Ratios of line segments in a triangle
Not sure if this belongs here, it's pretty tough to solve (and I have the solution from the book!) I thought I would share it here in case others wanted to be challenged, or if anyone can provide the reasoning behind the solution.
We have a triangle ABC as shown in the figure, with point E selected on side AB in such a way that AE : EB = 1:3, in other words, the ratio of the length of segment AE to the length of segment EB is 1 to 3. Draw the line segment CE.
We then have a point D selected on side BC so that CD : DB = 1:2. Then draw a line segment from A to D. This gives a point F, which is the intersection of segments AD and CE.
Solve for (EF/FC) + (AF/FD).