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

(N/A) Applying the Pythagoras theorem in $\triangle ACE$,we obtain:
$AC^{2} + CE^{2} = AE^{2}$ $...(1)$
Applying the Pythagoras theorem in $\triangle BCD$,we obtain:
$BC^{2} + CD^{2} = BD^{2}$ $...(2)$
Adding equation $(1)$ and equation $(2)$,we obtain:
$AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2}$ $...(3)$
Applying the Pythagoras theorem in $\triangle CDE$,we obtain:
$DE^{2} = CD^{2} + CE^{2}$ $...(4)$
Applying the Pythagoras theorem in $\triangle ABC$,we obtain:
$AB^{2} = AC^{2} + BC^{2}$ $...(5)$
Substituting the values from equation $(4)$ and equation $(5)$ into equation $(3)$,we obtain:
$AB^{2} + DE^{2} = AE^{2} + BD^{2}$
Hence,$AE^{2} + BD^{2} = AB^{2} + DE^{2}$.