# Equilateral Triangles on Sides of a Quadrilateral

What is it?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander BogomolnyThe applet may suggest the following statement [Yaglom, p. 39]:

On the sides of an arbitrary quadrilateral ABCD equilateral triangles ABP, BCQ, CDR, DAS are constructed, so that the orientation of the first and the third is different from that of the second and the fourth. Then the quadrilateral PQRS is a parallelogram.

(In [Yaglom] quadrilateral ABCD is unnecessarily required to be convex, whereas both the theorem and the proof remain valid even for self-intersecting quadrilaterals.)

What if applet does not run? |

Rotations will play an important role in the following. So let's introduce a convenience notation. For a point M and an angle α, let *identical transformation*, i.e. the transformation that leaves all the points unmoved.

Consider the four rotations: _{P} = (P, 60°),_{Q} = (Q, -60°),_{R} = (R, 60°),_{S} = (S, -60°).

For the same reason, the sum of the first two is a translation, as is the sum of the remaining two rotations. The two translations add up to the identity, or "zero" translation:

(T_{P} + T_{Q}) + (T_{R} + T_{S}) = 0,

which simply means that they are equal as vectors but point in opposite directions.

How can we visualize the two translations? Under a translation, all points move by the same vector. What happens, for example, to the point P under the sum _{P} + T_{Q}?_{P}: _{P}(P) = P._{Q}(P) = P'. Under the sum of the two rotations, P maps onto P'. The sum is, therefore, equivalent to a translation by vector PP'. Note that ΔPQP' is equilateral.

Similarly, if R' is the image of R under the sum _{R} + T_{S},

(This theorem has a natural generalization.)

### References

- I. M. Yaglom,
*Geometric Transformations I*, MAA, 1962

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