In an equilateral triangle ABC if AD ⊥ BC, then

$\triangle \mathrm{ABC}$ is an equilateral triangle and $\mathrm{AD} \perp \mathrm{BC}$.

In ∆ABD, applying Pythagoras theorem, we get

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2}$

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{BC}\right)^{2}\left(\because \mathrm{BD}=\frac{1}{2} \mathrm{BC}\right)$

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\left(\frac{1}{2} \mathrm{AB}\right)^{2}(\because \mathrm{AB}=\mathrm{BC})$

$\mathrm{AB}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AB}^{2}$

$3 \mathrm{AB}^{2}=4 \mathrm{AD}^{2}$

We got the result as $(b)$.