The Carpets Theorem With Parallelograms

The Carpets Theorem With Parallelograms: What is it about?
A Mathematical Droodle

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In 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.

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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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 ΔABN ΔADP.

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):

(Area(ABN) + Area(ADP)) / (½Area(ABCD))	= (BM + DM)/BD
	= 1.

References

T. Andreescu, B. Enescu, Mathematical Olympiad Treasures, Birkhäuser, 2004

Carpets Theorem

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