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

Answer:

yes it true that two and three prove that 9AD2= 7 AB2

Answer 2

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$\bf{ \: AB=BC=AC}$

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$\Large{➥}$ 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,

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$\bf{ ↦ AD^2 =AE^2 +ED^2.....(2)}$

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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,

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$\bf{\implies EB= \frac{1}{2} BC= \frac{AB}{2} = \frac{3AB}{6}}$ 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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⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀${\underline{\overline{\huge{\rm {\pink{\bold {Hence \: Proved}}}}}}}$