[asy]
draw((3,-13)--(21.5,-5)--(19,-18)--(9,-18)--(10,-6)--(23,-14.5)--cycle);
label("A",(3,-13),W);label("C",(21.5,-5),N);label("E",(19,-18),E);label("F",(9,-18),W);label("B",(10,-6),N);label("D",(23,-14.5),E);
//Credit to Zimbalono for the diagram[/asy]

If the sum of the measures in degrees of angles $A,~B,~C,~D,~E$ and $F$ in the figure above is $90n$, then $n$ is equal to

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad\textbf{(E) }6$
1. Let \( M \) and \( N \) be the intersection points of \( AD \) with \( BF \) and \( CE \) respectively. 
2. Define the angles at these intersections as follows:
   - Let \(\angle CNA = x\)
   - Let \(\angle BMD = y\)
3. We know that the sum of the angles around a point is \(360^\circ\). Therefore, we can write the following equations:
   \[
   x + \angle C + \angle A = 180^\circ
   \]
   \[
   y + \angle B + \angle D = 180^\circ
   \]
   \[
   \angle F + \angle E + (180^\circ - x) + (180^\circ - y) = 360^\circ
   \]
4. Simplify the third equation:
   \[
   \angle F + \angle E + 180^\circ - x + 180^\circ - y = 360^\circ
   \]
   \[
   \angle F + \angle E + 360^\circ - x - y = 360^\circ
   \]
   \[
   \angle F + \angle E - x - y = 0
   \]
   \[
   \angle F + \angle E = x + y
   \]
5. Add all the equations together:
   \[
   (x + \angle C + \angle A) + (y + \angle B + \angle D) + (\angle F + \angle E + 360^\circ - x - y) = 720^\circ
   \]
   \[
   x + \angle C + \angle A + y + \angle B + \angle D + \angle F + \angle E + 360^\circ - x - y = 720^\circ
   \]
   \[
   \angle A + \angle B + \angle C + \angle D + \angle E + \angle F + 360^\circ = 720^\circ
   \]
6. Subtract \(360^\circ\) from both sides:
   \[
   \angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^\circ
   \]
7. Given that the sum of the measures of the angles is \(90n\), we have:
   \[
   90n = 360^\circ
   \]
8. Solve for \( n \):
   \[
   n = \frac{360^\circ}{90^\circ} = 4
   \]

The final answer is \(\boxed{4}\)