Let $\omega_1$ be the circle centered at $E$ passes through $A$ and $B$. Similarly define $\omega_2$. We want to show $D$ has the same power wrt $\omega_1$ and $\omega_2$, and hence $EF\perp AD$. Let $f(\bullet)=P(\bullet, \omega_1)-P(\bullet, \omega_2)$. We know this function is linear over the plane to real number. It is sufficient to show $f(D)=0$. Let $BC$ intersects $\omega_1, \omega_2$ at $G, H$ respectively. So $\angle AGD=90^\circ-\angle GDE=\angle GAD$, which means $DG=DA$. Similarly we get $DH=DA$. Now \begin{align*} f(D)&=\frac{f(B)+f(C)}{2}\\ &=\frac{0-BC\cdot BH+CB\cdot CG-0}{2}\\ &=0 \end{align*}Which means $D$ lies on the radical axis of $\omega_1$ and $\omega_2$. And we are done.

I strongly dislike my proof, but I guess it's correct. Let $DE$ and $DF$ meet $AB$ and $AC$ at $U$ and $V$, respectively. It is well-known that $DU \cdot DE = DB \cdot DA$ and $DV \cdot DF = DC \cdot DA$ (this is essentially the basis for $\sqrt{bc}$ inversion.) Thus as $DB = DC$ we obtain that $EUVF$ is cyclic. Now note that as $\frac{BU}{UA} = \frac{DB}{DA} = \frac{DC}{CA} = \frac{CV}{VA}$, $UV$ and $BC$ are parallel. If $O$ is the circumcenter of triangle $ABC$, we have that $DO \perp UV$. Moreover, $UV$ and $EF$ are antiparallel in $\angle EDF$, so it suffices to demonstrate that $DA$ and $DO$ are isogonal in $\angle EDF$. We have $\angle ODC = 90^{\circ} - \angle HDC = 90^{\circ} - \frac{1}{2} \angle ADC = \angle EDA$, as $\angle EDF = \frac{1}{2}\angle BDC = 90^{\circ}$. Thus $DA$ and $DO$ are isogonal in $\angle EDF$, so we obtain $DA \perp EF$, as desired. Would love to see a less degenerate proof.

Let $O$ denote the center of $(ABC)$, and $O_b$ and $O_c$ denote the centers of $(ABM)$ and $(ACM)$. Then, we wish to show that $EF\parallel O_bO_c$. Note that $OO_bE$ are collinear on the perpendicular bisector of $AC$, and similarly with $OO_cF$. Note that $$\frac{O_bE}{O_CF}=\frac{R_{(ABM)}}{R_{(ACM)}}=\frac{c}{b}$$since $\sin\angle AMB=\sin\angle AMC$. Furthermore, let $O_b'$ be the reflection of $O$ across $O_b$. Note that $O_b'B\perp BC$ as $O_b$ was on the perpendicular bisector of $BM$. Thus, $$OO_b'=\frac{BM}{\sin\beta},$$so $$\frac{OO_b}{OO_c}=\frac{OO_b'}{OO_c'}=\frac{\frac{BM}{\sin\beta}}{\frac{BM}{\sin\gamma}}=\frac{c}{b},$$as desired.