Question

In an equilateral triangle $ABC$, $AD$ is the altitude drawn from $A$ on side $BC$. Prove that $3{AB}^2=4{AD}^2$.

Answer 1

Given, $ABC$ is a equilateral triangle

So, $AB=BC=CA=a$ (let)

In $\Delta ABD$ and $\Delta ACD$

$AB=AC$ and $AD=AD$

$\angle ADB=\angle ADC$

$\therefore \Delta ABD\cong \Delta ACD$

Thus $BD=CD=\frac{a}{2}$

Now in $\Delta ABD\angle D={90}^0$

$\Rightarrow A{B}^2=B{D}^2+A{D}^2={\left[\frac{CD}{2}\right]}^2=A{D}^2$

$\Rightarrow 3A{B}^2=4A{D}^2$