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

Solution :

We know that AD is a median of triangle ABC, then

\({AB}^2\) + \({AC}^2\) = 2\({AD}^2\) + \({1\over 2}{BC}^2\)

Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively.

\(\therefore\)  \({AB}^2\) + \({BC}^2\) = 2\({BO}^2\) + \({1\over 2}{AC}^2\)                 ………..(1)

and      \({AD}^2\) + \({CD}^2\) = 2\({DO}^2\) + \({1\over 2}{AC}^2\)              …………(2)

Adding (1) and (2), we get

\({AB}^2\) + \({BC}^2\) + \({AD}^2\) + \({CD}^2\) = 2(\({BO}^2\) + \({DO}^2\)) + \({AC}^2\)

\(\implies\)  \({AB}^2\) + \({BC}^2\) + \({AD}^2\) + \({CD}^2\) = 2(\({1\over 4}{BD}^2\) + \({1\over 4}{BD}^2\)) + \({AC}^2\)

\(\implies\)   \({AB}^2\) + \({BC}^2\) + \({AD}^2\) + \({CD}^2\) = \({AC}^2\) + \({BD}^2\)

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