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4 points are concyclic [X, F, G, H]

Source: Japan Mathematical Olympiad Finals 2005, Problem 4

April 16, 2005
geometrycircumcircleperpendicular bisector

Problem Statement

Given two points AA and BB on a circle Γ\varGamma. Let the tangents to this circle Γ\varGamma at the points AA and BB meet at a point XX. Let CC and DD be two points on the circle Γ\varGamma such that the points CC, DD, XX are collinear in this order and such that the lines CACA and BDBD are perpendicular. Let the line CACA intersect the line BDBD at a point FF. Let the line CDCD intersect the line ABAB at a point GG. Let HH be the point of intersection of the segment BDBD and the perpendicular bisector of the segment GXGX. Prove that the four points XX, FF, GG, HH lie on one circle.
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