Squares on Sides of a Quadrilateral
What is this about?
A Mathematical Droodle

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Discussion

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Let M be the midpoint of AC. By the Finsler-Hadwiger theorem,

and also

Therefore, triangles MO₁O₃ and MO₂O₄ are equal and one is a rotation of the other through 90°. Hence, (1) and (2) hold.

Remark

1. If any two adjacent vertices of a quadrilateral coalesce into a point, the statement just proven becomes a property of the Bride's Chair configuration.

2. The theorem we just proved is attributed to Van Aubel (Von Aubel in [Gardner, p. 176-178]) could also be found in [de Villiers, Yaglom, Finney] among others.

3. The argument falls through in case one of O₁O₃ or O₂O₄ is of zero length. This case is considered elsewhere. There it is shown that the two distances are either both 0 or not. The former case occurs iff the diagonals of the quadrilateral are equal and perpendicular.

4. The problem has a simple cosequence.

5. There is an additional exploration of this configuration.

6. As a consequence, the quadrilateral with vertices at the midpoints of the sides of quadrilateral O₁O₂O₃O₄ is a square. (This was problem 11328, Am Math Monthly 114 December 2007.)

References

1. E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #320b
2. M. de Villiers, The Role of Proof in Investigative, Computer-based Geometry: Some Personal Reflections, in Geometry Turned On, MAA Notes 41, 1997, pp. 15-24
3. R. L. Finney, Dynamic Proofs of Euclidean Theorems, Math Magazine, 43, pp. 177-185.
4. M. Gardner, Mathematical Circus, Vintage, 1981
5. I. M. Yaglom, Geometric Transformations I, MAA, 1962

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