This question is designed for high school students, so advanced linear algebra solution might not be needed.
Suppose I have a $\vec{G}$ as a meet point between vector $\vec{DB}$ and $\vec{CE}$ as illustrated here:
Note that the intersection point is labeled with G and not E.
If $\vec{AD}:\vec{DC}=3:1$ and $\vec{AE}:\vec{EB}=1:2$, then how to find the proportion $\vec{DG}:\vec{GB}$ using a vector concept? I don't know where to start. I was thinking to write $\vec{DB}$ and $\vec{EC}$ using this formula below
$$\vec p = \dfrac{m \vec c + n \vec a}{m + n}$$
When $p$ is $\vec{DB}$, $\vec a = A$, $\vec b =C$, $m=3$ and $n=1$ (The same formula goes for $\vec{CE}$ as well)
, but I'm not sure if that would help. The reason I'm thinking that way, because I expect to find $\vec G$ first, and then I may be able to use the formula above. However, again that may not be possible if I don't know the vector position yet. I'm not interested in solving this using trigonometry since the topic we were discussing in class is vector, but it's okay if that's the only choice, .
Complete the triangle to a parallelogram with vertices ACBH. Extend CE to intersect AH at I and to intersect BH at J.
Note that
$$\frac{AI}{BC}=\frac{AE}{BE}=\frac12,$$
so
$$\frac{HJ}{AC}=\frac{HI}{AI}=\frac{BC-AI}{AI}=1.$$
Now we know
$$\frac{DG}{GB}=\frac{DC}{BJ}=\frac{DC}{BH+HJ}=\frac18.$$