In $\Delta ABC$,$AB > AC$ and $D$ is the midpoint of $\overline{BC}$. $\overline{AM} \perp \overline{BC}$ and $M \in \overline{BC}$. Prove that $AB^{2} - AC^{2} = 2 \cdot BC \cdot DM$.

(N/A) In $\Delta ABM$,by Pythagoras theorem: $AB^{2} = AM^{2} + BM^{2}$ $(1)$.
In $\Delta ACM$,by Pythagoras theorem: $AC^{2} = AM^{2} + MC^{2}$ $(2)$.
Subtracting $(2)$ from $(1)$: $AB^{2} - AC^{2} = BM^{2} - MC^{2} = (BM - MC)(BM + MC)$.
Since $D$ is the midpoint of $BC$,$BD = DC = \frac{BC}{2}$.
We can express $BM$ as $BD + DM$ and $MC$ as $DC - DM = BD - DM$.
Substituting these: $BM - MC = (BD + DM) - (BD - DM) = 2DM$.
Also,$BM + MC = BC$.
Therefore,$AB^{2} - AC^{2} = (2DM)(BC) = 2 \cdot BC \cdot DM$.