Prove that $AM = MF$ where $M$ is the intersection between the perpendicular of circle $ω$ and line $AF$ 
Given is circle $ω$ and line $k$ disjoint with circle $ω$. Point $M$ is
the foot of the perpendicular from the center of the circle onto line $k$. A line
intersects line $k$ at point $A$ and circle $ω$ at points $B$ and $C$. Lines $\overline{BM}$ and
$\overline{CM}$ intersect circle $ω$ for the second time at points $D$ and $E$. Line $\overline{DE}$ meets
line $k$ at point $F$. Prove that $\overline{AM} = \overline{MF}$.

Once again, I've been assigned another set of geometry problems, and this is one of the harder ones that I haven't been able to figure out yet. I've asked my instructor what I could use for this problem, and he told me that Power of a Point would be useful here. But so far, I've tried applying Power of a Point, but I'm not getting anywhere with the few points that currently exist. Can anyone help out?
 A: 
Menelaus for triangle $AFP$ and transversal $M-E-C$: $${AM\over MF}\cdot {EF\over EP}\cdot {CP\over CA}=1$$ and Menelaus for triangle $AFP$ and transversal $M-B-D$: $${AM\over MF}\cdot {DF\over DP}\cdot {BP\over BA}=1$$
IF we multiply these equation we get:
$$\Big({AM\over MF}\Big)^2\cdot {DF\cdot EF\over \color{red}{EP\cdot DP}}\cdot {\color{red}{CP\cdot BP}\over CA\cdot BA}=1$$
By the power of the point $P$ with respect to circle $\omega $ we have:
$$\color{red}{ BP\cdot CP  = DP\cdot EP}$$ so we have
$$\Big({AM\over MF}\Big)^2\cdot {DF\cdot EF\over CA\cdot BA}=1$$
By the power of the points $A$ and $F$ we have: $$CA \cdot BA = AO^2-r^2 = x^2+d^2-r^2$$
and $$DF\cdot EF = FO^2-r^2 = y^2+d^2-r^2$$ so we have now $${x^2\over y^2}\cdot {y^2+d^2-r^2\over x^2+d^2-r^2} = 1\implies x^2= y^2$$
and we are done.
A: Indeed, this is just one of generalizations of the Butterfly theorem (see my comment above).
Consider point $B'$ and $C'$ such that $M$ is the midpoint of the segments $BB'$ and $CC'$. Clearly, $BCB'C'$ is a parallelogram with center at $M$. Thus, in order to prove that $AM=MF$ (that is, $A$ and $F$ are symmetric with respect to $M$), we only need to prove that $B'C'$ passes through $F$ (since $A\in BC$).
Firstly, it's easy to see that $B'C'DE$ is a cyclic quadrilateral (because $BC\parallel B'C'$ and $BDCE$ is cyclic).
Secondly, denote the circle symmetric to $\omega$ with respect to $M$ as $\omega'$. Then, $B',C'\in\omega'$ and $k$ is the radical axis of $\omega$ and $\omega'$ (due to the symmetry; note that $OM\perp k$). As we have proved before, the quadrilateral $B'C'DE$ is cyclic. Thus, the radical axes of these three circles are concurrent, i. e. lines $DE$, $k$ and $B'C'$ are concurrent.
Finally, recall that $F=DE\cap k$, so $F\in B'C'$, as desired.
A: COMMENT.-  The asked property is verified with any good function plotter taking arbitrary values and a suitable proof is geometric in nature. On the other hand, an analytical proof involves the consideration of seven unknowns (six by the coordinates of points C, D and E and one by the first coordinate of point F in the attached figure). We want to do it but this is a challenge that without trickery like the one we will use can be too annoying.

$\hspace{9 cm}$****************
Line $ABC\hspace{1cm}\dfrac{y-d}{x-a}=\dfrac{d-b_2}{a-b_1}\iff y=\dfrac{d-b_2}{a-b_1}x+\dfrac{ab_2-db_1}{a-b_1}=mx+h$ where $m=\dfrac{d-b_2}{a-b_1}$
Line $MBD\hspace{1cm}\dfrac{y-d}{x}=\dfrac{d-b_2}{-b_1}\iff y=\dfrac{b_2-d}{b_1}x+d=m_1x+d$ where $m_1=\dfrac{b_2-d}{b_1}$
Line $MEC\hspace{1cm}\dfrac{y-d}{x}=\dfrac{d-y_1}{-x_1}\iff y=\dfrac{y_1-d}{x_1}x+d=m_2x+d$ where $m_2=\dfrac{y_1-d}{x_1}$
►Point $C=(x_1,y_1)\in w$ in line $ABC$
$x_1^2+y_1^2=r^2\space\space$ then $\begin{cases}(m^2+1)x_1^2+2hmx_1+(h^2-r^2)=0\\(m^2+1)y_1^2-2hy_1+(h^2-m^2r^2)=0\end{cases}$
It follows (Vieta) $b_1x_1=\dfrac{h^2-r^2}{m^2+1}$ and $b_2y_1=\dfrac{h^2-m^2r^2}{m^2+1}$.
Similarly we have
►Point $D=(x_2,y_2)\in w$ in line $MBD$
$b_1x_2=\dfrac{d^2-r^2}{m_1^2+1}$ and $b_2y_2=\dfrac{d^2-m_1^2r^2}{m_1^2+1}$
►Point $E=(x_3,y_3)\in w$ in line $MEC$
$x_1x_3=\dfrac{d^2-r^2}{m_2^2+1}$ and $y_1y_3=\dfrac{d^2-m_2^2r^2}{m_2^2+1}$
$\hspace{9 cm}$****************
Now line $DEF$ is given by $\dfrac{y-y_2}{x-x_2}=\dfrac{y_3-y_2}{x_3-x_2}$ and because point $F$ belongs to this line one has
$\dfrac{d-y_2}{\alpha-x_2}=\dfrac{y_3-y_2}{x_3-x_2}$ so making $\alpha=-a\iff\overline{AM}=\overline{MF}$, after substitutions we obtain an equality $$f(a,d,b_1,b_2)=0$$ whose verification completes the solution. We leave this exercise for whoever wants to do it.
