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Let ABC be a triangle with AB=AC and let D be the midpoint of AC. The angle bisector of \angle BAC intersects the circle through D,B and C at the point E inside the triangle ABC. The line BD intersects the circle through A,E and B in two points B and F. The lines AF and BE meet at a point I, and the lines CI and BD meet at a point K. Show that I is the incentre of triangle KAB.

Proposed by Jan Vonk, Belgium and Hojoo Lee, South Korea

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