Question

In an equilateral triangle $ABC$, a point $D$ is taken on base $BC$ such that $BD:DC=2:1$. Prove that $9A{D}^2=7A{B}^2$.

Answer 1

Let A be origin and $B\left(\overrightarrow{b}\right)$ and $C\left(\overrightarrow{c}\right)$

As triangle is equilateral

$\Rightarrow |\overrightarrow{b}|=|\overrightarrow{c}|=|\overrightarrow{b}-\overrightarrow{c}|$

$\frac{A{D}^2}{A{B}^2}=\frac{{\left|\frac{2\overrightarrow{c}+\overrightarrow{b}}{3}\right|}^2}{|\overrightarrow{b}{|}^2}=\frac{1}{9}\frac{(2\overrightarrow{c}+\overrightarrow{b})\cdot (2\overrightarrow{c}+\overrightarrow{b})}{|\overrightarrow{b}{|}^2}$

$=\frac{1}{9}\frac{(4|\overrightarrow{c}{|}^2+4\overrightarrow{b}\cdot \overrightarrow{c}+|\overrightarrow{b}{|}^2)}{|\overrightarrow{b}{|}^2}$

$\{\overrightarrow{b}\cdot \overrightarrow{c}=|\overrightarrow{b}||\overrightarrow{c}|\cos \theta =|\overrightarrow{b}{|}^2\cos 60=\frac{|\overrightarrow{b}{|}^2}{2}\}$

$\Rightarrow \frac{A{D}^2}{A{B}^2}=\frac{1}{9}\frac{|\overrightarrow{b}{|}^2}{|\overrightarrow{b}{|}^2}(4+2+1)=\frac{7}{9}$.