In $\Delta ABC$,$\overline{AC}$ is the hypotenuse and $\overline{BE}$ is a median. Prove that $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.

(N/A) Given: In $\Delta ABC$,$\angle B = 90^{\circ}$ (since $\overline{AC}$ is the hypotenuse),and $\overline{BE}$ is a median to the hypotenuse $\overline{AC}$.
Since $\overline{BE}$ is a median,$E$ is the midpoint of $\overline{AC}$. Thus,$AE = EC = \frac{1}{2} AC$,which implies $AC = 2AE$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AB^{2} + BC^{2} = AC^{2}$.
We need to prove: $AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$.
Substitute $AB^{2} + BC^{2} = AC^{2}$ into the expression:
$AC^{2} + AC^{2} = 2AC^{2}$.
Since $AC = 2AE$,substitute this into the expression:
$2(2AE)^{2} = 2(4AE^{2}) = 8AE^{2}$.
Thus,$AB^{2} + BC^{2} + AC^{2} = 8AE^{2}$ is proved.