Consider arcs $AF, ED$. If we show that $AF + DE = 180^{\circ}$, we're done. But since $\angle{ABF} = \frac{B}{2}$ covers $AF$, we have $AF=B$. Similarly, $ED = EB+BD = A+C$, so $AF+DE = A+B+C = 180^{\circ}$, as desired.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=445661

Also $AF+DE=180^{\circ}$ follows immediately because $D=\hat S_A$, etc.

A rather tricky solution for someone who isn't used to the "big picture" configuration; but here goes. 1. Complete the entire "configuration" by drawing all the external angle bisectors of the angles A,B,C; thereby obtaining a big triangle formed by the external bisectors and excentres as vertices; and the original triangle being it's pedal triangle, and the incentre being the big triangle's orthocentre. 2. Then just observe that these intersection points D,E,F are just midpoints of segments connecting the incentre with the excentres. 3. Now it's just straightforward; if X,Y,Z are the excentres, and I being the incentre; consider the triangles IXY (and cyclically the other "component" triangles formed thanks to I). Observe that DE is the midline of IXY; parallel to XY. And since CF is perpendicular to XY; we must have CF perpendicular to DE; and cyclically the other arguments follow. Bonus: Just in case you want a "wording" overkill; there's always the good old "take the homothety map" centred at I with the ratio 0.5, implying I is indeed the orthocentre of DEF as well; hence the hypothesis is proved. (Apologies for not using LaTeX, but for some reason I can't "insert pictures" in my post; and LaTeX counts as inserting pictures for some reason)