Point P lies on the side AB of triangle ABC, and the ratio between AP and PB is 1 : 3
The median line from vertex A intersects line segment CP at point Q and side BC at point D
What is the ratio between AQ and QD
The answer should be 2 : 3 but I can't reach it! A vector approach would be optimal and greatly appreciated. Thanks in advance!
(The most direct solution would use Menelaus' theorem, as pointed out in a comment already. The following is an alternative way to solve it, per the "vector approach" part of OP's question.)
By construction:
$$ \begin{cases} \begin{align} \overrightarrow{AP} &= \frac{\overrightarrow{AB}}{4} \\[5px] \overrightarrow{AD} &= \frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} \end{align} \end{cases} \tag{1} $$
Let $\,\color{blue}{\dfrac{AQ}{QD} = \lambda} \iff \dfrac{AQ}{AD} = \dfrac{\lambda}{\lambda+1} \,$, then:
$$\overrightarrow{AQ} = \dfrac{\lambda \, \overrightarrow{AD}}{\lambda+1} \tag{2}$$
Let $\,\dfrac{CQ}{QP} = \mu\,$, then:
$$\overrightarrow{AQ} = \overrightarrow{AC}+\overrightarrow{CQ}=\overrightarrow{AC}+\dfrac{\mu\,\overrightarrow{CP}}{\mu+1}=\overrightarrow{AC}+\dfrac{\mu\,\big(\overrightarrow{AP}-\overrightarrow{AC}\big)}{\mu + 1}=\dfrac{\overrightarrow{AC}+\mu\,\overrightarrow{AP}}{\mu+1} \tag{3} $$
Equating $\,(2)\,$ and $\,(3)\,$, then using $\,(1)\,$:
$$ \dfrac{\lambda \, \overrightarrow{AD}}{\lambda+1} = \dfrac{\overrightarrow{AC}+\mu\,\overrightarrow{AP}}{\mu+1} \;\;\iff\;\; \frac{\lambda}{\lambda+1} \cdot \frac{\overrightarrow{AB}+\overrightarrow{AC}}{2} = \frac{1}{\mu+1}\cdot \overrightarrow{AC}+\frac{\mu}{\mu+1}\cdot\frac{\overrightarrow{AB}}{4} \tag{4} $$
Since $\,\overrightarrow{AB}, \overrightarrow{AC}\,$ are linearly independent, the component-wise coefficients must match in $\,(4)\,$:
$$ \begin{cases} \begin{align} \frac{\lambda}{2(\lambda+1)} &= \frac{\mu}{4(\mu+1)} \\[5px] \frac{\lambda}{2(\lambda+1)} &= \frac{1}{\mu+1} \\[5px] \end{align} \end{cases} $$
It then easily follows that $\,\mu=4\,$ and $\,\color{blue}{\lambda=\dfrac{2}{3}}\,$.