In an equilateral triangle $\triangle ABC$, $AD$ is drawn perpendicular to $BC$ meeting $BC$ in $D$. Prove that $AD^{2} = 3 BD^{2}$.

Question

In an equilateral triangle $\triangle ABC$, $AD$ is drawn perpendicular to $BC$ meeting $BC$ in $D$. Prove that $AD^{2} = 3 BD^{2}$.

Answer 1

It is fairly simple.

Note that $ABD$ is a right triangle, thus:

$AB^2=AD^2+BD^2$

You have also $BC=AB=2BD$

Then $4BD^2=AD^2+BD^2$

That is

$AD^2=3BD^2$

Answer 2

we have applying the Pythagorean theorem $$AD^2+BD^2=AB^2$$ and $$BD=\frac{1}{2}AB$$ can you finish?

Answer 3

Just some trigonometry: $$\tan\frac\pi3=\sqrt 3=\frac{AD}{BD}.$$