In the figure,$AD$ is a median of a triangle $ABC$ and $AM \perp BC$. Prove that $AC^2 + AB^2 = 2AD^2 + \frac{1}{2} BC^2$.

(N/A) Applying Pythagoras theorem in $\Delta ABM$,we obtain:
$AM^2 + MB^2 = AB^2 \quad \dots(1)$
Applying Pythagoras theorem in $\Delta AMC$,we obtain:
$AM^2 + MC^2 = AC^2 \quad \dots(2)$
Adding equations $(1)$ and $(2)$,we obtain:
$2AM^2 + MB^2 + MC^2 = AB^2 + AC^2$
Since $AD$ is a median,$BD = DC = \frac{BC}{2}$.
We can write $MB = BD - DM$ and $MC = DC + DM = BD + DM$.
Substituting these in the equation:
$2AM^2 + (BD - DM)^2 + (BD + DM)^2 = AB^2 + AC^2$
$2AM^2 + (BD^2 + DM^2 - 2BD \cdot DM) + (BD^2 + DM^2 + 2BD \cdot DM) = AB^2 + AC^2$
$2AM^2 + 2BD^2 + 2DM^2 = AB^2 + AC^2$
$2(AM^2 + DM^2) + 2BD^2 = AB^2 + AC^2$
In $\Delta ADM$,by Pythagoras theorem,$AM^2 + DM^2 = AD^2$.
So,$2AD^2 + 2(\frac{BC}{2})^2 = AB^2 + AC^2$
$2AD^2 + 2(\frac{BC^2}{4}) = AB^2 + AC^2$
$2AD^2 + \frac{BC^2}{2} = AB^2 + AC^2$