In $\triangle ABC$ , $\angle BAC = 60^{\circ} , AB = 2 , BC = \sqrt{6}$ . $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$ . Then $AD =$ $\_\_\_\_$ .
In $\triangle ABC$ , $\angle BAC = 60^{\circ} , AB = 2 , BC = \sqrt{6}$ . $AD$ bisects $\angle BAC$ and intersects $BC$ at point $D$ . Then $AD =$ $\_\_\_\_$ .