Let ABCD be a rectangle and AD and BC its diagonals.Let K be the point of intersection of the diagonals and P be the midpoint of AB.CP and DP intersect the diagonals at E ,F respectively.How do we find the area of PEKF if the area of ABCD is 20?I tried to eliminate the area of other parts and then subtract from the total,but there is always somekind of intersection.I also tried to use the fact that triangle PCD is half the area of the total rectangle.I even divided the rectangle into four smaller rectangles and then calculate each part separately.Some kind of hint will be appreciated.

• I have made a rough approximation of 2.33 but not sure if it is correct. – rah4927 Jun 13 '13 at 17:33
• $PCD$ is half of the area of which rectangle? – Riccardo Jun 13 '13 at 17:39
• @Ric ped,The rectangle ABCD of course. – rah4927 Jun 13 '13 at 17:40
• Ah, it's a rectangle so :) – Riccardo Jun 13 '13 at 17:44
• @Ric Ped,Ah I was stupid.How could I have forgotten to mention that? – rah4927 Jun 13 '13 at 17:46

The yellow part is $\dfrac{3}{4}$ of the whole rectangle and for the green part, I present THIS to you as a hint.
$\hspace{4.6cm}$
Hint: $\triangle KFP$ is similar to $\triangle BFD$ and $|KP|=\frac12|BD|$. . Furthermore, $|BP|=\frac12|BA|$