Question

In an equilateral triangle ABC, if AD ⊥ BC, then which of the following is true?


(a) 2AB² = 3AD²
(b) 4AB² = 3AD²
(c) 3AB² = 4AD²
(d) 3AB² = 2AD²

Answer 1

(c) 3AB² = 4AD²

Applying Pythagoras theorem in right-angled triangles ABD and ADC, we get:

$$\begin{gathered} A{B}^2=A{D}^2+B{D}^2 \\ \Rightarrow A{B}^2={\left(\frac{1}{2}AB\right)}^2+A{D}^2(\because △\mathrm{ABC}\hspace{0.17em}\mathrm{is}\mathrm{equilateral}\mathrm{and}AD=\frac{1}{2}\mathrm{AB}) \\ \Rightarrow A{B}^2=\frac{1}{4}A{B}^2+A{D}^2 \\ \Rightarrow A{B}^2-\frac{1}{4}A{B}^2=A{D}^2 \\ \Rightarrow \frac{3}{4}A{B}^2=A{D}^2 \\ \Rightarrow 3A{B}^2=4A{D}^2 \end{gathered}$$