Let $ M,M'$ be two conjugates point in triangle $ ABC$ (in the sense that $ \angle MAB=\angle M'AC,\dots$). Let $ P,Q,R,P',Q',R'$ be foots of perpendiculars from $ M$ and $ M'$ to $ BC,CA,AB$. Let $ E=QR\cap Q'R'$, $ F=RP\cap R'P'$ and $ G=PQ\cap P'Q'$. Prove that the lines $ AG, BF, CE$ are parallel.
From the well-known fact that the feet from isogonal conjugates to the sides are all collinear, we have that $P, Q, R, P', Q', R'$ are all concyclic, say on $\Omega.$ So by Brokard's Theorem, $AG$ is just the polar of the point $X = PQ' \cap P'Q.$ Let $P_1, Q_1, R_1$ be $MP \cap\Omega, MQ \cap \Omega, MR \cap \Omega.$ Then, since $\angle MPP' = 90$, we have that $P_1P$ is a diameter of $\Omega$, and similarly for $QQ_1, RR_1.$ Now, Pascal's Theorem on $PP_1P'QQ_1Q' \Rightarrow M, O, X$ are collinear. Hence, we have that $AG$ is perpendicular to $OM$, and similarly so are $BF, CE.$ Done!