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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HeYaAa♡all BrAiNlY ShiNeStArS_______♡My qUeStiØn is :--
===========================
● 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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Verified answer
[tex]\large{\underline{\underline{\mathfrak{\red{\sf{Explanation-}}}}}}[/tex]
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[tex]\orange{\boxed{\pink{\underline{\red{\mathfrak{Given-}}}}}}[/tex]
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[tex]\orange{\boxed{\pink{\underline{\red{\mathfrak{To\:Prove-}}}}}}[/tex]
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[tex]\orange{\boxed{\pink{\underline{\red{\mathfrak{Construction-}}}}}}[/tex]
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[tex]\orange{\boxed{\pink{\underline{\red{\mathfrak{Proof-}}}}}}[/tex]
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In ∆AED, by using Pythagoras theorem,
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[tex]\sf{\red{AD^2=AE^2+ED^2}}[/tex]
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[tex]\implies[/tex] [tex]\sf{AD^2=AE^2+(BE-BD)^2}[/tex]
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By using :
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[tex]\large{\underline{\mathfrak{\sf{\blue{(a-b)^2=a^2+b^2-2ab}}}}}[/tex]
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[tex]\implies[/tex] [tex]\sf{AD^2=AE^2+(BE^2+BD^2-2.BE.BD)}[/tex]
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By using given and construction,
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[tex]\implies[/tex] [tex]\sf{AD^2=AB^2+(\dfrac{BC}{3})^2-\cancel{2}.\dfrac{BC}{\cancel{2}}.\dfrac{BC}{3}}[/tex]
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[tex]\implies[/tex] [tex]\sf{AD^2=AB^2+\dfrac{BC^2}{9}-\dfrac{BC^2}{3}}[/tex]
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By taking LCM,
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[tex]\implies[/tex] [tex]\sf{AD^2=\dfrac{9AB^2+AB^2-3AB^2}{9}}[/tex]
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[tex]\implies[/tex] [tex]\sf{AD^2=\dfrac{7AB^2}{9}}[/tex]
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By cross multiplying,
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[tex]\huge{\underline{\underline{\boxed{\mathfrak{\sf{\purple{9AD^2=7AB^2}}}}}}}[/tex]
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Hence proved!
Step-by-step explanation:
Refer to the attachment........