In $\Delta ABC$,$m\angle B = 90^{\circ}$,$N \in \overline{AB}$ and $M \in \overline{BC}$. Prove that $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$. $N$ is a point on $AB$ and $M$ is a point on $BC$.
To prove: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Proof:
$1$. In right-angled $\Delta ABM$,by Pythagoras theorem: $AM^{2} = AB^{2} + BM^{2}$.
$2$. In right-angled $\Delta CBN$,by Pythagoras theorem: $CN^{2} = CB^{2} + BN^{2}$.
$3$. Adding these two equations: $AM^{2} + CN^{2} = AB^{2} + BM^{2} + CB^{2} + BN^{2}$.
$4$. Rearranging the terms: $AM^{2} + CN^{2} = (AB^{2} + BC^{2}) + (BM^{2} + BN^{2})$.
$5$. In $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
$6$. In right-angled $\Delta MBN$,by Pythagoras theorem: $MN^{2} = BM^{2} + BN^{2}$.
$7$. Substituting these into the equation from step $4$: $AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the statement is proved.