Question

HeYaAa♡all BrAiNlY ShiNeStArS_______♡My qUeStiØn is :--
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● In An EqÜilaTeRaL tRiAnGle ABC, D iS a pØint oN siĐe BC sŮcH that BD = 1/3 BC. PrØvE ThAt 9AD^2=7AB^2.
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Answer 1

$\large{\underline{\underline{\mathfrak{\red{\sf{Explanation-}}}}}}$

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$\orange{\boxed{\pink{\underline{\red{\mathfrak{Given-}}}}}}$

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• ∆ABC is an equilateral triangle.

• BD = $\dfrac{1}{3}$BC

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$\orange{\boxed{\pink{\underline{\red{\mathfrak{To\:Prove-}}}}}}$

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• 9AD² = 7AB²

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$\orange{\boxed{\pink{\underline{\red{\mathfrak{Construction-}}}}}}$

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• Draw AE ⊥ BC such that BE = EC = $\dfrac{BC}{2}$

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$\orange{\boxed{\pink{\underline{\red{\mathfrak{Proof-}}}}}}$

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In ∆AED, by using Pythagoras theorem,

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$\sf{\red{AD^2=AE^2+ED^2}}$

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$\implies$ $\sf{AD^2=AE^2+(BE-BD)^2}$

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By using :

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$\large{\underline{\mathfrak{\sf{\blue{(a-b)^2=a^2+b^2-2ab}}}}}$

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$\implies$ $\sf{AD^2=AE^2+(BE^2+BD^2-2.BE.BD)}$

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By using given and construction,

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$\implies$ $\sf{AD^2=AB^2+(\dfrac{BC}{3})^2-\cancel{2}.\dfrac{BC}{\cancel{2}}.\dfrac{BC}{3}}$

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$\implies$ $\sf{AD^2=AB^2+\dfrac{BC^2}{9}-\dfrac{BC^2}{3}}$

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By taking LCM,

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$\implies$ $\sf{AD^2=\dfrac{9AB^2+AB^2-3AB^2}{9}}$

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

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By cross multiplying,

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$\huge{\underline{\underline{\boxed{\mathfrak{\sf{\purple{9AD^2=7AB^2}}}}}}}$

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Hence proved!

Answer 2

Step-by-step explanation:

Refer to the attachment........