Balkan Mathematical Olympiad Shortlist

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 $ABC$.