Question

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

• A. True
• B. False

Answer 1

The correct option is A True

Let E be the perpendicular from A on BC.

Let $BD=x$

$BC=AC=AB=3x$

$AE=\sqrt{(3x{)}^2-(1.5x{)}^2}$

$DE=\frac{3x}{2}-\frac{x}{=}0.5x$
$AD=\sqrt{9x^2-2.25x^2+0.25x^2}=x\sqrt{7}$

$\frac{AD}{AB}=\frac{x\sqrt{7}}{3x}$

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