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Two circles \Omega_{1} and \Omega_{2} are externally tangent to each other at a point I, and both of these circles are tangent to a third circle \Omega which encloses the two circles \Omega_{1} and \Omega_{2}.
The common tangent to the two circles \Omega_{1} and \Omega_{2} at the point I meets the circle \Omega at a point A. One common tangent to the circles \Omega_{1} and \Omega_{2} which doesn't pass through I meets the circle \Omega at the points B and C such that the points A and I lie on the same side of the line BC.
Prove that the point I is the incenter of triangle ABC.

Alternative formulation. Two circles touch externally at a point I. The two circles lie inside a large circle and both touch it. The chord BC of the large circle touches both smaller circles (not at I). The common tangent to the two smaller circles at the point I meets the large circle at a point A, where the points A and I are on the same side of the chord BC. Show that the point I is the incenter of triangle ABC.

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