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Squares on Sides of a Quadrilateral: What is this about?
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Discussion
Erect similarly oriented squares on the sides of a quadrilateral ABCD. Denote their centers successively O1, O2, O3, and O4. Then
| (1) | O1O3 = O2O4 and | (2) |
O1O3  O2O4.
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Proof
Let M be the midpoint of AC. By the Finsler-Hadwiger theorem,
| MO1 = MO2 and |
MO1 MO2
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and also
| MO3 = MO4 and |
MO3 MO4.
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Therefore, triangles MO1O3 and MO2O4 are equal and one is a rotation of the other through 90o. Hence, (1) and (2) hold.
Remark
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.
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.
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.
References
- E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #320b
- 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
- R. L. Finney, Dynamic Proofs of Euclidean Theorems, Math Magazine, 43, pp. 177-185.
- M. Gardner, Mathematical Circus, Vintage, 1981
- I. M. Yaglom, Geometric Transformations I, MAA, 1962
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