# Reflections of the Orthocenter

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Copyright © 1996-2018 Alexander Bogomolny

### Reflections of the Orthocenter

The applet suggests the following theorem [F. G.-M., p. 142; Honsberger, pp. 17-18; Lalesco, 1.11] from triangle geometry:

The reflections of the orthocenter of a triangle in the side lines of the latter lie on its circumcircle.

Let H be the orthocenter of ΔABC with an altitude AH_{a}. Denote the second point of intersection of AH_{a} with the circumcircle as K_{a}. We wish to prove that

(1)

HH_{a} = H_{a}K_{a}.

In the following I assume that ΔABC is *acute*. If one of its angles is obtuse, the derivation is just a little different.

Let BH_{b} be another altitude of ΔABC. Since the right triangles CAH_{a} and CBH_{b} share angle C, their other acute angles (CAH_{a} and CBH_{b}) are equal:

(2)

∠CAH_{a} = ∠CBH_{b}.

On the other hand, the inscribed angles CAK_{a} and CBK_{a} are subtended by the same arc CK_{a}. Therefore,

(3)

∠CAH_{a} = ∠CBK_{a}.

From (2) and (3),

(4)

∠CBH_{b} = ∠CBK_{a}.

Thus BH_{a} is simultaneously an altitude and an angle bisector in ΔBHK_{a}. It's then also a median from B to HK_{a}, which immediately implies (1).

The circumcevian triangle of the orthocenter is known as the *circum-orthic* triangle.

For the sake of future references, let's observe that the points A, B, C, and K_{a}, K_{b}, and K_{c} divide the circumcircle into 3 pairs of arcs, the two arcs in a pair around any of the vertices A, B, or C being equal. The angular measure of the arcs next to A is twice

The circle through H, B, and C is the reflection of the circumcircle in BC and thus has the same radius. In addition, the reflection S of the vertex A lies on that circle: _{a} = H_{a}S.

The circumcircles of triangles ABC, ABH, BCH, and CAH are all equal.

The circumcircles of triangles AK_{b}K_{c}, BK_{c}K_{a}, CK_{a}K_{b} naturally coincide with the circumcircle of ΔABC. In a more general case where points K_{a}, K_{b}, K_{c} are the reflections of an arbitrary point in the side lines of ΔABC, the circumcircles of triangles AK_{b}K_{c}, BK_{c}K_{a}, CK_{a}K_{b} are concurrent and meet on the circumcircle of ΔABC.

**Note**: a different proof appears elsewhere.

### References

- F. G.-M.,
*Exercices de Géométrie*, Éditions Jacques Gabay, sixiéme édition, 1991 - R. Honsberger,
*Episodes in Nineteenth and Twentieth Century Euclidean Geometry*, MAA, 1995. - T. Lalesco,
*La Géométrie du Triangle*, Éditions Jacques Gabay, sixiéme édition, 1987

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny