Question

In an equilateral triangal ABC, D is a point on side BC such that $BD=\frac{1}{3}BC$ .Prove that $9A{D}^2=7A{B}^2$

Answer 1

We have the above figure according to the question.

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

And let $AF$ be perpendicular to $BC$

Now, since $ABC$ is equilateral,

$BF=FC$ and $\angle B=\angle C=\angle A={60}^o$

So, $AF=AC\cos 30$

$=AC\times \frac{\sqrt{3}}{2}$

Now,

$BF=\frac{1}{2}BC$

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

So, $DF=BF-BD$

$=\frac{1}{2}BC-\frac{1}{3}BC$

$DF=\frac{1}{6}\times BC$

So, by Pythagoras theorem in $△AFD$ we get,

$A{D}^2=A{F}^2+D{F}^2$

$A{D}^2={(AC\times \frac{\sqrt{3}}{2})}^2+{\left(\frac{1}{6}BC\right)}^2$

Now we know $AC=BC=AB$

So, we have,

$A{D}^2=\frac{3}{4}A{B}^2+\frac{1}{36}A{B}^2$

$\Rightarrow A{D}^2=(\frac{27}{36}+\frac{1}{36})A{B}^2$

$\Rightarrow A{D}^2=\frac{28}{36}A{B}^2$

$\Rightarrow A{D}^2=\frac{7}{9}A{B}^2$

$\Rightarrow 9A{D}^2=7A{B}^2$