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?


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.
