# D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $${AE}^2$$ + $${BC}^2$$ = $${AB}^2$$ + $${DE}^2$$.

## 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,  $${AE}^2$$ + $${BC}^2$$ = $${AB}^2$$ + $${DE}^2$$