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

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Question

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

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Accepted Answer

Step 1: Draw $AE \perp BC$. $\because 	riangle ABC$ is an equilateral triangle. $	herefore BE = \frac{BC}{2}$
Step 2: Now, $AD^2 = AE^2 + DE^2$ and $AB^2 = AE^2 + BE^2$
Step 3: $\Rightarrow AB^2 = AD^2 - DE^2 + BE^2$. $DE = BE - BD = \frac{BC}{2} - \frac{BC}{3} = \frac{BC}{6}$
Step 4: $\Rightarrow AD^2 + \frac{BC^2}{4} = AB^2 + \frac{BC^2}{36}$. Since $AB = BC$: $AD^2 = AB^2 - \frac{AB^2}{4} + \frac{AB^2}{36} = \frac{7AB^2}{9}$
Step 5: $\Rightarrow 9AD^2 = 7AB^2$

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