We have a paralelogram $ABCD$ and point $P$ inside it. Halflines $BP$ and $DP$ cuts respectively lines $AD$ and $AB$ at $E$ and $F$. Why are the area of $ABPD$ and $CEPF$ the same regardles of the position of $P$?
This is again some contest problem (I think), I had it in my notes for a long time (at least 10 years) and I thought it would be interesting for this site. I solved it only recently.