My solution: Let $ S=AI \cap BC $ and $ T=AI \cap EF $ be the midpoint of $ EF . $ From $ A(E,F;N,T)=\frac{EN}{FN}=\frac{DE}{DF}=\frac{CI}{BI} \cdot \frac{\sin \angle C }{\sin \angle B}=\frac{CM}{BM} \cdot \frac{BS}{CS}=A(C,B;M,S) \Longrightarrow A, M, N $ are collinear $ . $ Q.E.D

My solution: let $ AN$ intersect $ BC $ at $ R $ we must prove that $BR/CR=BI/CI$ note that $ BR/CR=\sin BAR/ \sin CAR \cdot c/b =FN/NE \cdot c/b=DF/DE \cdot c/b=\cos \frac{B}{2}/\cos \frac {C}{2} \cdot \sin C/\sin B=\sin \frac {C}{2}/\sin \frac{B}{2}$ so $ R=M $ DONE

Dear Mathlinkers, see also http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=24699 Sincerely Jean-Louis

Lemma: In $\Delta ABC $ with $O$ as circumcenter and $K $ as intersection of tangents to $\odot (ABC) $ at $B,C $. Let $OK $ $\cap $ $\odot (ABC) $ $=$ $M_A, M_{BC} $ with $KM_A < KM_{BC} $ . Let $AM_A $ $\cap $ $BC $ $=$ $L $. Let tangents to $\odot (ABC) $ at $A, M_{BC} $ intersect at $X $. Then, $X,L,K $ collinear Proof: Let $\odot (ABC) \cap AK $ $=$ $K_A$, then, $L $ $\in $ $K_AM_{BC} $. Now Apply Pascal on $AAM_AM_{BC}M_{BC}K_A $ $\qquad \blacksquare $ Now apply the Lemma on $\Delta DEF $ and conclude