Question

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

Answer 1

Given: $BD=\frac{1}{3}BC$

In $\Delta ABC$ and $\Delta ACE$

$AB=AC$…………(given)

$AE=AE$ ……….(common side)

$\angle AEB=\angle AEC$………(each ${90}^o$)

$\therefore \Delta ABC\cong \Delta AEC$ ……….(SAS test of congruence)

$\therefore BE=EC$ ……….(cpct)

Also, $BE=EC=\frac{BC}{2}$

Also, $\Delta ADE$

$A{D}^2=A{E}^2+D{E}^2$ ……….$\left(1\right)$

$\Delta ABE$

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

From $\left(1\right)$ and $\left(2\right)$

$A{D}^2-A{B}^2=D{E}^2-B{E}^2$

$A{D}^2-A{B}^2=(BE-BD{)}^2-B{E}^2$

$A{D}^2-A{B}^2={(\frac{BC}{2}-\frac{BC}{3})}^2-{\left(\frac{BC}{2}\right)}^2$

$A{D}^2-A{B}^2=\frac{B{C}^2}{36}-\frac{B{C}^2}{4}$ $(AB=BC=AC)$

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

$\therefore 9A{D}^2=7A{B}^2$.