Since $\angle BAD'=\angle BAD=\tfrac{1}{2}\angle ACB$ and $\angle CAE'=\angle CAE=\tfrac{1}{2}\angle ABC,$ we get $A,D',E'$ are collinear $\Longleftrightarrow$ $\angle BAD'+\angle CAE'=\angle BAC$ $\Longleftrightarrow$ $\angle ACB+\angle ABC=2\angle BAC$ $\Longleftrightarrow$ $\angle BAC=60^{\circ}.$ Let $I,N,Na$ denote the incenter, 9-point center and Nagel point of $\triangle ABC.$ By Furhmann theorem, $D'$ and $E'$ lie on the circle with diameter $\overline{HNa}$ for any $\triangle ABC,$ hence $O,H,D',E'$ are concyclic $\Longleftrightarrow$ $\angle HONa=90^{\circ}.$ As $I,N$ are the Nagel point and circumcenter of the medial triangle, then $ONa \parallel IN,$ so $\angle HONa=90^{\circ}$ $\Longleftrightarrow$ $IN \perp OH$ $\Longleftrightarrow$ $IH=IO$ $\Longleftrightarrow$ one of the angles of $\triangle ABC$ equals $60^{\circ}$ (well-known). From the previous discussion we get: If $D',E'$ are collinear, then $\angle BAC=60^{\circ}$ $\Longrightarrow$ $AH=AO$ $\Longrightarrow$ $IN \perp OH$ $\Longrightarrow$ $\angle HONa=90^{\circ}$ $\Longrightarrow$ $O,H,D',E'$ are concyclic. But the converse is not necessarily true; for instance the concyclity yields $IH=IO$ $\Longrightarrow$ one of the angles of $\triangle ABC$ equals $60^{\circ},$ say $\angle ACB=60^{\circ}$ $\Longrightarrow$ $D' \equiv O$ $\Longrightarrow$ $D'E'$ is perpendicular bisector of $AC$ $\Longrightarrow$ $A \notin D'E'.$

See http://www.artofproblemsolving.com/community/c6h567389p3324618