In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $D$ is a point on $\overline{BC}$. Prove that $AD^{2} + BC^{2} = AC^{2} + BD^{2}$.

(N/A) In $\Delta ABC$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AC^{2} = AB^{2} + BC^{2}$. (Equation $1$)
In $\Delta ABD$,since $m \angle B = 90^{\circ}$,by the Pythagorean theorem,we have $AD^{2} = AB^{2} + BD^{2}$. (Equation $2$)
From Equation $2$,we can write $AB^{2} = AD^{2} - BD^{2}$.
Substitute this value of $AB^{2}$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + BC^{2}$.
Rearranging the terms,we get $AC^{2} + BD^{2} = AD^{2} + BC^{2}$.
Hence,the statement is proved.