Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB² = AD² + BD² …………….(i)

AC² = AD² + DC² ………………..(ii)

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

AB² – AC² = BD² – DC²

= 9CD² – CD² [Since, BD = 3CD]

= 8CD²

= 8(BC/4)² [Since, BC = DB + CD = 3CD + CD = 4CD]

AB² – AC² = BC²/2

⇒ 2(AB² – AC²) = BC²

⇒ 2AB² – 2AC² = BC²

∴ 2AB² = 2AC² + BC².