In triangle $ABC$, $\angle BAC = 72^\circ$.  The incircle of triangle $ABC$ touches sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively.  Find $\angle EDF$, in degrees.

[asy]
import geometry;

unitsize(2 cm);

pair A, B, C, D, E, F, I;

A = (1,2);
B = (0,0);
C = (3,0);
I = incenter(A,B,C);
D = (I + reflect(B,C)*(I))/2;
E = (I + reflect(C,A)*(I))/2;
F = (I + reflect(A,B)*(I))/2;

draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(F--D--E);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
[/asy]
Since $BD$ and $BF$ are tangents from the same point to the same circle, $BD = BF$.  Hence, triangle $BDF$ is isosceles, and $\angle BDF = (180^\circ - \angle B)/2$.  Similarly, triangle $CDE$ is isosceles, and $\angle CDE = (180^\circ - \angle C)/2$.

Hence, \begin{align*}
\angle FDE &= 180^\circ - \angle BDF - \angle CDE \\
&= 180^\circ - \frac{180^\circ - \angle B}{2} - \frac{180^\circ - \angle C}{2} \\
&= \frac{\angle B + \angle C}{2}.
\end{align*} But $\angle A + \angle B + \angle C = 180^\circ$, so \[\frac{\angle B + \angle C}{2} = \frac{180^\circ - \angle A}{2} = \frac{180^\circ - 72^\circ}{2} = \boxed{54^\circ}.\]