Reflections of a Point on the Circumcircle: What Is This About?
A Mathematical Droodle

Explanation

Copyright © 1996-2009 Alexander Bogomolny

The applet suggests the following theorem [Coxeter & Greitzer, p. 45, Honsberger, pp. 43-44]:

Let P be a point on the circumcenter of ABC. T_a (T_b, T_c) is the foot of the perpendicular from P to BC and P_a (P_b, P_c) is the reflection of P in BC (AC, AB). We wish to prove that the four points P_a, P_b, P_c, and H are collinear.

Note that the points T_a, T_b, T_c lie on the simson of P with respect to ABC. Also, the points P_a, P_b, P_c are the images of points T_a, T_b, T_c under the homothety with center P and coefficient 2. Thus we immediately see that the points P_a, P_b, P_c are collinear. It remains to be shown that the orthocenter H lies on the same straight line.

In the diagram, T_aT_b is the simson line ABC. To prove the assertion, we shall show that HP_b||T_aT_b.

First, since PT_a is perpendicular to BC and PT_b is perpendicular to AC, the quadrilateral PT_aT_bC is cyclic. The inscribed angles T_aT_bP and T_aCP stand on the same arc T_aP and are, therefore equal:

(1)	TaTbP = TaCP.

HP_b is the reflection of K_bP in AC, while HK_b is perpendicular to AC. Therefore,

(2)	BKbP = KbHPb.

Angles BK_bP and BCP (= T_aCP) are equal as inscribed into the same circle and being subtended by the same arc BP:

(3)	BKbP = BCP.

The combination of (1)-(3) gives a crucial for the proof identity:

(4)	TaTbP = PbHHb.

Since one pair of the sides of these angles (PT_b and HH_b are both perpendicular to AC) are parallel, the same holds for the other pair of the sides: HP_b||T_aT_b.

1. H.S.M. Coxeter, S.L. Greitzer, Geometry Revisited, MAA, 1967
2. R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.

Copyright © 1996-2009 Alexander Bogomolny