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# In an equilateral triangle ABC, D is a point on side BC such that BD = $$\frac{1}{3}$$ BC. Prove that $$9AD^2 = 7AB^2$$.

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = $$\frac{1}{3}$$ ×BC

Let the side of the equilateral triangle be a, and AE be the altitude of $$\triangle$$ ABC.

BE = EC = $$\frac{BC}{2}$$ = $$\frac{a}{2}$$
And, $$AE = {{a \sqrt{3}} \over {2}}$$
Given, BD = $$\frac{1}{3}$$ ×BC
$$\Rightarrow$$ BD = $$\frac{a}{3}$$

$$\therefore$$ DE = BE – BD
= $$\frac{a}{2} – \frac{a}{3} = \frac{a}{6}$$

In $$\triangle$$ ADE, by Pythagoras theorem, $$AD^2 = AE^2 + DE^2$$ $$AD^2 = ({{a\ sqrt{3}} \over {2}})^2 + ({{a} \over {6}})^2$$
$$= ({{3a^2} \over {4}}) + ({{a^2} \over {36}})$$
$$={{28a^2} \over {36}}$$
$$={{7} \over {9}} AB^2$$
$$9 AD^2 = 7 AB^2$$