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.
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.
