We will prove for one condition, as the other can be 'flipped'. Suppose $\angle BAC = 90^{\circ}$. Let $I$ be the incenter. It suffices to prove for one side (say, $B$ and $N$) by symmetry. Now let $N'=BI \cap EF$. It's well known that (I forgot the lemma name) $\angle BN'C = 90^{\circ}$. Hence ${B,A,N',C}$ is concyclic. This implies $N'=BI \cap (ABC)$ is indeed the arc midpoint.

Part I : If $\angle BAC =90^{\circ}$, then $M,F,E,N$ are collinear. It is well know that $B,I,N$ and $C,I,M$ are collinear. Observe that $B,M,F,I,D$ and $C,N,E,I,D$ are concyclic. Thus, $$\measuredangle BMF = \measuredangle BIF = \measuredangle DIB = \measuredangle DIN = \measuredangle DCN = \measuredangle BCN = \measuredangle BMN$$So $M,F,N$ are collinear, analogously $N,E,M$ are also collinear which yields the desired conclusion. Part II : If $M,F,E,N$ are collinear, then $\angle BAC= 90^{\circ}$. By some angle chasing with concyclicity of $B,M,A,N,C$, we get that $MN$ is the angle bisector of $\angle AMI$ and $\angle ANI$. Thus $A$ is the reflection of $I$ over $MN$. Since $M,F,E,N$ are collinear, then $\angle FAE = \angle FIE$. But $\angle AFI = \angle AEI = 90^{\circ}$. Hence $\angle BAC = \angle FAE =90^{\circ}$, as desired.

excuse me, what do you mean by "acute triangle "?

By the Iran lemma, $IB, IC$ intersect $EF$ on $(BC)$, which immediately implies the problem.

gnoka wrote: excuse me, what do you mean by "acute triangle "? All the angles are less than $90^\circ$.

KI_HG wrote: gnoka wrote: excuse me, what do you mean by "acute triangle "? All the angles are less than $90^\circ$. Thanks,that's what I used to know. But I get confused when I finally find the ACUTE triangle ABC is a Right triangle

gnoka wrote: KI_HG wrote: gnoka wrote: excuse me, what do you mean by "acute triangle "? All the angles are less than $90^\circ$. Thanks,that's what I used to know. But I get confused when I finally find the ACUTE triangle ABC is a Right triangle Well don't ask me about that, it is from the problem itself