Three Circles through the Incenter

Let I be the incenter of a triangle ABC in which AC ≠ BC. Let Γ be the circle passing through A, I and B. Suppose Γ intersects the line AC at A and X and intersects the line BC at B and Y . Show that AX = BY.

Let's complicate the picture by drawing two other circles through the incenter and the remaining pairs of vertices. Let's draw the internal and external angle bisectors of ΔABC. We do not need that many extra elements for the proof, but having them all together may make the argument more transparent.

The internal bisector of angle C and the two external bisectors of angles at A and B intersect at the excenter I_c of ΔABC. The internal and external bisectors at any vertex are perpendicular, implying, in particular, that BI⊥BI_c and AI⊥AI_c. Thus we have two right triangles IAI_c and IBI_c. Put together they form a cyclic quadrilateral IAI_cB. The circumcircle of IAI_cB passes through the points A, B and I and is, therefore, the circle stipulated by the problem.

Since triangles IAI_c and IBI_c are right, their shared hypotenuse II_c is a diameter of the circle at hand. It follows that the center of the circle lies on the (internal) bisector of angle at C. This means that the whole circle is symmetric in the bisector as are the side lines AC and BC. We conclude that the circle intercepts the two side lines in two mutual reflections in the bisector. But this are AX and BY, with A corresponding to Y and B to X under the reflection. The two segments are therefore equal.

There are five solutions in all: