In the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that $AC^{2} = AD^{2} + BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$.

(N/A) Applying Pythagoras theorem in $\Delta AMD$,we obtain:
$AM^{2} + MD^{2} = AD^{2} \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^{2} + MC^{2} = AC^{2}$
Since $MC = MD + DC$,we have:
$AM^{2} + (MD + DC)^{2} = AC^{2}$
$AM^{2} + MD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Substituting $AM^{2} + MD^{2} = AD^{2}$ from equation $(1)$:
$AD^{2} + DC^{2} + 2MD \cdot DC = AC^{2}$
Since $AD$ is a median,$DC = \frac{BC}{2}$. Substituting this value:
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + 2MD \cdot \left(\frac{BC}{2}\right) = AC^{2}$
$AD^{2} + \left(\frac{BC}{2}\right)^{2} + BC \cdot DM = AC^{2}$
Thus,$AC^{2} = AD^{2} + BC \cdot DM + \left(\frac{BC}{2}\right)^{2}$.