[alogo] Three collinear points

Consider a triangle t = ABC and its circumcircle c. Draw line A*C* parallel to AC, intersecting the circle at DF. Draw also a parallel to AC from B intersecting the circle at K. Let further H be the intersection point of the diagonals of the quadrilateral q = ACA*C*. Then K, H and the middle G of EF are on a line.

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The proof is an application of the basic figure for the harmonic ratio (see Harmonic.html ). In fact, considering the complete quadrilateral ABCH we see that (KHIG)=-1, where I is the intersection point of KH with AC and G is the intersection point of KH with DF. If follows that also (LHAA*)=-1. If follows that the bundle of lines at K: (KL, KH, KA, KA*) is a harmonic one. Thus, the parallel A*C* to the ray KL of this bundle determines equal segments by the other three: MG = GA*. By the symmetry of the figure MD = FA*, hence G is the middle of DF.

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