Estonian Mathematical Olympiad

Question

Let D be a point on the side BC of an acute triangle ABC and let E be a point on the line segment AD. Let F and G be the feet of the altitudes drawn from the vertex D in triangles ABD and ACD, respectively. The line BE intersects the circumcircle of the triangle DEG at point H E. Prove that points B, F, G and H are concyclic. FIGURE_0 FIGURE_1 FIGURE_2

Accepted Answer

Since AFD = AGD = 90^, the quadrilateral AFDG is cyclic (Fig. 28). Using inscribed angles in the circumcircle of the quadrilateral AFDG, we get BFG = 180^ - AFG = 180^ - ADG = 180^ - EDG. FIGURE_0 Fig. 28 FIGURE_1 Fig. 29 FIGURE_2 Fig. 31 BHG = EHG = EDG. Thus BFG = 180^ - BHG, so the points B, F, G and H are concyclic. * If the order of the points is D, E, H, G (Fig. 30), then opposite angles of the cyclic quadrilateral DEHG give BHG = EHG = 180^ - EDG. So BFG = BHG, hence B, F, H and G are concyclic. * If the order of the points is D, H, E, G (Fig. 31), then we similarly get BHG = 180^ - EHG = 180^ - EDG. So again BFG = BHG, hence B, F, H and G are concyclic.