Question

In an equilateral ∆ ABC , the side BC is trisected at D. Prove that 9AD² = 7AB²
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Answer 1

Answer:

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Step-by-step explanation:

hence proved

Answer 2

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Let side = a

Draw AE⊥BC So, as triangle ABC is equilateral.

BE= a / 2 and as BD= a / 3

DE=BE−BD= a / 6

In △ADE by Pythagoras theorem

AD² = AE² + DE²

⇒( √3/2a ) + a/36² = 7a²/9 = 7AB²/9

⇒ 9AD² = 7AB²