Question

In $△ABC$, $C$ is a point on $BD$ such that $BC:CD=1:2$ and $△ABC$ is an equilateral triangle. Prove that ${AD}^2=7{AC}^2$

Answer 1

Data: In $△ABD$, $BC:CD=1:2$
In $△ABC$, $AB=BC=CA$
To Prove: ${AD}^2=7{AC}^2$
Construction: Draw $AE\perp BC$
Proof: In $△ABC$,
$BE=EC=\frac{a}{2}$ and
$AE=\frac{a\sqrt{3}}{2}$
In $△ADE$, $\angle AED={90}^o$ [$\because$ Construction]
$\therefore$ ${AD}^2={AE}^2+{ED}^2$ [$\because$ Pythagoras theorem]
${AD}^2={\left(\frac{a\sqrt{3}}{2}\right)}^2+{(2a+\frac{a}{2})}^2$
${AD}^2=\frac{3a^2}{4}+{\left(\frac{5a}{2}\right)}^2$
${AD}^2=\frac{3a^2}{4}+\frac{25a^2}{4}$
${AD}^2=\frac{3a^2+25a^2}{4}$
${AD}^2=\frac{28a^2}{4}$
${AD}^2=7a^2$
$\therefore$ ${AD}^2=7{AC}^2$