The incircle of ABC touches sides at I and J. D, E, F are the bases of altitudes from C, B, and A. The incenters of BDF of CEF are N and M, respectively. Show that JIMN is a parallelogram and that IM is perpendicular to BC.
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Solution
Since DEF is the orthic triangle, the triangle EFG is similar to BAC with a ratio of cos C. Then IM is perpendicular to BC since IYC is a right triangle. Also, MC is a bisector, so IM = r, the inradius of ABC. A similar argument for JN shows IM and JN are equal and parallel.