Question

In an equilateral triangle $ABC,D$ is a point on the side $BC$ such that $BD=\frac{1}{3}BC$. Prove that $9A{D}^2=7A{B}^2$

Answer 1

$DrawAE\perp BC$

$AB=BC=AC=x$

$BD=\frac{x}{3}$

$BE=EC=\frac{BC}{2}=\frac{x}{2}$

$BE=\frac{x}{2}$

$BD+DE=\frac{x}{3}+DE=\frac{x}{2}$

$DE=\frac{x}{6}$

$\Delta AEB$

$A{B}^2=A{E}^2+{\left(BE\right)}^2$

$x^2={\left(AE\right)}^2+{\left(\frac{x}{2}\right)}^2$

$x^2={\left(AE\right)}^2+\frac{12}{4}$

$A{E}^2=x^2-\frac{x^2}{4}=\frac{3x^2}{4}$

$\Delta AED$

$A{D}^2=A{E}^2+D{E}^2$

$A{D}^2=\frac{3x^2}{4}+{\left(\frac{x}{6}\right)}^2$

$A{D}^2=\frac{3x^2\times 9+x^2}{36}=\frac{28x^2}{36}$

$A{D}^2=\frac{7x^2}{9}$

$9A{D}^2=7x^2$