In an equilateral triangle ABC, D is a point on side BC such that BD = \dfrac{1}{3BC}. Prove that

For Competitive Exams

Question 15 In an equilateral triangle ABC, D is a point on side BC such that BD = $\dfrac{1}{3BC}$. Prove that $9 AD^2 = 7 AB^2$.

Solution:

Given, ABC is an equilateral triangle

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

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC

BE = EC = $\dfrac{BC}{2} = \dfrac{a}{2}$

And, AE = $a\dfrac{\sqrt{3}}{2}$

Given, BD = $\dfrac{1}{3BC}$

BD =$\dfrac{a}{3}$

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

In ΔADE, by Pythagoras theorem

$AD^2 = AE^2 + DE^2$

$AD^2 = (\dfrac{a\sqrt{3}}{2})^2 + (\dfrac{a}{6})^2$

$AD^2 = \dfrac{3a^2}{4} + \dfrac{a^2}{36}$

$AD^2 = \dfrac{28a^2}{36}$

$AD^2 = \dfrac{7}{9} AB^2$

$9 AD^2 = 7 AB^2$

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