Never thought converse of Orthic triangle would come in a contest Anyways,this one felt easy for a G8 and purely projective. As we already know that $AD,BE,CF$ concur at a point $H$, Let $EF \cap BC=T$, by the definition of harmonics $$-1=(T,D;B,C)\stackrel{A}{=}(T,AD \cap FE;F,E)$$As,$AD$ is the angle bisector and $(T,AD \cap FE;F,E)$ it follows that $\measuredangle TDA=90^\circ$ and we can do similarly for other two sides and we are done!