詳解
因為 $\overline{BD}:\overline{DC}=1:2$，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）：
$$
\left|\overset{\large\rightharpoonup}{AD}\right|^2
= \left|\dfrac{2}{3}\overset{\large\rightharpoonup}{AB}
+ \dfrac{1}{3}\overset{\large\rightharpoonup}{AC}\right|^2.
$$
因此
$$
\begin{aligned}
s^2
&= \dfrac{4}{9}c^2
+ \dfrac{4}{9}\left(\overset{\large\rightharpoonup}{AB}\cdot\overset{\large\rightharpoonup}{AC}\right)
+ \dfrac{1}{9}b^2 \\
&= \dfrac{4}{9}c^2+\dfrac{4}{9}bc\cos60^\circ+\dfrac{1}{9}b^2 \\
&\implies s^2=\dfrac{4}{9}c^2+\dfrac{2}{9}bc+\dfrac{1}{9}b^2
= \dfrac{1}{9}\left(b^2+4c^2+2bc\right).
\end{aligned}
$$
故答案為選項 $(2)$。
解析圖：三角形分點示意圖