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 ABD or BCD. 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 ABD 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 ABD than for ABN ADP.
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):