In the triangle ABC the angle bisector AD is drawn, E is the point of tangency of the inscribed circle to the side BC, I is the center of the inscribed circle ΔABC. The point A1 on the circumscribed circle ΔABC is such that AA1∣∣BC. Denote by T - the second point of intersection of the line EA1 and the circumscribed circle ΔAED. Prove that IT=IA. geometryequal segmentsincirclecircumcircle