Question

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

Answer 1

Step-1:Construction:

Draw altitude $AE$ perpendicular to $BC$

Step-2: Express $AE$ in terms of $AB$:

Consider right angled $△AEB$ with right angle at vertex $E$.

By applying Pythagoras theorem

$A{E}^2+E{B}^2=A{B}^2$

$\Rightarrow$ $A{E}^2=A{B}^2-E{B}^2$

In an equilateral triangle, the altitude bisects the opposite side i.e. $EB=\frac{BC}{2}=\frac{AB}{2}\left[\because AB=BC\right]$

$\Rightarrow$$A{E}^2=A{B}^2-{\left(\frac{AB}{2}\right)}^2$

$\Rightarrow$$A{E}^2=\frac{3}{4}A{B}^2$ $...\left(i\right)$

Step-3: Express $ED$ in terms of $AB$:

Consider right angled $△AED$ with right angle at vertex $E$.

By applying Pythagoras theorem

$A{E}^2+E{D}^2=A{D}^2...\left(ii\right)$

From the figure we know that

$ED=EB-BD$

$\Rightarrow$$ED=\frac{BC}{2}-\frac{BC}{3}=\frac{BC}{6}$

$\Rightarrow$$ED=\frac{AB}{6}$ $\left[\because BC=AB\right]$ $...\left(iii\right)$

Step-4: Substitute the obtained values to prove the required result:

Substituting $\left(i\right)$ and $\left(iii\right)$ in $\left(ii\right)$ we get,

$\Rightarrow$$\frac{3}{4}A{B}^2+{\left(\frac{AB}{6}\right)}^2=A{D}^2$

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

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

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

Dividing throughout by $4$ we get

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

Hence,it is proved that $9A{D}^2=7A{B}^2$.