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Explanation

A 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 AU = AV.

First, as the two tangents from the same point to the same circle, BN = BM. So ΔBMN is isosceles. And, since ΔBMN is similar to ΔAMU, the latter is also isosceles:

 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 AU = AV.

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.