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

$△ABC$ is an Equilateral triangle such that,
$AB=BC=AC$
Construction:
Draw an Altitude $AE$ such that,

$E$ lies on $BC$ and

$AE$ perpendicular to $BC$

In $△AEB$

By Pythagoras Theorem,

$A{B}^2=A{E}^2+E{B}^2.....\left(1\right)$

In $△AED$

By Pythagoras Theorem,

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

From 1) and 2)

$A{D}^2=E{D}^2+A{B}^2-E{B}^2.....\left(3\right)$

Since, $EB=\frac{1}{2}BC=\frac{AB}{2}=\frac{3AB}{6}$ and

$ED=\frac{BC}{6}=\frac{AB}{6}$

$\therefore$ $A{D}^2=\frac{A{B}^2}{36}+A{B}^2-\frac{9A{B}^2}{36}$

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