Midpoints from Gergonne Triangle: What is this about?
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Copyright © 1996-2012 Alexander BogomolnyA problem has been posted at the Mathforum:
| The incircle of triangle ABC touches sides AB, BC, AC at M, N, K, respectively. The line through A parallel to KN meets MN at D. The line through A parallel to MN meets KN at E. Show that the line DE bisects sides AB and AC of triangle ABC. |
We recognize ΔKMN as Gergonne triangle of ΔABC. The argument used to explain the existence of Adams's Circle is very relevant for the above problem.
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Let the line through A parallel BC intersects MN in U and KN in V. We want to show that
First, as the two tangents from the same point to the same circle,
| AM = AU. |
Similarly,
| AK = AV. |
But since AK = AM (again as the two tangents from a point to a circle), we indeed obtain the required
In ΔNUV, A is the midpoint of side UV and lines AD and AE are parallel to the other two sides. Therefore, D is the midpoint of NU and E is the midpoint of NV. Since DE||BC||UV, any transversal of the three is bisected by DE. In particular, this is true of AB and AC.
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