### The Carpets Theorem With Parallelograms: What is it about?

A Mathematical Droodle

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Copyright © 1996-2018 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.

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 $\Delta ABD$ or $\Delta BCD.$ 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 $\Delta ABD$ and as the second carpet the union of triangles $ABN$ and $ADP.$ Since both have areas equal to $\displaystyle \frac{Area(ABCD)}{2},$ the theorem applies immediately. However, the fact is more obvious for $\Delta ABD$ than for $\Delta ABN\cup \Delta ADP.$

Draw the diagonal $AC.$

(1)

$\displaystyle \frac{Area(ABN)}{Area(ABC)} = \frac{BN}{BC} = \frac{BM}{BD}.$

(The latter identity follows from the similarity of triangles $BCD$ and $BNM.)$

Similarly,

(2)

$\displaystyle \frac{Area(ADP)}{Area(ACD)} = \frac{DP}{CD} = \frac{DM}{BD}.$

Since $\displaystyle Area(ABC) = Area(ACD) = \frac{1}{2}Area(ABCD),$ we get the required result by adding (1) and (2):

$\displaystyle \begin{align} \frac{Area(ABN) + Area(ADP)}{\displaystyle \frac{1}{2}Area(ABCD)}&= \frac{BM + DM}{BD}\\ &= 1. \end{align}$

### References

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

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Copyright © 1996-2018 Alexander Bogomolny