We know that $AD,BE,CF$ concur at the orthocenter of $DEF$, denoted by $H$. Let $X'$ denote the isogonal conjugate of $H$ wrt $ABC$, If $DEF$ isn't the pedal triangle of $X'$, then its pedal triangle is homothetic with $DEF$ which is impossible since the lines that connect the corresponding points of the two homothetic triangles don't concur. Hence $X'\equiv X$. It's a well known property of isogonal conjugates that $O,X,H$ are collinear and $RST$ is the pedal triangle of $H$. We will prove that $BS\cap CT=Y'$ lies on line $OXH$, in which it's clear that the concurrence at $Y$ follows. Apply Pappus' theorem on $CES,BFT$ and we have $H,Y',ET\cap FS$ are collinear. Let $HS,HT\cap (O)=S',T'$ again and since $SS'\cap TT'=O$, by Pascal theorem we have $O,H,ET\cap FS$ are collinear and hence $O,X,H,ET\cap FS,Y'$ are collinear and we are done.

My solution: Since the lines through $ A, B, C $ and perpendicular to $ EF, FD, DE $ are concurrent at the orthocenter of $\triangle DEF $ , so $ \triangle DEF $ and $ \triangle ABC $ are orthologic $ \Longrightarrow $ the perpendiculars through $ D, E, F $ to $ BC, CA, AB $ are concurrent at $ X $ . Since $ AD, BE, CF $ are concurrent , so by Terquem theorem we get $ AR, BS, CT $ are concurrent at $ Y $ . Since $ \triangle RST $ is the pedal triangle of $ X' $ ($ X' $ is the isogonal conjugate of $ X $ ) , so the line passing $ A, B, C $ and perpendicular to $ ST, TR, RS $ are concurrent at $ X $ , hence by Sondat theorem (for $\triangle ABC $ and $\triangle RST $ ) we get $ Y, X', X $ are collinear. ie. $ X, O, Y $ are collinear Q.E.D Remark: (1) Easy to see $ X' = AD \cap BE \cap BF $ which is the orthocenter of $\triangle DEF $ (2) Another interesting property in this configuration : Denote $ O' $ as the circumcenter of $ \triangle ABC $ , then the isogonal conjugate of $ Y $ lie on $ O'H $ . (see http://www.artofproblemsolving.com/community/c6h366335 )

Here is my solution. Let $H$ be the orthocenter of $DEF$, which clearly lies on $AD,BE,CF$. By Desargues on $EDF,CH_DB$, where $H_D$ is the foot of the perpendicular from $D$ to $EF$ (define $H_E,H_F$ similarly), we get that $H_EH_F,EF,CB$ are concurrent and therefore $D(T,E,H,F)$ is a harmonic pencil, and so $TEH_DF$ is a harmonic quadrilateral. Thus if $H_D'$ is the reflection of $H_D$ across the perpendicular bisector of $EF$ (which is also clearly the point diametrically opposite to $D$), since $TH_D$ is symmedian of $TEF$ we get that $TH_D'$ is a median of $TEF$. Thus $T$ lies on $MH_D'$, where $M$ is the midpoint of $EF$, but $H$ also lies here, and so these 4 points are collinear. Therefore, $\boxed{\angle HTF=90}$ (since $DH_D'$ is a diameter). Reflecting $\ell$, the line perpendicular to $CB$ that passes through $D$, through $O$, it is easily seen that it transforms into line $TH$. The analogous happens for $E$ and $F$. Thus the three original lines are concurrent at the reflection of $H$ through $O$. Therefore, $X$ is the reflection of $H$ through $O$. We must now prove $Y$ exists and lies on the Euler line of $\triangle DEF$. To this end, we invert through $H$ with any radius. Let $Z'$ denote the inverse of $Z$ for any point $Z \neq H$. Lines $HD,HE,HF$ invert to themselves. Notice that the circle $HEH_DH_FS$ has a diameter $HE$, and hence its inverse is a line perpendicular to $HE$ (the analogous applies to $HD,HF$). From this we gather that $D',E',F'$ are the feet of the altitudes in a triangle, and that $S,R,T$ lie on the sides. Since $DEFSRT$ are concyclic, so are their inverses and so $S', R', T'$ are the midpoints of said triangle, and their circumcenter is the 9-point circle. Finally, since $\angle HSA=\angle HTA=90$, we gather that $STAH$ are cyclic, so $S',T',A'$ are collinear. From this, $A'=T'S' \cap HD'$. The analogous occurs for $B',C'$. From now on, for convenience, let $Z$ denote what we previously denoted by $Z'$, for any letter $Z$. Let the big triangle be $XYZ$. The question is now to prove that the circumcenters of $HTC,HRA,HSB$ have the Euler line of $XYZ$ as a common radical axis (since $O$ inverted to a point on this line, because it also contains the circumcenter of $DEF$, and $H$ and $O$ are collinear with this circumcenter by symmetry). Let $G$ be the barycenter of $XYZ$, let's prove that $G$ has the same power of a point to all 3 circles. Let $Z_0$ be the intersection of $ZT$ and the 9-point circle. Then $(ZZ_0)(ZT)=(ZS)(ZE)=(ZC)(ZH)$ since $HCSE$ is cyclic because $\angle HCS= \angle HES=90$. Then $THCZ_0$ are concyclic. But $ZT$ is the median of $XYZ$, and so $G$ lies on it. Therefore $G$ has the same power of a point to the circumcircle of $THCZ_0$ than to the 9-point circle, and by analogy we are done. $\blacksquare \text{ Q.E.D}$.

ELMOSL 2014 G10 wrote: We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear. Proposed by Sammy Luo Clearly $AD,BE,CF$ are concurrent at the Orthocenter $U$ of $\Delta DEF$. So, $AR,BS,CT$ are concurrent at the Cyclocevian Conjugate $(Y)$ of $U$ WRT $\Delta ABC$ $(\bigstar)$. Also Clearly $\{\Delta ABC,\Delta DEF\}$ are Orthologic, so the perpendiculars to $BC,CA,AB$ through $D,E,F$ concurs at $X$. Now notice that the Perpendiculars from $R,S,T$ to $BC,CA,AB$ respectively concurs at a point $X^*$ which is the Isogonal Conjugate of $X$. Also the perpendiculars from $A,B,C$ to $ST,RT,SR$ are concurrent at $X$. Hence, $\{\Delta ABC,\Delta RST\}$ are Orthologic with centers of Orthology $\{X,X^*\}$. Also from $(\bigstar)$ we get that $\{\Delta ABC,\Delta RST\}$ are perspective. So, by Sondat's Theorem $Y\in\overline{XX^*}$ but $O\in\overline{XX^*}$. Hence. $\overline{O-X-Y}$. $\blacksquare$