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

## Solution :

Let ABC be an equilateral triangle and let D be a point on BC such that BD = $$1\over 3$$ BC.

Draw AE $$\perp$$ BC. Join AD.

In $$\triangle$$ AEB and AEC, we have :

AB = AC            (ABC is equilateral)

$$\angle$$ AEB = $$\angle$$ AEC

and AE = AE

$$\therefore$$  By SAS criteria of congruence, we have :

$$\triangle$$ AEB $$\cong$$ $$\triangle$$ AEC

So,  BE = EC

Now, we have :

BD = $$1\over 3$$BC, DC = $$2\over 3$$BC and BE = EC = $$1\over 2$$BC                ……….(1)

Since, $$\angle$$ C = 60, therefore

$$\triangle$$ ADC is an acute triangle.

$$\therefore$$  $${AD}^2$$ = $${AC}^2$$ + $${DC}^2$$ – $$2DC \times EC$$

= $${AC}^2$$ + $$({2\over 3}BC)^2$$ – $$2\times {2\over 3}BC\times {1\over 2}BC$$

= $${AC}^2$$ + $$({4\over 9}BC)^2$$ – $$({2\over 3}BC)^2$$

= $${AB}^2$$ + $$({4\over 9}AB)^2$$ – $$({2\over 3}AB)^2$$             (AB = AB = AC)

= $$(9 – 4 – 6){AB}^2\over 9$$ = $${7\over 9}{AB}^2$$

So,  $$9{AD}^2$$ = $$7{AB}^2$$