Let $A'$ be the antipode of $A$ on $\odot(ABC)$, $G = \overline{DF} \cap \odot(ABC)$, $J = \overline{AH} \cap \odot(ABC)$, $I = \overline{JF} \cap \odot(ABC)$, $U = \overline{JF} \cap \overline{GA'}$, $V = \overline{GA'} \cap \overline{BC}$, $W = \overline{AM} \cap \odot(ABC)$. [asy][asy] size(9cm); defaultpen(fontsize(9pt)); pair A = dir(130), B = dir(210), C = dir(-30), AA = -A; pair H = A+B+C; pair M = (B+C)/2; pair d = (2*B*C/(B+C)), D = intersectionpoints(circumcircle(A, B, C), A--d)[1]; pair F = foot(D, B, C); pair G = intersectionpoints(circumcircle(A, B, C), D--(D+100*(F-D)))[0]; pair J = intersectionpoints(circumcircle(A, B, C), A--(H+100*(H-A)))[1]; pair W = intersectionpoints(circumcircle(A, B, C), A--(M+100*(M-A)))[1]; pair I = intersectionpoints(circumcircle(A, B, C), J--(J+100*(F-J)))[0]; pair U = extension(J, I, G, AA); pair V = extension(I, W, B, C); pair N = extension(H, F, A, M); pair R = 2*N - H; pair E = intersectionpoints(circumcircle(A, B, C), I--R)[1]; draw(A--B--C--cycle, red+linewidth(1)); draw(circumcircle(A, B, C), heavymagenta); draw(I--W^^I--R, dashed); draw(A--J^^A--D--G^^J--I^^G--R--H--AA^^A--W); string[] names = {"$A$", "$B$", "$C$", "$H$", "$J$", "$D$", "$M$", "$W$", "$F$", "$I$", "$G$", "$A'$", "$U$", "$V$", "$N$", "$R$", "$E$"}; pair[] points = {A, B, C, H, J, D, M, W, F, I, G, AA, U, V, N, R, E}; pair[] ll = {A, B, C, H, J, D, M, W, F, I, G, AA, U, V, N, R, E}; int pt = names.length; for (int i=0; i<pt; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Easy angle chasing gives us $\overline{GA'} \parallel \overline{AM}$ (For instance by noting that $\measuredangle DGA' = \measuredangle JAW$ due to isogonality of $(AD, AW)$ and $(AJ, AA')$ w.r.t. $\angle BAC$, and the fact that $GD \parallel AJ$). Moreover, since $M$ is the midpoint of $\overline{HA'}$, therefore $MN \parallel A'R$. Therefore, $R \in \overline{GA'}$. Claim 01 $(G, V; U, R)$ is harmonic. Proof. The pencil $F(D, C; J, H)$ is harmonic because $\overline{FD} \perp \overline{FC}$ and $\overline{FH}, \overline{FJ}$ are reflection of each other w.r.t. $\overline{FC}$ since $H, J$ are reflection of each other w.r.t. $BC$. So, $(G, V; U, R) \stackrel{F}{=} F(D, C; J, H) = -1$. $\blacksquare$ Claim 02 $(G, W; J, E)$ is harmonic. Proof. The pencil $D(G, W; J, E)$ is harmonic because $\overline{DG} \perp \overline{DW}$) and $\overline{DJ}, \overline{DE}$ are reflection of each other w.r.t. $\overline{DG}$ since $\angle JDG = \angle JA'G = 90^{\circ} - \angle DGA' = \angle GDE$, since $AM \parallel GA'$. So, $(G, W; J, E) = D(G, W; J, E) = -1$. $\blacksquare$ Claim 03 $I, V, W$ are collinear. Proof. Ratio Lemma Bash. (Denote $f(\star) = \pm \frac{B\star}{C\star}$, negative if $\star \in $ segment $\overline{BC}$ or below $\overline{BC}$, and positive otherwise) \begin{align*} f(V) = f(G)f(A') &= \frac{f(G)}{f(J)} \\ &= \frac{f(G)}{f(F)} \cdot \frac{f(F)}{f(J)} \\ &= \frac{1}{f(D)} \cdot f(I) \\ &= f(W) \cdot f(I), \end{align*}which means $V \in \overline{IW}$, as desired. $\blacksquare$ Now, by considering the projection of $I(G, W; J, E)$ on $\overline{GA'}$, we obtain that $I, E, R$ are collinear. Therefore: \[ 180^{\circ} - \angle AER = \angle AEI = \angle AJI = \angle FHJ = \angle RFD, \]as wanted. $\square$

@above very nice solution. Other way of proving $I-V-W$ collinearity is as the following: $GIFV$ is cyclic because $FV \parallel JA'$. Then $\angle FIV=\angle FGV=\angle JAW=\angle JIW \implies I-V-W$

Let $AM$ meet $\Gamma$ at $P$ and Let $D_a$ be reflection of $P$ across $M$. Let $DF$ meet $\Gamma$ at $S$. Let reflection of $D_a$ across $N$ be $D_a'$. Claim $: \angle HD_aN = \angle 90$. Proof $: CM^2 = AM.MP = AM.DM \implies \angle CD_aM = \angle C$ with same approach we have $\angle BD_aM = \angle B$ so $BD_aC = \angle 180 - \angle A = \angle BHC \implies BHD_aC$ is cyclic. $\angle HBC + \angle HD_aC = \angle 180 \implies \angle HD_aN = \angle 90$. Claim $: D_a$ is reflection of $D$ across $BC$. Proof $: \frac{CD_a}{AC} = \frac{D_aM}{CM} = \frac{D_aM}{BM} = \frac{BD_a}{AB}$ and $\frac{BD}{AB} = \frac{CD}{AC}$ so $D_a$ is reflection of $D$ across $BC$. Claim $: REDD_a'$ is cyclic. Proof $:$ Let $H'$ be reflection of $H$ across $D_a$. $\angle H'ED = \angle AH'D_a = \angle AHD_a = \angle 90 - \angle HAD_a = \angle 90 - \angle DAN = \angle D_aDE$ so $D_aH'ED$ is isosceles trapezoid. Note that $D_aH' = D_aH = PD_a'$ and $D_aH'RD_a'$ is rectangle so $DERD_a'$ is cyclic. Note that for proving $\angle AED + \angle DER + \angle DFR = \angle 180$ we must prove $\angle D_aDD_a' = \angle AED + \angle DER$ cause $FN || DD_a'$ and by last claim we have $\angle D_a'DE = \angle DER$ so we need to prove $\angle EDD_a = \angle AED$ or $AS || DE$ which is true cause it's well known that $HD_a || AS$ and $DE || H'D_a || HD_a$. we're Done By the way does anyone know what's $D_a$'s original name in English?

Mahdi_Mashayekhi wrote: we're Done By the way does anyone know what's $D_a$'s original name in English? I think it’s the A-Humpty Point w.r.t. to the triangle ABC, some also call it the HM point.

I'll borrow Seicchi28's diagram It's enough to show that $\angle AER = \angle DFH$. Let $A'$ be the anti-pod of $A$ wrt $(ABC)$. Since $M$ was the midpoint of $HR$ and $M$ is the midpoint of $HA'$ , we have: $A'R||AM$ and by easy angle chasing , $AR,DF$ intersects on $(ABC)$ ; name that intersection $G$. Note that $\angle AEA' =90$ so it's enough to show that $\angle RFD=\angle A'ER$ since $FJ,FH$ are symmetric wrt $FD$.So since $AD,AM$ were isogonal wrt $\angle BAC$ , it's enough to show that $RE,FJ$ are concurrent on $(ABC)$. Redefine $JF \cap (ABC) = I$ and $IE \cap A'G = R'$ Now , by Pascal on $EGJA'ID$ , we have :$\overline{F,R',A'J \cap ED=P}$ are collinear. Now reflect $D$ about $BC$ to get $X_A$ , the $A$-Humpty point. It's well-known that $\angle HX_AM = 90$.So if $P'$ is the reflection of $H$ wrt $F$ , $P' \in DE$ and also $P'A'||BC$. So $P'=P$ and $H,F,P$ would be collinear so $R=R'$ and we're done.