Solution :
From triangle ACE,
\({AE}^2\) = \({EC}^2\) + \({AC}^2\) ……….(1) (By Pythagoras Theorem)
From triangle DCB,
\({BD}^2\) = \({BC}^2\) + \({DC}^2\) ………(2)
Adding (1) and (2), we get
\({AE}^2\) + \({BD}^2\) = \({EC}^2\) + \({AC}^2\) + \({BC}^2\) + \({DC}^2\)
By Pythagoras Theorem in right triangle ECD and ABC, \({DE}^2\) = \({EC}^2\) + \({DC}^2\) and \({AB}^2\) = \({BC}^2\) + \({AC}^2\)
Hence, \({AE}^2\) + \({BC}^2\) = \({AB}^2\) + \({DE}^2\)