Question

In an equilateral triangle $ABC$, the side $BC$ is trisected at $D$, then $9A{D}^2$ is _________.

• A. $7A{B}^2$
• B. $8B{C}^2$
• C. $4A{C}^2$
• D. $\frac{3}{2}A{B}^2$

Answer 1

Answer: (A)

$7A{B}^2$

Let $M$ be the midpoint of side $BC$

$AM=\frac{\sqrt{3}}{2}AB$

$MD=\frac{1}{2}BC-\frac{1}{3}BC=\frac{1}{6}BC=\frac{1}{6}AB$

Applying Pythagoras theorem in $△AMD$ ,

$A{D}^2=A{M}^2+M{D}^2$

$\Rightarrow A{D}^2=\frac{3}{4}A{B}^2+\frac{1}{36}A{B}^2$

$\Rightarrow A{D}^2=\frac{7}{9}A{B}^2$

Or $\Rightarrow 9A{D}^2=7A{B}^2$

So, Option (A) is the correct answer.