TST

First, we present two following lemmas. Let $ABC$ be a triangle inscribed in $(O)$ and $N$ be the nine-point center of $△ABC$.

Lemma 1. Let $K$ be the center of $(BOC)$, then $AN$, $AK$ are isogonal with respect to $\angle BAC$.

Proof. Let $X$ be the reflection of $O$ through $BC$. Let $H$ be the orthocenter of $△ABC$. We have $AH=OX$, and $AH$ is parallel to $OX$. Hence, $AX$ cuts $OH$ at the midpoint of $OH$, which is $N$, and thus $N$ is the midpoint of $AX$. We have $\angle OXB=\angle KOB=\angle KBO$, so $△OKB\sim △OBX$, implying that $\frac{OK}{OB}=\frac{OB}{OX}$, or $$OX\cdot OK=O{B}^2=O{A}^2.$$ Hence, we have $\frac{OK}{OA}=\frac{OA}{OX}$, implying that $△OKA\sim △OAX$. Therefore, $\angle OKA=\angle XAO=\angle DAK$. Since $AH$, $AO$ are isogonal with respect to $\angle BAC$, we have $\angle BAK=\angle NAC$, implying that $AN$, $AK$ are isogonal with respect to $\angle BAC$.

Lemma 2. Let $B^{\prime }$, $C^{\prime }$ be the reflections of $B$, $C$ through $AC$, $AB$, respectively, then $AK$ is perpendicular to $B^{\prime }{C}^{\prime }$. Proof. Let $Y$, $Z$ be the reflection of $O$ through $CA$, $AB$. Similar to the argument above, we have $N$ is the midpoint of $BY$ and $CZ$. Let $N_b$, $N_c$ be the reflection of $N$ through $CA$, $AB$. Since $O{C}^{\prime }$ is the reflection of $ZC$ through $AB$, and $N$ is the midpoint of $ZC$, then $N_c$ is the midpoint of $O{C}^{\prime }$. Similarly, $N_b$ is the midpoint of $O{B}^{\prime }$. Hence, $N_b{N}_c$ is parallel to $B^{\prime }{C}^{\prime }$. We have $$\angle KA{N}_b=\angle N_{b}AC+\angle KAC=\angle NAC+\angle NAB=\angle BAC,$$ and similarly, $\angle KA{N}_c=\angle BAC$. On the other hand, we have $A{N}_b=AN=A{N}_c$, and since $\angle KA{N}_b=\angle KA{N}_c$, we get $AK$ is perpendicular to $N_b{N}_c$. So $AK\perp B^{\prime }{C}^{\prime }$, as desired. $\square$

Back to the main problem, let $O_a$, $O_b$, $O_c$ be the circumcenters of $△NBC$, $△NCA$, $△NAB$ and $K_a$, $K_b$, $K_c$ be the circumcenters of $△O_{a}BC$, $△O_{b}CA$, $△O_{c}AB$. By Lemma 2, we have $$m_a\equiv N{K}_a,m_b\equiv N{K}_b,m_c\equiv N{K}_c$$



a) If $N$ is the orthocenter $H$, let $m_a^{\prime }$, $m_b^{\prime }$ and $m_c^{\prime }$ be the reflections of $m_a$, $m_b$ and $m_c$ through the bisectors of angles $BHC$, $CHA$ and $AHB$. Let $J$ be the nine-point center of $ABC$. Note that $(J)$ is also the nine-point circle of $BHC$. By Lemma 1, we have $H{K}_a$, $KJ$ are isogonal with respect to $\angle BHC$, so $m_a^{\prime }$ goes through $J$. Similarly, $m_b^{\prime }$, $m_c^{\prime }$ also go through $J$, and thus $m_a^{\prime }=m_b^{\prime }=m_c^{\prime }=HJ$.

b) Assume that $N$ is the nine-point center of $△ABC$. Let $N_a$, $N_b$ and $N_c$ be the reflections of $N$ through $BC$, $CA$ and $AB$. Let $(X)$, $(Y)$ and $(Z)$ be the nine-point circles of $△NBC$, $△NCA$ and $△NAB$. We will show that the reflections of $m_a$, $m_b$ and $m_c$ through $BC$, $CA$ and $AB$ concur on $N_a{N}_b{N}_c$. Denote $(N)$ the nine-point circle of $△ABC$. Let $P$ be the Poncelet point of the four points $A$, $B$, $C$ and $N$; $E$, $F$ be the midpoints of $AC$, $AB$. We have $P$, $E$ are the intersections of $(N)$ and $(Y)$, and $P$, $F$ are the intersections of $(N)$ and $(Z)$. Hence, $PE\perp NY$, $PF\perp NZ$, and we have $$\angle YNZ=180^{\circ }-\angle FPE=180^{\circ }-\angle BAC.$$ Let $S$, $T$ be the intersections of $N{K}_b$, $N{K}_c$ with $CA$, $AB$, respectively. Let $K$ be the center of $N_a{N}_b{N}_c$. Denote the intersection of the reflection line of $m_b$, $m_c$ through $AC$, $AB$ as $L$. We have $BK\perp N_a{N}_c$, $CK\perp N_b{N}_c$, and $K$ is the isogonal conjugate of $N$ in $△ABC$. We have then $$\begin{array}{rl}\angle L{N}_b{N}_a& =180^{\circ }-\angle RS{N}_b-\angle N_{b}RA\\ & =180^{\circ }-\angle NSC-(90^{\circ }-\angle ACK)\\ & =90^{\circ }-\angle NSC+\angle ACK.\end{array}$$ And similarly, $\angle L{N}_c{N}_a=90^{\circ }-\angle NTB+\angle ABK$. It suffices to show that $L\in (N_a{N}_b{N}_c)$, and to show that, we need to show $\angle ACK+\angle ABK=\angle NSC+\angle NTB$. Since $N{K}_b$, $NY$ are isogonal with respect to $\angle ANC$, we have $$\angle NSC=180^{\circ }-\angle K_{b}NC-\angle NCA=\angle YNA-\angle NCA,$$ and similarly, $\angle NTB=\angle ZNA-\angle NBA$. Thus, $$\angle NSC+\angle NTB=\angle YNZ-(\angle NBA+\angle NCA).$$ So we need to show that $$\angle B+\angle C=\angle NBA+\angle NCA+\angle ABK+\angle ACK,$$ but this is true since, $\angle B=\angle ABN+\angle CBN=\angle ABN+\angle ABK$ and $\angle C=\angle ACN+\angle BCN=\angle ACN+\angle ACK$. The proof is completed. $\square$