A question on equilateral triangles

Question

A point $D$ is on the side $BC$ of an equilateral $\triangle ABC$ such that $DC = \frac{1}{4} BC$ .

Then $AD^2 = ?$

Image I drew...

Options are $13 CD^2$, $9 AB^2$, $6 CD^2$, $12 BC^2$.

Answer 1

Let $E$ be the mid point of $BC$. Then $$ AE^2 = AB^2-BE^2 = 16-4 = 12$$

$$ AD^2 = AE^2 + ED^2 = 12 + 1^2 = 13$$

$$CD^2 = 1$$

So the answer is $13 CD^2$