Question

In an equilateral triangleABC, D is a point on side BC such that $BD=\frac{1}{3}$BC. prove that 9AD=7AB.

Answer 1

Give that, ABC is an equilateral triangle.
Given $BD=\frac{1}{3}BC$
Let, BD = x
$\therefore x=\frac{1}{3}BC$
$\Rightarrow BC=3x$
Construct perpendicular AM.

$AM=\frac{\sqrt{3}}{2}\left(3x\right)$
$AM=\frac{3\sqrt{3}}{2}x$
Consider triangle ADM,
$AD=\sqrt{M{D}^2+A{M}^2}$
$=\sqrt{(\frac{x}{2}{)}^2+(\frac{3\sqrt{3}}{2}x{)}^2}$
$=\sqrt{\frac{x^2}{4}+\frac{27x^2}{4}}$
$AX=\sqrt{7}x$
$\Rightarrow 3\times AD=3\times \sqrt{7}x$
$\Rightarrow 3AD=\sqrt{7}\times \left(3x\right)$
$\Rightarrow 3AD=\sqrt{7}AB$
$\Rightarrow 9AD=3\sqrt{7}AB$