In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = AD^{2} + 3BD^{2}$.

(A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ and $\overline{AD}$ is a median to side $\overline{BC}$.
Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$. Therefore,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.
Substituting $BC = 2BD$,we get $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$. (Equation $1$)
In right-angled $\Delta ABD$,by Pythagoras theorem: $AD^{2} = AB^{2} + BD^{2}$,which implies $AB^{2} = AD^{2} - BD^{2}$. (Equation $2$)
Substitute Equation $2$ into Equation $1$:
$AC^{2} = (AD^{2} - BD^{2}) + 4BD^{2}$
$AC^{2} = AD^{2} + 3BD^{2}$.
Hence,the result is proved.