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Q15 In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that
$9 AD^2 = 7 AB^2$

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Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC.

To prove : $9 AD^2 = 7 AB^2$

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

In $\triangle$ AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

$a^2=AE^2+(\frac{a}{2})^2$

$\Rightarrow a^2-(\frac{a^2}{4})=AE^2$

$\Rightarrow (\frac{3a^2}{4})=AE^2$

$\Rightarrow AE=(\frac{\sqrt{3}a}{2})$

Given : BD = 1/3 BC.

$BD=\frac{a}{3}$

$DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$

In $\triangle$ ADE, by Pythagoras theorem,

$AD^2=AE^2+DE^2$

$\Rightarrow AD^2=(\frac{\sqrt{3}a}{2})^2+(\frac{a}{6})^2$

$\Rightarrow AD^2=(\frac{3a^2}{4})+(\frac{a^2}{36})$

$\Rightarrow AD^2=(\frac{7a^2}{9})$

$\Rightarrow AD^2=(\frac{7AB^2}{9})$

$\Rightarrow 9AD^2=7AB^2$

Posted by

seema garhwal

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