** 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.**

Proof: The center of circle Γ lies on the intersection of the circumcircle and the perpendicular bisector of side AB . Indeed, let I_{c} be that point, ∠AI_{c}B + γ = 180° (ACBI_{c} is a cyclic quadrilateral) or ∠AI_{c}B = 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 = ∠AI_{c}B, so point I_{c} is really the sought center.

Further, we have that ∠AOI_{c} = ∠BOI_{c}, since point O lies on the perpendicular bisector of side AB, so ∠ACI_{c} = ∠BCI_{c} as the inscribed angles of ∠AOI_{c} and ∠BOI_{c}. In conclusion: Point I_{c} lies on the bisector of the angle at point C.

Since line CI_{c} passing through the center of circle Γ and ∠ACI_{c} = ∠ABI_{c}, according to the line reflection property, we conclude that AX = BY.

There are five solutions now: