Solving geometry problem, in a triangle, using vectors P is the middle of the median line from vertex A, of ABC triangle. Q is the point of intersection between lines AC and BP.

 A: Starting you off in more detail....
With $\mathbb{a, b}$ denoting the vertices $A, B$ and $C$ being the Origin, one gets $$\mathbb m = \tfrac12\mathbb b, \quad \mathbb p = \tfrac12(\mathbb{m+a})=\tfrac12\mathbb a+\tfrac14 \mathbb b$$
Now $Q$ is located on the intersection of $\vec{BP} = t\mathbb b+(1-t)\mathbb p$ and $\vec {CA} = s\mathbb a$, so we solve to get $t = -\frac13, s = \frac23$, giving $\mathbb q = \frac23 \mathbb a$.
Actually, we have all that is needed to answer the questions by now...
A: Let $\vec a=\vec{CA},\vec b=\vec{CB}$. Then, we have
$$\vec{CM}=\frac 12\vec b$$$$\vec{CP}=\frac 12\vec{CA}+\frac 12\vec{CM}=\frac 12\vec a+\frac 14\vec b\tag 1$$
Also, setting $QC:AQ=s:1-s,BP:PQ=t:1-t$ gives
$$\vec{CP}=t\vec{CQ}+(1-t)\vec{CB}=ts\vec a+(1-t)\vec b\tag2$$
Now comparing $(2)$ with $(1)$ will give you the answer.
A: Hint: 
Part 1: Apply Menelaus' theorem on triangle $AMC$ and line $BPQ$.
Part 2: Apply Menelaus' theorem on triangle $BCQ$ and line $APM$ (needs the answer form part 1).
I'm giving only hints but you should be able to figure it out after learning the theorem (you will learn more!). If you need further help, feel free to comment.
