Take a look at this

Let $N$ be the intersection of $CM$ and the perpendicular line through $A$ on $AM$ Define $J=CN\cap AB$ So, $(N,M;J,C)=-1$ So, taking the pencil from point $B$ we get, $BN||AM$ $N$ lies on the circle with diameter $AB$ Consequently, $BN||AM||CH$, where $H=NA\cap BM$ $H$ lies on the circle with diameter $CH$ Let $K$ be the second intersection of the two circles, which lies on $BC$ Then, $\angle{KFM}=\angle{KAN}$ And $\angle{KEM}=\angle{KAH}$ So, $\angle{KEM}=\angle{KFM}$ So, $KEMF$ is cyclic. Then simple angle chasing gives $\triangle MED$ and $\triangle MKB$ similar. So, $MK^2=ME.MB$ Similarly, $MK^2=MF.MC$ We are done.

Double BM to S, SG to J. Now BH/GS=AH/AG=MH/MG so B,S,M collinear so ABM=MSJ=EAJ so MD^2=ME.MB similar for F hence BEFC cyclic

Let $\omega_1 \cap \omega_2 \cap \overline{BC}=K$. Redefine $E$ as $E^*$ which is Humpty-Point WRT $\angle B$ in $\triangle ABD$. Let $H_B$ be the orthocenter of $\triangle ABD$. Let $X_B$ be the perpendicular foot from $B$ on $\overline{AD}$. Now, notice that \begin{align*} \angle AE^*K+\angle ACK &= \angle AE^*H_B+\angle H_BE^*K+\angle ACB=180^\circ-\angle ADH_B+\angle H_BBK+\angle C \\ &=90+\frac{\angle A}{2}+\angle X_BBD+\angle C=90+\frac{\angle A}{2}+90-\frac{\angle A}{2}-\angle C+\angle C=180^\circ \end{align*} which gives that $\angle AE^*C=\angle AKC=90^\circ$. So $E \equiv E^*$. Analogously, we have $F$ Humpty-Point of $\triangle ACD$ WRT $\angle C$. It's clear that $\angle MAE=\angle MBA$ which gives $MA^2=ME\cdot MB$. Similarly $MA^2=MF\cdot MC$. So $ME\cdot MB=MF\cdot MC,$ which finishes the problem. $\blacksquare$

See my solution on my Youtube channel in the link below. https://www.youtube.com/watch?v=zokW7q-ocgc

Let External Angle Bisector of $\angle BAC$ intersect $(AC)$, $(AB)$ at $X$ and $Y$. Let $BX$ intersect $AD$ at $R$. As, $(AB,AC;AR,AX)=-1$. Also, $CX//AD$, Let $BA$ intersect $CX$ at $T$. So, Projecting from $A$$(AB,AC;AR,AX)$ on $CX$ $\implies (C,T;X,\infty)=-1$. This means $BX$ bisects $AD$.So, $R=M$. $\angle MEA=\angle XEA=\angle ACX=\angle CAD=\angle BAD$. From here we get , $\triangle MAE \sim \triangle MBA$ . $\implies MA^2=MD^2=ME.MB$ , similarily, $\triangle MAF \sim \triangle MCA$, So, $MA^2=MC.MF$. $\implies ME.MB=MF.MC$. Thus, $BEFC$ is cyclic.

Beautiful problem! I just love the result! Me being such dumb nub, I fakesolved this initially and was jumping around until I realized that it was a fakesolve. This is why I gave up on synthetic tools and started applying stewart on this cuz I no longer had any motivation to continue... [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ /* A few re-additions are done using bubu-asy.py. This adds the dps, xmin, linewidth, fontsize and directions. */ pair A = (-57.30375,53.83617); pair B = (-83.34998,-39.17961); pair C = (43.19882,-39.61374); pair H = (-57.62315,-39.26787); pair D = (-31.07334,-39.35895); pair L = (-70.32686,7.32827); pair K = (-7.05246,7.11121); pair P = (-20.20788,-77.96332); pair X = (-22.03030,7.16259); pair Y = (61.56501,6.87581); pair M = (-44.18855,7.23860); pair I = (-40.28433,-6.63285); pair E = (-69.06976,-22.25320); pair F = (-23.36015,-3.92844); import graph; size(12cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); real xmin = -130, xmax = 85, ymin = -100, ymax = 85; draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(C--A, linewidth(0.5)); draw(circle((-19.96366,-6.77439), 71.18934), linewidth(0.5)); draw(circle(L, 48.29685), linewidth(0.5) + blue); draw(circle(K, 68.61788), linewidth(0.5) + blue); draw(circle(P, 74.10196), linewidth(0.5) + linetype("4 4") + red); draw(B--M, linewidth(0.5) + red); draw(M--C, linewidth(0.5) + red); draw(A--P, linewidth(0.5)); draw(H--A, linewidth(0.5)); draw(L--Y, linewidth(0.5)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$H$", H, dir(270)); dot("$D$", D, SE); dot("$L$", L, NW); dot("$K$", K, NE); dot("$P$", P, dir(270)); dot("$X$", X, NW); dot("$Y$", Y, NE); dot("$M$", M, NW); dot("$I$", I, NW); dot("$E$", E, NW); dot("$F$", F, NE); clip((xmin,ymin)--(xmax,ymin)--(xmax,ymax)--(xmin,ymax)--cycle); [/asy][/asy] Let $P$ be the midpoint of the minor arc $\widehat{BC}$ and $I$ be the incenter of $\triangle ABC$. Also, $H$ be the foot of the $A$-altitude. Let $X$ and $Y$ be the intersections of $A$-midline with $\odot(AHB)$ and $\odot(AHC)$ respectively such that $X$ and $Y$. Also, let $K$ and $L$ be the midpoints of $AB$ and $AC$ respectively. All the lengths used in this solution are directed. Firstly, we prove the following main claim. $\textbf{CLAIM: }$ The inversion $\mathbf{I}(\odot(M,MA))$ swaps $\odot(AHB)\leftrightarrow\odot(AHC)$. $\textbf{PROOF: }$ Firstly, apply a suitable scale such that $LK=1$. Also let $x=AL$ and $y=KA$. Now from the angle bisector condition, we get that $LM=\dfrac x{x+y}$ and $MK=\dfrac{y}{x+y}$ Now we multiply everything by a scale factor of $(x+y)$. So we finally have $AL=x(x+y)$, $KA=y(x+y)$, $LM=x$ and $MK=y$. Thus it suffices to prove that $MA^2=MX\cdot MY$. We now have, $MX=LX-LM=LA-LM=x(x+y)-x=x(x+y-1)$ and $MY=MK+KY=y+y(x+y)=y(x+y+1)$. So $MX\cdot MY=xy(x+y+1)(x+y-1)$. Finally by applying Stewart's Theorem on $\triangle ALK$, we get that $MA^2=xy(x+y+1)(x+y-1)$ and we are done. $\square$ Firstly, we have $PD\cdot PA=PB^2$. So the inversion $\mathbf{I}(\odot(P,PB^2)$ swaps $D\leftrightarrow A$. This gives us that $\left\{\odot(BIC),\odot(AD)\right\}$ are orthogonal. So the inversion $\mathbf{I}(\odot(AD))$ fixes $\odot(BIC)$. Now redefine $E=\odot(BIC)\cap\odot(ACH)$ and $F=\odot(BIC)\cap\odot(ABH)$. Now we have that the inversion $\mathbf{I}(\odot(M,MA))$ fixes $\odot(BIC)$. It also swaps $\odot(ABH)\leftrightarrow\odot(ACH)$. So the intersection $E=\odot(BIC)\cap\odot(ACH)\leftrightarrow\odot(BIC)\cap\odot(ABH)=B$ and similarly $F\leftrightarrow C$. So we get that $\overline{M-E-B}$ and $\overline{M-F-C}$ are collinear and we are done.