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# 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
$$DE^2 + AB^2 = AE^2 + BD^2$$.