Embarrassing to the Solver

First notice that $F$ is the circumcenter of the triangle. Also notice that $D$ is the midpoint of arc $BC$ and $E$ is the midpoint of arc $AC$ .

Connect $FB = FE$ since they are both radii of the circle. Therefore $\bigtriangleup BFE$ is isosceles and $\angle FBE = \angle BEF = 10$

Similarly, $\angle FAD = 5$

Now let $\angle BAF = \angle ABF = z$ , $\angle FBC = \angle FCB = x$ , and $\angle FCA = \angle FAC = y$

Notice that the angles of the triangles are $x+y , y+z, z+x \implies x+y+z = 90$

Now using the property of the angle bisectors, we see that

$z-5 = y+5$ and $z-10 = x+10$ which using all three equations yields the solutions of $(x,y,z) = (20 , 30 , 40)$ which implies that $\angle ABC = 60$

Finding the length of $DF$ is equivalent to finding the circumradius.

By extended law of sines,

$\frac{AC}{\sin 60 } = 2R , AC = 3 \implies R = \sqrt{3} \implies a = \boxed{3}$