Bisectal Circle

Besides the definitive ∠AOC = ∠COB, the angle bisector OC of an angle AOB, has other characteristic properties. For example, the sides OA and OB of the angle are symmetric in the angle bisector OC. As another example, circles centered on OC are isogonal with respect to the rays OA and OB, i.e., they form equal angles with the sides of the angle. This latter property admits a different formulation: all circles orthogonal to the angle bisector, are isogonal with respect to the sides of the angle.

Applying inversion in a point different from the vertex of the angle transforms the sides of the angle into two intersecting circles (say, (O1) and (O2) centered at O1 and O2, respectively) and the angle bisector into a circle through their points of intersection. The latter circle plays a role similar to that of the angle bisector. Let's denote it (Q), where Q, its center, may be either finite or infinite. It's infinite if (O1) and (O2) have the same radii; it is finite otherwise. In the latter case, Q is the center of homothety of (O1) and (O2), while (Q) is the circle of inversion with respect to which (O1) and (O2) are symmetric. Further, all circles orthogonal to (Q) are fixed under this inversion. The ones that intersect (O1) and (O2) are isogonal with respect to the latter: they intersect (O1) and (O2) under equal angles.

In order to possess this property the circles need not intersect. Any two circles (O1) and (O2), not within one another, have an external center of homothety, which also serves as the center of inversion that maps one circle on the other. A circle is orthogonal to the circle of inversion (Q) iff it is isogonal with respect to (O1) and (O2).


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The French terms for (Q) is cercle bissecteur ([Hadamard, #321]), for which bisectal circle should be an adequate translation, although I must admit to have never seen this term before.

Two intersecting straight lines form four angles with two angle bisectors. Two circles, too, can be inverse images of each other in two ways. Thus, in general, two circles possess a pair of bisectal circles, centered at the external and internal homothety centers of the two given circles. The power of inversion in the latter case is negative. (The applet above only presents the "external" case.)

Let Si on (Oi), i = 1, 2, be antihomologous with respect to the inversion in (Q). Any circle (S) through S1, S2 is fixed under this inversion because the power of Q with respect to (S) is exactly the power of inversion in (Q). It is then orthogonal to (Q) and, since inversion preserves angles, forms equal angles with (O1) and (O2).

The argument is reversible. If a circle (S) cuts (O1) and (O2) under equal angles, then the four points of intersection can be split into two pairs of antihomologous points with respect to the external center of homothety Q. The points in each pair are inverse images of each other under in version in (Q), thus making (S) fixed under this inversion. It is then also orthogonal to (Q).

The argument holds even when (S) is tangent to (O1) and (O2). But, additionally, in this case we may note that the points of contact of (S) with (O1) and (O2) serve as the (internal) centers of homothety of (S) with (O1) and (O2). Which gives an extra reason for these two points to be collinear with the external center of homothety S of (O1) and (O2). This is now a consequence of Monge's Three Circles Theorem.

References

  1. J. Hadamard, Leçons de géométrie élémentaire, tome I, 13e édition, reprint 1988, #227-228
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