## Simsons of Diametrically Opposite Points

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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 _{a}_{c}:

∠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 _{a} and Q_{a} whose distances from A' are A'M. From them we construct two diagonally opposite pairs P, Q and P', Q' on the circumcircle of ABC, where, in the diagram, the other candidate for P' for P would be above the line BC. The Simson lines of each pair intersect at right angles and pass through P_{a} and Q_{a} and so their points of intersection are the points of intersection of the circle with diameter P_{a}Q_{a} with the nine point circle. One of these must be the given M. We have ignored the interchange of P with Q that gives the same point M and the fact that the nine point circle is described twice as P goes round the circumcircle. Q(uite) E(asily) D(one)

### 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