Question

In an equilateral triangle $ABC$, the side $BC$ is trisected at point $D$. Prove that $9{AD}^2=7{AB}^2$. (Hint : Draw $AE\perp BC$).

Answer 1

Let side $=a$

Draw $AE\perp BC$ So, as triangle $ABC$ is equilateral.

$BE=\frac{a}{2}$ and as $BD=\frac{a}{3}$

$DE=BE-BD=\frac{a}{6}$

In $△ADE$ by Pythagoras theorem

${AD}^2={AE}^2+{DE}^2$

$={\left(\frac{\sqrt{3}}{2}a\right)}^2+{\frac{a}{36}}^2={\frac{7a}{9}}^2={\frac{7AB}{9}}^2$

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