In Figure, AD is a median of a triangle ABC and AM \(\perp\) BC. Prove that :
\( (i) AC^2 = AD^2 + BC.DM + 2 (\frac{BC}{2} )^2\)
\( (ii) AB^2 = AD^2 – BC.DM + 2 (\frac{BC}{2} )^2\)
\( (iii) AC^2 + AB^2 = 2 AD^2 + {{1} \over {2}} BC^2\)

(i) By applying Pythagoras Theorem in ?AMD, we get,
\(AM^2 + MD^2 = AD^2\) ………………. (i)

Again, by applying Pythagoras Theorem in \(\triangle\) AMC, we get,
\(\Rightarrow AM^2 + MC^2 = AC^2\)
\(\Rightarrow AM^2 + (MD + DC)^ 2 = AC^2\)
\(\Rightarrow (AM^2 + MD^2 ) + DC^2 + 2MD.DC = AC^2\)

From equation(i), we get,
\(AD^2 + DC^2 + 2MD.DC = AC^2\)

Since, DC=\(\frac{BC}{2} \) , thus, we get,
\(\Rightarrow AD^2 + (BC/2)^ 2 + 2MD.(BC/2)^ 2 = AC^2\)
\(\Rightarrow AD^2 + (BC/2)^ 2 + 2MD × BC = AC^2\)
Hence, proved.


(ii) By applying Pythagoras Theorem in -ABM, we get;
\(\Rightarrow AB^2 = AM^2 + MB^2\)
\(\Rightarrow (AD^2 - DM^2 ) + MB^2\)
\(\Rightarrow (AD^2 - DM^2 ) + (BD - MD)^2\)
\(\Rightarrow AD^2 - DM^2 + BD^2 + MD^2 - 2BD × MD\)
\(\Rightarrow AD^2 + BD^2 - 2BD × MD\)
\(\Rightarrow AD^2 + (\frac{BC}{2} )^2 – 2 \frac{BC}{2} MD\)
\(\Rightarrow AD^2 + (\frac{BC}{2} )^2 – BC.MD\)
Hence, proved.


(iii) By applying Pythagoras Theorem in \(\triangle\) ABM, we get,
\(AM^2 + MB^2 = AB^2\) ………………….… (i)

By applying Pythagoras Theorem in \(\triangle\) AMC, we get,
\(\Rightarrow AM^2 + MC^2 = AC^2\) …………………..… (ii)

Adding both the equations (i) and (ii), we get,
\(\Rightarrow 2AM^2 + MB^2 + MC^2 = AB^2 + AC^2\)
\(\Rightarrow 2AM^2 + (BD - DM)^2 + (MD + DC)^2 = AB^2 + AC^2\)
\(\Rightarrow 2AM^2+BD^2 + DM^2 - 2BD.DM + MD^2 + DC^2 + 2MD.DC = AB^2 + AC^2\)
\(\Rightarrow 2AM^2 + 2MD^2 + BD^2 + DC62 + 2MD (- BD + DC) = AB^2 + AC^2\)
\(\Rightarrow 2(AM^2+ MD^2) + (\frac{BC}{2} )^2 + (\frac{BC}{2} )^2+ 2MD(\frac{-BC}{2} + \frac{BC}{2} )^ 2 = AB^2 + AC^2\)
\(\Rightarrow 2AD^2 + (\frac{BC}{2} )^2 = AB^2 + AC^2\)