Let the point $M -$ bisector middle $AD$ acute triangle $ABC$. A circle $\omega_1$ with a diameter of $AC$ intersects the segment $BM$ at point $E$, and a circle $\omega_2$ with a diameter of $AB$ intersects the segment $CM$ at point $F$. Prove that the points $B, E, F$ and $C$ lie on a circle.
My work so far:
Let $\omega_1 \cup \omega_2 = H$. Then $\angle AHB = \angle AHC = 90^{\circ} \Rightarrow AH \perp BC$ and $H \in BC$
Construction:- Produce $AE$ and $AF$ to cut $BC$ at $P$ and $Q$ respectively.
$X$ is the radical center of the 3 can-be-proven-circles ($EPHX$, $HQFX$ and $EAFX$). [Reference: http://paulscottinfo.ipage.com/PLC/4/4.html#anchor4 ]
By angles in the same segment, we have $\theta = \phi = \omega$.
Result follows.