Question

In an equilateral $△ABC,D$ is a point on the side BC, such that BD trisects BC. If $mA{D}^2=7A{B}^2.$. Find $m$

Answer 1

$letlengthofeachsideof△ABC=a$

$In△ABC$

$Usingcosinerule$

$\cos B=\frac{{AB}^2+{BD}^2-{AD}^2}{2AB.BD}$

$\cos \left(60\right)=\frac{a^2+\frac{a^2}{9}-{AD}^2}{\frac{{2a}^2}{3}}$

$\frac{1}{2}\times \frac{{2a}^2}{3}=\frac{{10a}^2}{9}-{AD}^2$

${AD}^2=\frac{{10a}^2}{9}-\frac{{2a}^2}{6}$

${AD}^2=\frac{{10a}^2}{9}-\frac{a^2}{3}$

${AD}^2=\frac{{10a}^2-{3a}^2}{9}$

${AD}^2=\frac{{7a}^2}{9}$

$9{AD}^2={7a}^2$

$\because AB=a$

${9AD}^2={7AB}^2$

$Given:m{AD}^2={7AB}^2$

$m=9$