Given an acute triangle DFG, let A, B, and C denote the feet of the altitudes of DFG through D, F, and G. Prove that AD, BF, and CG bisect the angles of the triangle ABC.
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It should be $ABC$ in the second line . Yeah this is a well known thig that the othocenter of the main triangle is the incenter of the orthic triangle. This is because let H be the othocenter of DFG. Then we see that $ \angle DFH = \angle DGH $. Now $ HCFA , HBGA$ are cyclic . Thus $ \angle CAH = \angle BAH $. Hence proved.
Chirantan Chowdhury
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