Simsons of Diametrically Opposite Points
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Copyright © 1996-2018 Alexander Bogomolny
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 Pa, Pb, Pc and Qa, Qb, Qc be the feet of the perpendiculars from P and Q onto BC, AC, AB.
In the diagram, angles CPaP and CPbP are right which makes quadrilateral CPPaPb cyclic. It follows that the inscribed angles CPaPb and PPaPb are equal. A similar argument applies to show that quadrilateral CQQbQa is also cyclic and that angles QCQb and QQaQb are equal.
Now, angles CPPb and QCQb have perpendicular sides and are therefore equal. It follows that
∠CPaPb = ∠QQaQb
But one pair of their sides, viz., CPa and QQA are orthogonal which implies that this is also true for the second pair:
PaPb ⊥ QaQb.
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 PaQa and the right angle gives that
∠MPcC' = ∠MC'Pc.
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
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
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