, 2021

Solution:
The first solution is based on the main Lemma. We present this Lemma with two different proofs.
Lemma: Let $ABC$ be an acute triangle with $AB = BC$. Let $P$ be any point on $AC$. Line passing through $P$ perpendicular to $AB$, intersects ray $BC$ in point $T$. If the line $AT$ intersects the circumscribed circle of the triangle $ABC$ the second time at point $K$, then $\angle AKP = \angle ABP$.

Proof 1:
Let $H$ be the orthocenter of the triangle $ABP$. Then
$$\angle BHP = 180^\circ - \angle BAC = 180^\circ - \angle BCP$$
So $BHPC$ is cyclic. Then we get
$$TK \cdot TA = TC \cdot TB = TP \cdot TH$$
So, $AHPK$ is also cyclic. But then
$$\angle AKP = 180^\circ - \angle AHP = \angle ABP$$

Proof 2:
Consider the symmetric points $B'$ and $C'$ of $B$ and $C$, respectively, with respect to the line $PT$. It is clear that
$$TC' \cdot TB' = TC \cdot TB = TK \cdot TA$$
So $B'C'KA$ is cyclic. Also, because of the symmetry we have
$$\angle PC'B' = \angle PCB = \angle PAB$$
So $B'C'PA$ is also cyclic. Therefore, the points $B', C', K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have
$$\angle PKA = \angle PB'A = \angle PBA.$$

So, lemma is proved and now return to the problem.

Let $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M'$ and $N'$ the intersection points of the circumcircle of the triangle $HEF$ with the segments $EC$ and $FB$, respectively. We are going to show that $M = M'$ and $N = N'$ and it will prove the points $E, F, N, M$ lie on a common circle.
Of course, $A$ is an orthocenter of the triangle $BCH$. Therefore $\angle BHA = \angle BCA$, $\angle CHA = \angle CBA$ and $\angle HBA = \angle HCA$. Thus
$$\angle HEF = \angle HBA + \angle EAB = \angle HCA + \angle FAC = \angle HFE$$
So, the triangle $HEF$ is isosceles, $HE = HF$.
By using lemma, we get
$$\angle AM'E = \angle AHE = \angle ACB$$
and
$$\angle AN'F = \angle AHF = \angle ABC.$$

Therefore $M = M'$ and $N = N'$ and we are done.

Solution:
Let $X, Y$ be projections of $B$ on $AC$, and $C$ on $AB$, respectively. Let $\omega$ be circumcircle of $BXYC$. Let $Z$ be intersection of $EC$ and $\omega$ and $D$ be projection of $E$ on $BA$.
$$\angle MAC = \angle AME - \angle MCA = \angle XCB - \angle XCE = \angle ZCB = \angle ZXB$$
Since $BXYC$ is cyclic $\angle ACY = \angle XBA$, and since $DEXA$ is cyclic
$$\angle EXD = \angle EAD = \angle FAC$$
Therefore, we get that the quadrangles $BZXD$ and $CMAF$ are similar. Hence $\angle FMC = \angle DZB$. Since $ZEDB$ is cyclic,
$$\angle DZB = \angle DEB = \angle XAB.$$
Thus $\angle FMC = \angle XAB$. Similarly, $\angle ENB = \angle YAC$. We get that $\angle FMC = \angle ENB$ and it implies that the points $E, F, N, M$ lie on a circle.