Thread: Areas of triangles

In ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of PEC=4 and area of PFB=8 and area of PBC=10. Find the area of quadrilateral AFPE.
I tried using reallyyyy many many approaches, even basic-most properties, but i couldn't get to it.

Originally Posted by amey

In ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of PEC=4 and area of PFB=8 and area of PBC=10. Find the area of quadrilateral AFPE.
I tried using reallyyyy many many approaches, even basic-most properties, but i couldn't get to it.

I'm not sure that I understand your question correctly ...(?)

If F is the midpoint of AB and if E is the midpoint of AC

then the area of triangle BCE is as large as the area of triangle ABE.

Therefore




Solve for

Sorry, but A-F-B means points A, F and B are collinear. F is not midpoint of AB

Originally Posted by amey

Sorry, but A-F-E means points A, F and E are collinear. F is not midpoint of AB

Have you omitted some other bit of information?

No. Atleast the problem is that way only.

Easy solution, once one realises this is a mini "crossing ladder" classic problem; go here:
Crossed Ladder Puzzle, Triangle Area Puzzle

Thanks. I have got it now.