The diagram below shows a circle with center $F$. The angles are related with $\angle BFC = 2\angle AFB$, $\angle CFD = 3\angle AFB$, $\angle DFE = 4\angle AFB$, and $\angle EFA = 5\angle AFB$. Find the degree measure of $\angle BFC$.
[asy]
size(4cm);
pen dps = fontsize(10);
defaultpen(dps);
dotfactor=4;
draw(unitcircle);
pair A,B,C,D,E,F;
A=dir(90);
B=dir(66);
C=dir(18);
D=dir(282);
E=dir(210);
F=origin;
dot("$F$",F,NW);
dot("$A$",A,dir(90));
dot("$B$",B,dir(66));
dot("$C$",C,dir(18));
dot("$D$",D,dir(306));
dot("$E$",E,dir(210));
draw(F--E^^F--D^^F--C^^F--B^^F--A);
[/asy]
1. Let the measure of $\angle AFB$ be $x$.
2. According to the problem, the angles are related as follows:
   - $\angle BFC = 2\angle AFB = 2x$
   - $\angle CFD = 3\angle AFB = 3x$
   - $\angle DFE = 4\angle AFB = 4x$
   - $\angle EFA = 5\angle AFB = 5x$
3. The sum of all angles around point $F$ in a circle is $360^\circ$. Therefore, we can write the equation:
   \[
   x + 2x + 3x + 4x + 5x = 360^\circ
   \]
4. Simplify the equation:
   \[
   15x = 360^\circ
   \]
5. Solve for $x$:
   \[
   x = \frac{360^\circ}{15} = 24^\circ
   \]
6. Now, we need to find the measure of $\angle BFC$:
   \[
   \angle BFC = 2x = 2 \times 24^\circ = 48^\circ
   \]

The final answer is $\boxed{48^\circ}$