Discussion

In the applet two squares OABC and OPQR share a common vertex. M and N are centers of the two squares. K and L are the midpoints of AP and CR, respectively. It appears that the quadrilateral KMLN is a square. This is indeed so and is known as the Finsler-Hadwiger theorem.

Consider triangles AOR and COP. OR = OP. Moreover one may be obtained from the other by a suitable rotation through 90^o around O. OA = OC. These too map on each other by the same rotations. Finally, the same is true of triangles AOR and COP, which are thus equal. From here, AR = CP and the two line segments are perpendicular.

In APR, KN that connects midpoints of sides AP and PR is parallel and equal to half AR. Similarly, in ACP, KM is parallel and equal to half CP. Therefore, KM = KN and the two are perpendicular.

In an absolutely similar fashion, from the geometry of triangles CPR and ACR, LM = LN and the two are perpendicular. This proves that KMLN is in fact a square.

Here is another proof. Let M₉₀, N₉₀, and K₁₈₀ denote the rotations around M through 90^o, around N through 90^o, and around K through 180^o. For example, M₉₀(A) = O, N₉₀(O) = P, K₁₈₀(P) = A. The composition of the three rotations has a fixed point at A and is a rotation through 360^o = 0^o (mod 360). It is therefore a translation with a fixed point, i.e. the identity transformation.

Denote the image of M under N₉₀ as M': M' = N₉₀(M). (MNM' is isosceles with angle N being 90^o.) Then

It follows that M = K₁₈₀(M'). But K₁₈₀(M') is a reflection in K. Therefore K is the midpoint of MM'. Therefore MNK is isosceles with angle K being 90^o.

Remark