The perpendicular from A on side BC of a \(\triangle\) ABC intersects BC at D such that DB = 3CD (see Figure). Prove that \(2AB^2 = 2AC^2 + BC^2\).

Given, the perpendicular from A on side BC of a \(\triangle\) ABC intersects BC at D such that;

DB = 3CD.

In \(\triangle\) ABC,
AD \(\perp\) BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,
\(AB^2 = AD^2 + BD^2\) ……………………….(i)
\(AC^2 = AD^2 + DC^2\) ……………………………..(ii)

Subtracting equation (ii) from equation (i), we get

\(AB^2 – AC^2 = BD^2 – DC^2\)
\(= 9CD^2 – CD^2\) [\(\because \) BD = 3CD]
\(= 9CD^2 = 8(BC/4)^2 \)
[\(\because \) BC = DB + CD = 3CD + CD = 4CD]

\(\therefore AB^2 – AC^2 = {{BC^2} \over {2}}\)
\(\Rightarrow 2(AB^2 – AC^2) = BC^2\)
\( \Rightarrow 2AB^2 – 2AC^2 = BC^2\)
\(\Rightarrow 2AB^2 = 2AC^2 + BC^2\).