1. ## Areas of triangles

In $\displaystyle \triangle$ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of $\displaystyle \triangle$PEC=4 and area of $\displaystyle \triangle$PFB=8 and area of $\displaystyle \triangle$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.

2. Originally Posted by amey
In $\displaystyle \triangle$ABC, E and F are such that A-F-B and A-E-C. Segments BE and CF intersect at P. Area of $\displaystyle \triangle$PEC=4 and area of $\displaystyle \triangle$PFB=8 and area of $\displaystyle \triangle$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
$\displaystyle \Delta(BCE) = \Delta(ABE)$

$\displaystyle (10+4) = 8+area(AFPE)$

Solve for $\displaystyle area(AFPE)$

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

4. 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?

5. No. Atleast the problem is that way only.

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

7. Thanks. I have got it now.