D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that \(AE^2 + BD^2 = AB^2 + DE^2\).

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 \(\triangle\) ACE, we get
\(AC^2 + CE^2 = AE^2\) ………………………………………….(i)

In \(\triangle\) BCD, by Pythagoras theorem, we get
\(BC^2 + CD^2 = BD^2\) ………………………………..(ii)

From equations (i) and (ii), we get,
\(AC^2 + CE^2 + BC^2 + CD^2 = AE^2 + BD^2\) …………..(iii)

In \(\triangle\) CDE, by Pythagoras theorem, we get
\(DE^2 = CD^2 + CE^2\)

In \(\triangle\) ABC, by Pythagoras theorem, we get
\(AB^2 = AC^2 + CB^2\)

Putting the above two values in equation (iii), we get
\(DE^2 + AB^2 = AE^2 + BD^2\).