Question

In an equilateral triangle ABC,D is a point on BC such that CD:BC

= 2/3 . Prove that 9AD2 = 7 AB2​

Answer 1

Given : $\bf{BD = \frac{1}{3} BC}$

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To Prove : $\bf{9AD^2 =7AB^2}$

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Construction : Draw an Altitude AE such that E lies on BC and AE perpendicular to BC

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Proof : $\bf{\triangle ABC}$ is an Equilateral triangle such that$\bf{ \: AB=BC=AC}$

‎

$\Large{➥}$ In $\bf{ \triangle AEB}$

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By Pythagoras Theorem,

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$\bf{ ↦AB^2 =AE^2+EB^2.....(1)}$

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$\Large{➥}$ In $\bf{\triangle AED}$

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By Pythagoras Theorem,

‎

$\bf{ ↦ AD^2 =AE^2 +ED^2.....(2)}$

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From (1) and (2)

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$\Large{\leadsto}$ $\bf{ AD^2=ED^2 +AB^2 −EB^2 }$

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Since,

‎

$\bf{\implies EB= \frac{1}{2} BC= \frac{AB}{2} = \frac{3AB}{6}}$ and

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⠀⠀⠀⠀⠀⠀⠀⠀$\bf{\implies ED= \frac{BC}{6} = \frac{AB}{6}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\bf{\implies AD^2= \frac{AB^2}{36} +AB^2 − \frac{9AB^2}{36}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{{\bf{\red{\implies \therefore 9AD^2 = 7AB^2}}}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀${\underline{\overline{\huge{\rm {\pink{\bold {Hence \: Proved}}}}}}}$

Answer 2

GIVEN :

$\bf{BD = \frac{1}{3} BC}$

‎

To Prove :

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

‎

Construction : Draw an Altitude AE such that E lies on BC and AE perpendicular to BC

‎

Proof : $\bf{\triangle ABC}$ is an Equilateral triangle such that$\bf{ \: AB=BC=AC}$

‎

$\sf{➥}$ In $\bf{ \triangle AEB}$

‎

By Pythagoras Theorem,

‎

$\sf{ ↦AB^2 =AE^2+EB^2.....(1)}$

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$\sf{➥}$ In $\bf{\triangle AED}$

‎

By Pythagoras Theorem,

‎

$\sf{ ↦ AD^2 =AE^2 +ED^2.....(2)}$

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From (1) and (2)

‎

${\mapsto}$ $\sf{ AD^2=ED^2 +AB^2 −EB^2 }$

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Since,

‎

$\sf{\implies EB= \dfrac{1}{2} BC= \dfrac{AB}{2} = \dfrac{3AB}{6}}$ and

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⠀⠀⠀⠀⠀⠀⠀⠀$\sf{\implies ED= \dfrac{BC}{6} = \dfrac{AB}{6}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf{\implies AD^2= \dfrac{AB^2}{36} +AB^2 − \dfrac{9AB^2}{36}}$

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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\boxed{{\sf{\pink{\implies \therefore 9AD^2 = 7AB^2}}}}$

‎

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀${\underline{\overline{\huge{\rm {\purple {Hence \: Proved}}}}}}$

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