jmc

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