In $\Delta ABC$,if $A(b, c)$,$B(-a, 0)$,and $C(a, 0)$,and $D$ is the midpoint of $\overline{BC}$,then prove that $AB^2 + AC^2 = 2(AD^2 + BD^2)$.

(N/A) Given coordinates are $A(b, c)$,$B(-a, 0)$,and $C(a, 0)$.
Since $D$ is the midpoint of $\overline{BC}$,the coordinates of $D$ are $(\frac{-a+a}{2}, \frac{0+0}{2}) = (0, 0)$.
Now,calculate the squares of the lengths:
$AB^2 = (b - (-a))^2 + (c - 0)^2 = (b+a)^2 + c^2 = b^2 + 2ab + a^2 + c^2$.
$AC^2 = (b - a)^2 + (c - 0)^2 = b^2 - 2ab + a^2 + c^2$.
Adding them: $AB^2 + AC^2 = (b^2 + 2ab + a^2 + c^2) + (b^2 - 2ab + a^2 + c^2) = 2(a^2 + b^2 + c^2)$.
Now calculate the right side:
$AD^2 = (b - 0)^2 + (c - 0)^2 = b^2 + c^2$.
$BD^2 = (-a - 0)^2 + (0 - 0)^2 = a^2$.
Thus,$2(AD^2 + BD^2) = 2(b^2 + c^2 + a^2)$.
Since both sides are equal to $2(a^2 + b^2 + c^2)$,the identity is proven.