Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC² + CE² = AE² ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC² + CD² = BD² ………………………………..(ii)

From equations (i) and (ii), we get,

AC² + CE² + BC² + CD² = AE² + BD² …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE² = CD² + CE²

In ΔABC, by Pythagoras theorem, we get

AB² = AC² + CB²

Putting the above two values in equation (iii), we get

DE² + AB² = AE² + BD².