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 $BC$.
Step $1$: Apply the Pythagorean theorem in $\Delta ABM$ and $\Delta CBN$.
In $\Delta ABM$,since $\angle B = 90^{\circ}$,we have $AM^{2} = AB^{2} + BM^{2}$.
In $\Delta CBN$,since $\angle B = 90^{\circ}$,we have $CN^{2} = CB^{2} + BN^{2}$.
Step $2$: Add the two equations.
$AM^{2} + CN^{2} = (AB^{2} + BM^{2}) + (CB^{2} + BN^{2})$.
Step $3$: Rearrange the terms.
$AM^{2} + CN^{2} = (AB^{2} + CB^{2}) + (BM^{2} + BN^{2})$.
Step $4$: Use the Pythagorean theorem for $\Delta ABC$ and $\Delta MBN$.
In $\Delta ABC$,$AC^{2} = AB^{2} + CB^{2}$.
In $\Delta MBN$,$MN^{2} = BM^{2} + BN^{2}$.
Step $5$: Substitute these into the equation from Step $3$.
$AM^{2} + CN^{2} = AC^{2} + MN^{2}$.
Hence,the result is proved.