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.


Figure 1

    Proof: The center of circle Γ lies on the intersection of the circumcircle and the perpendicular bisector of side AB . Indeed, let Ic be that point, ∠AIcB + γ = 180° (ACBIc is a cyclic quadrilateral) or ∠AIcB = 180° - γ = α + β + γ - γ = α + β. Now, ∠IAB = α/2 and ∠IBA = β/2, so the central angle of arc AIB = 2( α/2 + β/2) = α + β = ∠AIcB, so point Ic is really the sought center.
    Further, we have that ∠AOIc = ∠BOIc, since point O lies on the perpendicular bisector of side AB, so ∠ACIc = ∠BCIc as the inscribed angles of ∠AOIc and ∠BOIc. In conclusion: Point Ic lies on the bisector of the angle at point C.
    Since line CIc passing through the center of circle Γ and ∠ACIc = ∠ABIc, according to the line reflection property, we conclude that AX = BY.

There are five solutions now: