In the five-sided star shown, the letters $A$, $B$, $C$, $D$, and $E$ are replaced by the numbers 3, 5, 6, 7, and 9, although not necessarily in this order. The sums of the numbers at the ends of the line segments $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, $\overline{DE}$, and $\overline{EA}$ form an arithmetic sequence, although not necessarily in this order.  What is the middle term of the arithmetic sequence?

[asy]
pair A,B,C,D,E;
A=(0,10);
B=(5.9,-8.1);
C=(-9.5,3.1);
D=(9.5,3.1);
E=(-5.9,-8.1);
draw(A--B--C--D--E--cycle,linewidth(0.7));
label("$A$",A,N);
label("$B$",B,SE);
label("$C$",C,NW);
label("$D$",D,NE);
label("$E$",E,SW);
[/asy]
Each number appears in two sums, so the sum of the sequence is \[
2(3+5+6+7+9)=60.
\]The middle term of a five-term arithmetic sequence is the mean of its terms, so  $60/5=\boxed{12}$ is the middle term.

The figure shows an arrangement of the five numbers that meets the requirement.

[asy]
pair A,B,C,D,E;
A=(0,10);
B=(5.9,-8.1);
C=(-9.5,3.1);
D=(9.5,3.1);
E=(-5.9,-8.1);
draw(A--B--C--D--E--cycle,linewidth(0.7));
label("7",A,N);
label("6",B,SE);
label("5",C,NW);
label("9",D,NE);
label("3",E,SW);
label("14",(0,1.1),N);
label("13",(0.7,0),NE);
label("10",(-0.7,0),NW);
label("11",(0,-0.7),SW);
label("12",(0,-0.7),SE);
[/asy]