Problem: Let $\mathcal C_1$ and $\mathcal C_2$ be concentric circles, with
$\mathcal C_2$ in the interior of $\mathcal C_1$.
From a point $A$ on $\mathcal C_1$ one draws the tangent $AB$
to $\mathcal C_2$ ($B\in \mathcal C_2$).
Let $C$ be the second point of intersection of ray $AB$ and
$\mathcal C_1$, and let $D$ be the midpoint of $\ol{AB}$.
A line passing through $A$ intersects $\mathcal C_2$ at $E$ and $F$
in such a way that the perpendicular bisectors of $\ol{DE}$ and $\ol{CF}$
intersect at a point $M$ on line $AB$.
Find, with proof, the ratio $AM/MC$.

Solution: By power of a point we have
\[ AE \cdot AF = AB^2
  = \left( \half AB \right) \cdot \left( 2AB \right)
  = AD \cdot AC \]
and hence $CDEF$ is cyclic.
Then $M$ is the circumcenter of quadrilateral $CDEF$.

\begin{center}
\begin{asy}
pair A = dir(170);
pair C = dir(40);
pair B = midpoint(A--C);
pair D = midpoint(A--B);
pair M = midpoint(C--D);
pair O = origin;

filldraw(unitcircle, opacity(0.1)+lightcyan, heavycyan);
draw(CP(O, B), heavycyan);
pair E = IP(CP(M, D), CP(O, B));
pair F = OP(CP(M, D), CP(O, B));
filldraw(CP(M, D), opacity(0.1)+lightred, red);
draw(A--C, heavygreen);
draw(A--F, heavygreen);
draw(D--E, red);
draw(C--F, red);
pair U = midpoint(D--E);
pair V = midpoint(C--F);
draw(U--M--V, red+dotted);

dot("$A$", A, dir(A));
dot("$C$", C, dir(C));
dot("$B$", B, dir(B));
dot("$D$", D, dir(D));
dot("$M$", M, dir(M));
dot("$O$", O, dir(45));
dot("$E$", E, dir(220));
dot("$F$", F, dir(F));

/* TSQ Source:

A = dir 170
C = dir 40
B = midpoint A--C
D = midpoint A--B
M = midpoint C--D
O = origin R45

unitcircle 0.1 lightcyan / heavycyan
CP O B heavycyan
E = IP CP M D CP O B R220
F = OP CP M D CP O B
CP M D 0.1 lightred / red
A--C heavygreen
A--F heavygreen
D--E red
C--F red
U := midpoint D--E
V := midpoint C--F
U--M--V red dotted

*/
\end{asy}
\end{center}

Thus $M$ is the midpoint of $\ol{CD}$
(and we are given already that $B$ is the midpoint of $\ol{AC}$,
$D$ is the midpoint of $\ol{AB}$).
Thus a quick computation along $\ol{AC}$ gives $AM/MC = 5/3$.