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