Thébault's Problem II

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

On the sides AB and AD of the square ABCD erect equilateral triangles AED and AFB -- both either on the inside or the outside of the square. Then triangle CEF is also equilateral.

Proof 1

Consider, for example, the case where E and F are located outside the square ABCD. First of all, triangles CDE and and CBF are equal. Indeed, CD = CB and DE = BF. Also, CDE = 90^o + 60^o = CBF.

Next, DCE + BCF = DCE + DEC = 180^o - 150^o = 30^o. Therefore, in CEF, ECF = 90^o - 30^o = 60^o and also CE = CF. CEF is thus equilateral.

Proof 2

Let's make a use of complex variables. Assume the origin is at C and let u and v be two complex numbers corresponding to B and D, respectively. In fact, u = ir, v = r, for some positive real number r. Let a be the counterclockwise rotation through 60^o. 1/a = a^-1 then is the rotation through -60^o. a = 3/2 + i/2. a^-1 = 3/2 - i/2.

E corresponds to the complex number u + av, F corresponds to v + u/a = (av + u)/a. The ratio of the two numbers is a, which exactly means that the length of CE and CF are equal (this is because |a| = 1), while the angle between the two is 60^o. This completes the proof. But we can easily establish an additional fact.

The area of the equilateral triangle of side x equals 3·x²/4.

We just found that the side of the equilateral triangle CEF that was constructed outwardly equals u + av. For the one constructed inwardly the side is given by u + v/a. Omitting the coefficient 3/4, the sum of areas of the "outward" and the "inward" triangles CEF is |u + av|² + |u + v/a|².

Since |a| = |1/a| = 1, we obtain

|u + av|2 + |u + v/a|2 = |u + av|2 + |au + v|2   = |u|2 + |av|2 + |au|2 + |v|2   = 2(|u|2 + |v|2),

which means that the total area of two triangles CEF (one constructed outwardly, the other inwardly) equals twice the combined area of triangles ABF and ADE.

Proof 3

Proof 4