Simsons of Diametrically Opposite Points
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A Mathematical Droodle
28 November 2015, Created with GeoGebra
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Copyright © 1996-2017 Alexander Bogomolny
28 November 2015, Created with GeoGebra
Let P lie on the circumcircle of triangle ABC. Let Q be the point on the circumcircle diametrically opposite P. Then the Simson lines of P and Q are perpendicular. As P changes, the points of intersection of the two simsons traces the 9-point circle of ABC.
If you change one of the vertices A, B, or C, the intersection point still traverses a circle.
Hubert Shutrick sent me the following proof. Let P_{a}, P_{b}, P_{c} and Q_{a}, Q_{b}, Q_{c} be the feet of the perpendiculars from P and Q onto BC, AC, AB.
In the diagram, angles CP_{a}P and CP_{b}P are right which makes quadrilateral CPP_{a}P_{b} cyclic. It follows that the inscribed angles CP_{a}P_{b} and PP_{a}P_{b} are equal. A similar argument applies to show that quadrilateral CQQ_{b}Q_{a} is also cyclic and that angles QCQ_{b} and QQ_{a}Q_{b} are equal.
Now, angles CPP_{b} and QCQ_{b} have perpendicular sides and are therefore equal. It follows that
∠CP_{a}P_{b} = ∠QQ_{a}Q_{b} |
But one pair of their sides, viz., CP_{a} and QQ_{A} are orthogonal which implies that this is also true for the second pair:
P_{a}P_{b} ⊥ Q_{a}Q_{b}. |
This proves that the simsons of P and Q are indeed orthogonal.
Let A'B'C' be the medial triangle of ΔABC. Since the perpendicular to BC at A' passes through the circumcenter of ABC, A' is also the midpoint of P_{a}Qa and the right angle gives that
∠MP_{c}C' = ∠MC'P_{c}. |
However,
∠A'B'C' = ∠ABC = θ + φ = ∠A'MC', |
so that M is on the circumcircle of ΔA'B'C', the nine point circle of ABC. Conversely, given a point M such that
Note
Hubert Shutrick found a generalization of this property to the case of two arbitrary points P and Q.
References
- F. G.-M., Exercise de Géométrie, Éditions Jacques Gabay, 1991, p. 330
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