Question

In an equilateral triangle ABC, D is a point on side BC such that $BD=\frac{BC}{3}$. Then:

• A. $9A{D}^2=7A{B}^2$
• B. $7A{D}^2=9A{B}^2$
• C. $9A{D}^2=7A{B}^3$
• D. $9A{D}^3=7A{B}^4$

Answer 1

The correct option is A $9A{D}^2=7A{B}^2$


Let the side of the equilateral triangle be a, and AE be the altitude of $\Delta ABC$.
$\therefore BE=EC=\frac{BC}{2}=\frac{a}{2}$
And, $AE=a\frac{\sqrt{3}}{2}$
given that, $BD=\frac{BC}{3}$
$\therefore BD=\frac{a}{3}$
$DE=BE-BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$
Applying Pythagoras theorem in $\Delta ADE$, we get
$A{D}^2=A{E}^2+D{E}^2$
$A{D}^2={\left(\frac{a\sqrt{3}}{2}\right)}^2+{\left(\frac{a}{6}\right)}^2$
$=\left(\frac{3a^2}{4}\right)+\left(\frac{a^2}{36}\right)$
$=\frac{28a^2}{36}$
$=\frac{7}{9}A{B}^2$
$\Rightarrow 9A{D}^2=7A{B}^2$