因為 $BD : DC = 1 : 2$，由平面向量分點公式得： $$\overset{\large\rightharpoonup}{AD} = \dfrac{2}{3}\overset{\large\rightharpoonup}{AB} + \dfrac{1}{3}\overset{\large\rightharpoonup}{AC}$$ 對等式兩邊自我內積（即求平方長度）： $$\left|\overset{\large\rightharpoonup}{AD}\right|^2 = \left|\dfrac{2}{3}\overset{\large\rightharpoonup}{AB} + \dfrac{1}{3}\overset{\large\rightharpoonup}{AC}\right|^2$$ $$s^2 = \dfrac{4}{9}\left|\overset{\large\rightharpoonup}{AB}\right|^2 + \dfrac{4}{9}\left(\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}\right) + \dfrac{1}{9}\left|\overset{\large\rightharpoonup}{AC}\right|^2$$ $$s^2 = \dfrac{4}{9}c^2 + \dfrac{4}{9}bc\cos 60^\circ + \dfrac{1}{9}b^2$$ 將 $\cos 60^\circ = \dfrac{1}{2}$ 代入，整理得： $$s^2 = \dfrac{4}{9}c^2 + \dfrac{2}{9}bc + \dfrac{1}{9}b^2 = \dfrac{1}{9}\left(b^2 + 4c^2 + 2bc\right)$$ 故答案為選項 $(2)$。