In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{AD}$ is a median. Prove that $AC^{2} = 4AD^{2} - 3AB^{2}$.

(A) $1$. In $\Delta ABC$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AC^{2} = AB^{2} + BC^{2}$ (Equation $1$).
$2$. Since $\overline{AD}$ is a median to side $\overline{BC}$,$D$ is the midpoint of $\overline{BC}$. Thus,$BD = DC = \frac{1}{2} BC$,which implies $BC = 2BD$.
$3$. In $\Delta ABD$,since $\angle B = 90^{\circ}$,by Pythagoras theorem,$AD^{2} = AB^{2} + BD^{2}$.
$4$. Rearranging for $BD^{2}$,we get $BD^{2} = AD^{2} - AB^{2}$.
$5$. Substitute $BC = 2BD$ into Equation $1$: $AC^{2} = AB^{2} + (2BD)^{2} = AB^{2} + 4BD^{2}$.
$6$. Substitute $BD^{2} = AD^{2} - AB^{2}$ into the expression: $AC^{2} = AB^{2} + 4(AD^{2} - AB^{2})$.
$7$. Simplify: $AC^{2} = AB^{2} + 4AD^{2} - 4AB^{2} = 4AD^{2} - 3AB^{2}$.
$8$. Hence,$AC^{2} = 4AD^{2} - 3AB^{2}$ is proved.