Question

In equilateral $△$ABC, D$\in \overline{BC}$ such that BD : DC = 1:2. Prove that 3AD = $\sqrt{7}$ AB

Answer 1

REF.image

Let$A\left(\overrightarrow{o}\right),B\left(\overrightarrow{b}\right),C\left(\overrightarrow{c}\right)$

$\Rightarrow$ By Section's formula,

$\overrightarrow{D}=\frac{\overrightarrow{c}+2\overrightarrow{b}}{3}$

$\Rightarrow \overrightarrow{AD}=\frac{\overrightarrow{c}+2\overrightarrow{b}}{3}-\overrightarrow{0}=\frac{\overrightarrow{c}+2\overrightarrow{b}}{3}$

$\overrightarrow{AB}=\overrightarrow{b}:AD=\sqrt{\overrightarrow{AD}.\overrightarrow{AD}}=\frac{\sqrt{|\overrightarrow{c}{|}^2+4|\overrightarrow{b}{|}^2+4\overrightarrow{c}.\overrightarrow{b}}}{9}$

$|\overrightarrow{c}|=|\overrightarrow{b}|=|\overrightarrow{a}|$ (equilateral triangle);$\overrightarrow{b}.\overrightarrow{c}=|\overrightarrow{b}||\overrightarrow{c}|cos60=|\overrightarrow{b}{|}^2\left(\frac{1}{2}\right)$

$\Rightarrow AD=\frac{\sqrt{5|\overrightarrow{b}{|}^2+2|\overrightarrow{b}{|}^2}}{9}=\frac{\sqrt{7}}{3}|\overrightarrow{b}|;|\overrightarrow{AB}|=|\overrightarrow{b}|$

$\Rightarrow AD/AB=\sqrt{7}/3$