If ∆ABC is an equilateral triangle such that AD ⊥ BC, then AD2 =
Question:

If $\triangle \mathrm{ABC}$ is an equilateral triangle such that $\mathrm{AD} \perp \mathrm{BC}$, then $\mathrm{AD}^{2}=$

(a) $32 \mathrm{DC} 2$

(b) $2 \mathrm{DC}^{2}$

(c) $3 \mathrm{CD}^{2}$

(d) $4 \mathrm{DC}^{2}$

Solution:

Given: In an equilateral $\triangle \mathrm{ABC}, \mathrm{AD} \perp \mathrm{BC}$.

Since $\mathrm{AD} \perp \mathrm{BC}, \mathrm{BD}=\mathrm{CD}=\mathrm{BC} 2$

Applying Pythagoras theorem,

$\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}$

$\mathrm{BC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}($ Since $\mathrm{AC}=\mathrm{BC})$

$(2 \mathrm{DC})^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2}($ Since $\mathrm{BC}=2 \mathrm{DC})$

$4 D C^{2}=A D^{2}+D C^{2}$

$3 \mathrm{DC}^{2}=\mathrm{AD}^{2}$

$3 \mathrm{CD}^{2}=\mathrm{AD}^{2}$

We got the result as (c)