Question

Question 15
In an equilateral triangle ABC, D is a point on side BC such that $BD=\frac{1}{3}BC$. Prove that $9A{D}^2=7A{B}^2$.

Answer 1

Let the side of an equilateral triangle be a, and AE be the altitude of $\Delta ABC$.
$\therefore BE=EC=\frac{BC}{2}=\frac{a}{2}$
And, $AE=\frac{a\sqrt{3}}{2}$
given that, $BD=\frac{1}{3}BC$
$\therefore BD=\frac{a}{3}$
$DE=BE-BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$
Applying Pythagoras theorem in $\Delta ADE$, we get
$A{D}^2=A{E}^2+D{E}^2$
$\Rightarrow$ $A{D}^2$= $(\frac{a\sqrt{3}}{2}{)}^2+(\frac{a}{6}{)}^2$
$\Rightarrow$ $A{D}^2=\left(\frac{3a^2}{4}\right)+\left(\frac{a^2}{36}\right)$
$\Rightarrow$$A{D}^2=\frac{28a^2}{36}$
$\Rightarrow$ $A{D}^2=\frac{7}{9}A{B}^2$ [Since, the side of an equilateral triangle is a.i.e;AB=BC=AC=a]
$\Rightarrow 9A{D}^2=7A{B}^2$