Let $H$ and $O$ are orthocenter and circumcenter of $\triangle $$ABC$ $D$ is projection of $A$ to $BC$ $U= AB$ $ \cap $ circle passes through $A,C$ and tangent to $BC$ $W=AC $ $\cap$ circle passes through $A,B$ and tangent to $BC$ $A'=$reflection of $A$ respect to $B_1C_1$ $X$=$(AB_1C_1) \cap (AMN)$ $Y=UW \cap BC$ We want to prove that $X-O-H $ collinear. Take an inversion centered $A$ with radius $\sqrt{AD.AH}$ $M \rightarrow U$ $N \rightarrow W$ $O\rightarrow A'$ $X\rightarrow Y$ $H\rightarrow D$ We should prove $(AA'DY )$ is cyclic. $\iff$ $AA' \perp A'Y$ $\iff$ midpoint of $AY$ $\in$ $B_1C_1$ We will show that midpoints $BW$ and $CU$ $\in$ $B_1C_1$.Let $K$ and $L$ are midpoints of $BW$ and $CU$ $\angle C_1B_1B=\angle B_1BW=\angle KB_1B$ So $K-L-B-C$ collinear By gaussline we know that $K,L$ and midpoint of $AY$ are collinear so we are done.$\blacksquare$

If P = midpoint of BC then M, N = PB1 \cap AB, PC1 \cap AC. It suffices to show that the antipode of A wrt (AMN) is on OH, so if T, U are the midpoints of AB, AC then it suffices to show that C1T : TM = B1U : UN which is obvious by the Ratio Lemma on B1PN, C1PM since P, T, U, B1, C1 are concyclic on the 9-point circle and PU || TC1, PT || UB1.