The Carpets Theorem With Parallelograms: What is it about?
A Mathematical Droodle
What if applet does not run? 
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Copyright © 19962017 Alexander BogomolnyIn the applet, point M is located on the diagonal BD of the parallelogram ABCD. MN and MP are drawn parellel to the sides of the parallelogram so that another parallelogram  MNCP  is formed. The applet purports to illustrate a simple fact that the combined area of the blue triangles equals that of the red triangle. To see that, we need only use a property of the trapezoids twice. In two trapezoids  ABNM and ADPM  the areas of the triangles formed by the diagonals and adjacent to the nonparallel sides are equal.
What if applet does not run? 
In the applet, the point M is always located on the diagonal BD. But as the proof shows this is not necessary. It may be either in ΔABD or ΔBCD. In one case the red quadrilateral AEMF is a dart, in the other a kite. With M on BD, it's a triangle. In this case we also have a nice application of the Carpets Theorem. We may take as one carpet ΔABD and as the second carpet the union of triangles ABN and ADP. Since both have areas equal to Area(ABCD)/2, the theorem applies immediately. However, the fact is more obvious for ΔABD than for
Draw the diagonal AC.
(1)  Area(ABN)/Area(ABC) = BN/BC = BM/BD. 
(The latter identity follows from the similarity of triangles BCD and BNM.)
Similarly,
(2)  Area(ADP)/Area(ACD) = DP/CD = DM/BD. 
Since Area(ABC) = Area(ACD) = ½Area(ABCD), we get the required result by adding (1) and (2):

References
 T. Andreescu, B. Enescu, Mathematical Olympiad Treasures, Birkhäuser, 2004
Carpets Theorem
 The Carpets Theorem
 Carpets in a Parallelogram
 Carpets in a Quadrilateral
 Carpets in a Quadrilateral II
 Square Root of 2 is Irrational
 Carpets Theorem With Parallelograms
 Two Rectangles in a Rectangle
 Bisection of Yin and Yang
 Carpets in Hexagon
 Round Carpets
 A Property of Semicircles
 Carpets in Triangle
 Carpets in Triangle, II
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Copyright © 19962017 Alexander Bogomolny62073034 