In ABC, the points A', B', C' of tangency of the incircle with the sides ABC form the intouch A'B'C'. From the midpoints Am, Bm, Cm of the sides of the latter drop perpendiculars AmAh, BmBAh, CmCAh on the opposite sides of ABC. The three lines are AmAh, BmBAh, CmCAh. The proof is remarkably short and reminds one of that employed to prove the concurrency of maltitudes in an inscribed quadrilateral. Recollect that the medians in a triangle are divided in ratio 1:2 by the center of the triangle. We apply this fact to the intouch triangle A'B'C'. (I'll use G' to denote its center, although the point is not named in the applet.) This means that points A', B', C' can be obtained from points Am, Bm, Cm by a homothety with coefficient -2 and center G'. A homothety maps straight lines onto parallel lines. Therefore, say, A'Am is mapped onto a line through A' parallel to A'Am and thus perpendicular to BC. Since BC is tangent to the incircle of ABC, that line passes through its incenter I. The same holds of the images of B'Bm and C'Cm. The three images therefore concur (in the incenter I of ABC.) Since homothety is a reversible transformation, the lines themselves are also concurrent. Furthermore, the point of concurrency S and the incenter I are collinear with G' and 2·SG' = G'I.

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