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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 $P$?

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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.

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![enter image description here

The triangles ABD, CBD, CDF and CBE all have the same area. Subtracting CYD we get CYF = DYB. Subtracting BCX we get CXE = BXD. The mauve quadrilateral ECFB therefore has the same area as CBD + PBD which is the same as DBA + PBD which is the same as the green quadrilateral's.

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    $\begingroup$ Could you please crop the picture to remove all the unnecessary whitespace? $\endgroup$
    – bobble
    Apr 18 at 0:00

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