Let $AE$ intersect $\omega$ again at $P$, we have $\angle{AEB_1}=90^{\circ}\Rightarrow \angle{B_1EP}=90^{\circ}$, so $P$ is the antipode of $B_1$ wrt. $\omega$. Similarly, $CF$ also pass through $P$. Hence it's enough to show that $P\in DM$ Let $X$ and $Y$ be midpoints of $AB_1$ and $B_1C$, respectively. It's clear that $XE$ and $YF$ are tangent to $\omega$. this gives us $\omega$ is the $D$-excircle of $\triangle{DXY}$. We also have $AM=\frac{AC}{2}=XB_1+B_1Y=XY\Rightarrow MY=AX=XB_1$. Since $B_1$ is the contact point of $D$-excircle and $XY$, $M$ is the contact point of the incircle of $\triangle{DXY}$ and $XY$. Then $D,M,P$ collinear follow from simple homothety.