Area of triangle ABD is twice the area of triangle ACD

$$\angle\ ADB=\theta\ $$

$$\frac{1}{2}\left(AD\right)\left(BD\right)\sin\theta\ =2\left(\frac{1}{2}\left(AD\left(CD\right)\sin\left(180-\theta\ \right)\right)\right)$$

$$BD\ =2CD$$

Therefore, BD = 2 and CD = 1

$$\angle\ ABC=\angle\ ACB=60^{\circ\ }$$

Applying cosine rule in triangle ADC, we get

$$\cos\angle\ ACD=\ \frac{\ AC^2+CD^2-AD^2}{2\left(AC\right)\left(CD\right)}$$

$$\frac{1}{2}=\ \frac{\ 9+1-AD^2}{6}$$

$$AD^2=7$$

$$AD=\sqrt{\ 7}$$

The answer is option C.