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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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