Reflections of a Point on the Circumcircle
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Copyright © 1996-2017 Alexander Bogomolny
Reflections of a Point on the Circumcircle
The applet suggests the following theorem [Coxeter & Greitzer, p. 45, Honsberger, pp. 43-44]:
The reflections of a point on the circumcircle of a triangle in the side lines of the latter are collinear with the orthocenter of the triangle.
Let P be a point on the circumcenter of ΔABC. T_{a}
Note that the points T_{a}, T_{b}, T_{c} lie on the simson of P with respect to ΔABC. 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_{a}T_{b} is the simson line ΔABC. To prove the assertion, we shall show that HP_{b}||T_{a}T_{b}.
First, since PT_{a} is perpendicular to BC and PT_{b} is perpendicular to AC, the quadrilateral PT_{a}T_{b}C is cyclic. The inscribed angles T_{a}T_{b}P and T_{a}CP stand on the same arc T_{a}P and are, therefore equal:
(1) | ∠T_{a}T_{b}P = ∠T_{a}CP. |
HP_{b} is the reflection of K_{b}P in AC, while HK_{b} is perpendicular to AC. Therefore,
(2) | ∠BK_{b}P = ∠K_{b}HP_{b}. |
Angles BK_{b}P and BCP (= T_{a}CP) are equal as inscribed into the same circle and being subtended by the same arc BP:
(3) | ∠BK_{b}P = ∠BCP. |
The combination of (1)-(3) gives a crucial for the proof identity:
(4) | ∠T_{a}T_{b}P = ∠P_{b}HH_{b}. |
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_{a}T_{b}.
Corollary
The simson line of a point P on the circumcircle of a triangle bisects the segment PH joining the point to the orthocenter H.
References
- H.S.M. Coxeter, S.L. Greitzer, Geometry Revisited, MAA, 1967
- R. Honsberger, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, MAA, 1995.
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Copyright © 1996-2017 Alexander Bogomolny
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