, 2022

Let $X$ be the intersection point of $AM$ and $\Gamma$ other than $A$, then
$$\angle XDC = \angle XAC = \angle MAC = \angle BAD = \angle BCD,$$
that is, $DX \parallel BC$. Since $AD, AX$ are isogonal lines with respect to $\angle BAC$, $BDXC$ is an isosceles trapezoid, hence the reflection $D'$ of $D$ with respect to $BC$ is the reflection of $X$ with respect to $M$. Let $A^*$ be the antipode of $A$ with respect to $\Gamma$, and let $Y$ be the midpoint of $\overline{HE}$, then $M$ is the midpoint of $\overline{HA}^*$, so $\triangle A^*ER$ is the image of $\triangle MYN$ under a homothety centered at $H$ with ratio $2$. Since $AX$ is perpendicular to $DE$, and $AX, AA^*$ are isogonal lines with respect to $\angle DAE$, $DXA^*E$ is an isosceles trapezoid. Note that (mod $360^\circ$)
$$\angle FMA = \angle DXA = 90^\circ - \angle EDX = 90^\circ - \angle A^*ED = \angle (XA, A^*E) = \angle AMY,$$
so combined with $\overline{FM} = \frac{1}{2} \cdot \overline{DX} = \frac{1}{2} \cdot \overline{A^*E} = \overline{MY}$, we obtain that $Y$ is the reflection of $F$ with respect to $AM$.
Since $\angle MAE = \angle DAA^* = \angle HAM$, $AE$ is the reflection of line $AH$ with respect to $AM$. Therefore
$$180^\circ - \angle DFR = \angle (NF, HA) = \angle (AE, YN) = \angle AER.$$