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

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

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, $\angle DHG = \angle AHE,$ which implies $\angle GHE = \angle DHA = 90^{\circ}.$ Q.E.D.

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 $u + (v - u)/2 + i\cdot (v - u)/2 = (u + v)/2 + i\cdot (v - u)/2.$

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

$OE$ corresponds to $(v - u)/2 - i\cdot (u + v)/2.$

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 $v - iu = -i\cdot (u + iv),$ one is obtained from the other with the rotation through $90^{\circ}.$ In addition, $|v - i\cdot u| = |i\cdot (u + iv)| = |(u + iv)|.$ 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

$\begin{align} |u + iv|^{2} + |u - iv|^{2} &= 2(|u|^{2} + |iv|^{2})\\ &= 2(|u|^{2} + |v|^{2})\\ &= |u - v|^{2} + |u + v|^{2}, \end{align}$

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.

Consider $\Delta ABD$ 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 $\angle EOH = 90^{\circ}.$

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.

Grégoire Nicollier
University of Applied Sciences of Western Switzerland
Route du Rawyl 47, CH--1950 Sion, Switzerland
December 12, 2015

This is one of the series of spectral proofs.

Let $q=(1,\,i,\,-1,\,-i)$ be the positively oriented unit square of the complex plane. Any nontrivial parallelogram $Q$ centered at the origin of the complex plane can be expressed up to a spiral similarity as $Q=q+b\bar q$ for some $b=x+iy$ or $Q=\bar q$. We suppose $Q=q+b\bar q$. Let $T(Q)$ be the quadrilateral obtained by erecting isosceles right-angled right-hand ears on the sides of $Q$. The transformation $T$ is linear and deletes $\bar q$: thus $T(Q)=T(q+b\bar q)=T(q)=\sqrt2e^{i\pi/4}q$ is a square.

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