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

(N/A) Given: $AD$ is the median of $\Delta ABC$,so $BD = DC = \frac{BC}{2}$.
Also,$AM \perp BC$.
In right-angled $\Delta ABM$,by Pythagoras theorem:
$AB^{2} = AM^{2} + BM^{2}$
In right-angled $\Delta ADM$,by Pythagoras theorem:
$AD^{2} = AM^{2} + DM^{2} \implies AM^{2} = AD^{2} - DM^{2}$
Substituting $AM^{2}$ in the first equation:
$AB^{2} = (AD^{2} - DM^{2}) + BM^{2}$
Since $BM = BD - DM$,we have:
$AB^{2} = AD^{2} - DM^{2} + (BD - DM)^{2}$
$AB^{2} = AD^{2} - DM^{2} + BD^{2} + DM^{2} - 2 \cdot BD \cdot DM$
$AB^{2} = AD^{2} + BD^{2} - 2 \cdot BD \cdot DM$
Since $BD = \frac{BC}{2}$,substituting this value:
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - 2 \cdot \left(\frac{BC}{2}\right) \cdot DM$
$AB^{2} = AD^{2} + \left(\frac{BC}{2}\right)^{2} - BC \cdot DM$