a parallelogram in which diagonals $AC$ and $BD$ intersect each other at $O$.

This implies,

$BO$ and $DO$ are medians of triangles $ABC$ and $ADC$ respectively.

$\mathrm{AB}^{2}+\mathrm{BC}^{2}=2 \mathrm{BO}^{2}+\frac{1}{2} \mathrm{AC}^{2}$...........(i)

$\mathrm{AD}^{2}+\mathrm{CD}^{2}=2 \mathrm{OD}^{2}+\frac{1}{2} \mathrm{AC}^{2}$...........(ii)

Adding (i) and (ii), we get,

$\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{AD}^{2}=2(\mathrm{BO}^{2}+\mathrm{OD}^{2})+\mathrm{AC}^{2}$

$=2(\frac{1}{4} \mathrm{BD}^{2}+\frac{1}{4} \mathrm{BD}^{2})+\mathrm{AC}^{2}$ (Since $\mathrm{DO}=\mathrm{BO}=\frac{1}{2} \mathrm{BD}$)

$=2 \times \frac{1}{2} \mathrm{BD}^{2}+\mathrm{AC}^{2}$

$\mathrm{AB}^{2}+\mathrm{BC}^{2}+\mathrm{CD}^{2}+\mathrm{AD}^{2}=\mathrm{AC}^{2}+\mathrm{BD}^{2}$

Hence proved.

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