Let $P$ the symmetrical point of $A$ as far as $B$ is concerned. Obviously it is $BP=BA$. Let $c_1$ be the circumcircle of the triagle $BEF$ and $c_2$ be the circumcircle of the triangle $PDC$. The centers of these circles are $M$ and $M'$ respectively. Consider the homothetic transformation with center the point $A$ and ratio of $2$. The points $B, E, F$ "go" to $P, D, C$ respectively, so the circle $c_1$ "goes" to $c_2$. As a result $M$ "goes" to $M'$, so it is $AM=MM'$. Moreover the line $BM$ goes to $PM'$, so it is $BM//PM'$ (1). It is a well-known lemma that $BA^2=BD\cdot BC$. Due to the fact that $BA=BP$, we get that $BP^2=BD\cdot BC$. Consequetly $BP$ is tangent to the circle $c_2$, that is to say $BP\perp PM'$, therefore it is $PM'//AC$ (2). From (1) and (2) we get that $AC//BM$.

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As $AD,AC$ are antiparallels wrt $BA,BC$ and $E,F$ are both midpoints we get $E,F$ are isogonal wrt $\angle B$. Also by homothety $EF \parallel CD$ so: $$\angle EFB=\angle CBF=\angle EBA$$So $AB$ is tangent to circle $BEF$ hence $BM \perp AB \implies BM \parallel AC$

Let $K$ be the midpoint of $AB$. Since $AB$ tangent to $(AEF)$ we have $BK^2=AK^2=KE \cdot KF$ we deduce $BK$ tangent to $(BEF)$. So $BM\|AC$ because both are perpedicular to $AB$

Obviously, triangles $ABC$ and $ABD$ are similar, so $\frac{AB}{AD}=\frac{BC}{AC}$. Since $E, F$ are midpoints, $AE=\frac{AD}{2}$ and $CF=\frac{AC}{2}$, so: $\frac{AB}{AE}=\frac{BC}{CF} \implies AEB$ and $BCF$ are similar with $\angle ABE=\angle CBF, \angle BAE=\angle BCF, \angle AEB=\angle BFC$ $EF$ is midline, so it is parallel to $BC$ and $\angle CBF=\angle BFE=\angle ABE$ $\angle BFE=\frac{\angle BME}{2} -\angle BME$ is central angle. From isosceles triangle $BME: \angle BME=180^\circ-2\angle EBM$ and now $\angle BFE=\angle CBF=\angle ABE=90^\circ-\angle EBM$ $\angle ABM=\angle ABE+\angle EBM=90^\circ-\angle EBM+\angle EBM=90^\circ$. So, $BM \perp AB$ and $BM \parallel AC$

Coordinate bashing suffices. Let $A=(0,0),B=(b,0),C=(0,c).$ Then let the perpendicular bisector of $BF,$ or $y-\frac{c}{4}=\frac{2b}{c}(x-\frac{b}{2}),$ intersect $x=b$ at $M=(b,\frac{4b^2+c^2}{4c}).$ So the circle centered at $M$ through $B$ has equation $(x-b)^2+(y-\frac{4b^2+c^2}{4c})^2=(\frac{4b^2+c^2}{4c})^2.$ Now note $D=(\frac{bc^2}{b^2+c^2},\frac{b^2c}{b^2+c^2}),$ so $F=(\frac{bc^2}{2b^2+2c^2},\frac{b^2c}{2b^2+2c^2}).$ Now it's very easy to verify that $F$ lies on the circle as desired.