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# 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

$$AB^2 = AD^2 + BD^2$$ ……………………….(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]

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