Just Some Altitudes

Let $H$ be the orthocenter of $ABC$ . Then, since $\angle AEH = \angle AFH = 90^\circ,$ quadrilateral $AEHF$ is cyclic . Equivalently, we can find that $BFHD$ and $CEHD$ are cyclic as well.

Consider the circumcircle about $AEHF.$ We see that $\angle HEF = \angle HAF$ since they subtend the same arc. Similarly, $\angle FCD = \angle DEH.$ Finally, $\angle HAF = \angle FCD$ since $\triangle HFA \sim \triangle HDC.$ Thus, $\angle HEF = \angle FCD = \angle DEH$ ; in other words, $\overline{BE}$ bisects $\angle DEF.$

This gives $\angle DEF = \angle HEF + \angle DEH = 2\angle HEF = 2(90^\circ - \angle ABD) = 2(90^\circ - 40^\circ) = \boxed{100^\circ}.$

Remark: There might be a more obvious way to see that $\overline{BE}$ bisects $\angle DEF.$ Does anyone see such an approach?