Team Selection Test for JBMO

Question

In an equilateral triangle ABC, let D be a point on the side [BC] other than the vertices. Let I be the excenter of the triangle ABD opposite to the side [AB] and J be the excenter of the triangle ACD opposite to the side [AC]. Let the circumcircles of the triangles AIB and AJC intersect at the point E for the second time. Prove that A is the incenter of the triangle IEJ. FIGURE_0

Accepted Answer

First note that IBA = JCA = 60^. Then IEA = JEA = 60^ and IEJ = 120^. Therefore A is on the angle bisector of the angle IEJ. Next observe that IAD = 90^ + BAD/2 and JAD = 90^ + CAD/2. Hence IAJ = 360^ - 180^ - ( BAD + DAC)/2 = 150^ and therefore A is inside the triangle IEJ. Let X be the incenter of the triangle IEJ. We know that X is inside the triangle IEJ and IXJ = 90^ + IEJ/2 = 150^. Thus I, X, A, J are cyclic. Since E, X, A are collinear we get X = A as desired. FIGURE_0