D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB²+ DE².

In ΔACE, by pythagoes thoeorem
AC² + CE² AE² …(1)
In ΔBCD, by Pythagoras theorem
BC² + CD² = DB² …(2)
From the equation (1) and (2), we have
AC² + CE² + BC² + CD² = AE² + DB ² …(3)
In ΔBCD, by pythagoras theorem
DE² = CD² + CE² …(4)
In ΔABC, by pythagoras theorem
AB² = AC² + CB² …(5)
From the equation (3), (4) and (5), we have
DE² + AB² = AE² + DB²