Thébault's Problem I
What Is It?
A Mathematical Droodle

9 October 2015, Created with GeoGebra

Thébault's Problem I

The applet suggests the following theorem (Victor Thébault, 1937):

On the sides of parallelogram ABCD erect squares -- all either on the outside or the inside of the parallelogram. Their centers then form another square.

Proof 1

Consider the case of the squares erected on the outside of the parallelogram.

Triangles AEH, BFE, CGF, and DHG are obviously equal, so that the quadrilateral EFGH is a rhombus. Further, ∠DHG = ∠AHE, which implies ∠GHE = ∠DHA = 90°. Q.E.D.

Proof 2

Place the origin at the center of the parallelogram ABCD, and let u and v be complex numbers associated with D and A. Then DA is associated with v - u, while AB is corresponds to -(v + u). Further,

OH corresponds to u + (v - u)/2 + i·(v - u)/2 = (u + v)/2 + i·(v - u)/2.

Similarly we can compute OE, OF, OG. For example,

OE corresponds to (v - u)/2 - i·(u + v)/2.

The difference of the two numbers corresponds to the side EH of the quadrilateral EFGH and equals u + iv. If the squares were constructed inwardly, EH would correspond to u - iv.

Similarly, v - iu corresponds to GH. Since v - iu = -i·(u + iv), one is obtained from the other with the rotation through 90°. In addition, |v - i·u| = |i·(u + iv)| = |(u + iv)|. Therefore the quadrilateral EFGH is a square.

Recall now the Parallelogram Law: for any two complex numbers X and Y,

|X + Y|² + |X - Y|² = 2(|X|² + |Y|²),

from which we derive

|u + iv|² + |u - iv|² = 2(|u|² + |iv|²)

= 2(|u|² + |v|²)

= |u - v|² + |u + v|²,

which means that the sum of the areas of the two squares EFGH (one constructed inwardly, the other outwardly) equals the sum of the areas of two distinct squares constructed on the sides of the parallelogram.

Proof 3

Consider ΔABD and the squares on sides AD and AB. As was shown elsewhere, the square whose diagonal coincides with EH has one of its vertices in the midpoint of BD, i.e., O, such that ∠EOH = 90°.

The tesselation of the plane into parallelograms and squares that serves to prove the Law of Cosines also provides an additional demonstration of Thebault's theorem.

References

R. Nelsen, Proofs Without Words II, MAA, 2000, p. 19

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	Jack D'Aurizio Proof of Sawayama's Lemma
	Y. Sawayama's Theorem
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49552092

\|u + iv\|² + \|u - iv\|²	= 2(\|u\|² + \|iv\|²)
	= 2(\|u\|² + \|v\|²)
	= \|u - v\|² + \|u + v\|²,

Thébault's Problem I What Is It? A Mathematical Droodle