Let $U, V, W$ be the feet of the perpendiculars from $A, B, C$ to $EF, FD, DE$ respectively, and let $X, Y, Z$ be the feet of the perpendiculars from $D, E, F$ to $BC, CA, AB$ respectively. Since $U$ and $Z$ both subtend a right angle from $AF$, $AFUZ$ is concyclic and so (with an appropriate sign convention) $\angle UAB = \angle UAZ = \angle UFZ = \angle EFZ$. Combining this with five similar equalities of angles, we see that
$$\frac{\sin \angle UAB \sin \angle VBC \sin \angle WCA}{\sin \angle CAU \sin \angle ABV \sin \angle BCW} = \frac{\sin \angle EFZ \sin \angle FDX \sin \angle DEY}{\sin \angle ZFD \sin \angle XDE \sin \angle YEF}$$

Yet by the angle form of Ceva's theorem, the left hand expression is 1 if and only if the Cevians $AU, BV, CW$ concur, i.e. iff $ABC$ is perpendicular to $DEF$. Similarly, the right hand expression is 1 if and only if $DEF$ is perpendicular to $ABC$. Thus $ABC$ is perpendicular to $DEF$ if and only if $DEF$ is perpendicular to $ABC$.