Find the ratio of two line segments in a triangle

[Andraz Cepic]

1. The problem statement, all variables, and given/known data
Triangle ABC has a point D on the line segment AB which cuts the segment in ratio AD : DB = 2 : 1.
Another point E is on the line segment BC, cutting it in ratio BE : EC = 1 : 4.
Point F is the intersection of the line segments AE and DC.
Given this data find the ratio of line segments AF : FE.

Homework Equations

Vectors, ratios, triangle,...

The Attempt at a Solution

Studying for a high school final exam, I stumbled upon this task. I tried for an hour but could not solve it. Normally I would work hard until I solved the thing, but the exam is tomorrow, so I don't have the time to play around as I have to do other things.

Firstly, I tried to write AF and FE as vectors. Then I wrote them as linear combinations of vector AE, so that
AF = m*AE and
FE = n*AE,
or AF + FE = AE,
or AF + FE = (m+n)AE, it follows that m+n = 1.

I wrote the ratio as |AF| / |FE| = m / n.
And then as |AF| / |FE| = 1/n - 1.(Since m = 1-n. Wrote this just to see if I can find n somehow)
Now I tried to express the two vectors with base vectors, for which I chose AB and BC, however, I could never find a connection where variables would cancel out to get the ratio, always I got too many variables, or I tried to express AE with base vectors and then somehow find either n or m, but was always just going in circles.

So I have no idea what to do right now and I am sure I could solve it after playing around long enough, but I really want to understand this before the exam. I am sorry If it seems as if I didn't even try but I am so desperate to know since this is the only task in linear algebra that I could not solve. I am probably just missing sth BIG.

Thanks in advance for all answers!

 

[Charles Link]

I think I successfully solved it. I applied the law of sines about a half dozen times(editing: I rewrote the solution, and did a count, and it turned out I used the law of sines a total of 5 times=5 separate equations) and also used the fact that ## \sin(180-\theta)=\sin(\theta) ## for the angle at AFC and the angle at CFE. It took a couple of algebraic substitutions (e.g. dividing one equation by the other, etc.), but I think I got the correct answer in about 10-15 minutes. In any case, an interesting problem. :)