sorry,i don't understand part a,what does "are the same" mean?

They are the same line. I'll make that clear.

Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$

Groupsolved with L567, For part a) We claim that the desired line is the Euler line of triangle ABC. Let $E,F$ be the feet of the perpendicular from $B,C$ respectively. Proof. Let $l_1,l_2$ be the lines through $A$ parallel to the Euler line and the perpendicular from $A$ to $B_1C_1$. Now, if we can show that $l_1,l_2$ are isogonal in $\angle BAC$, then the desired result will follow as then we will have $\angle(l_2,BH)=\angle (l_2, AC)+90\angle (l_1,AB)+90=\angle (l_1, CH)$ and the isogonality follows. Now, to prove that these two lines are isogonal, we just need that $l_1$ passes through the circumcenter of $(AB_1C_1)$. Let this point be $O_A$. We claim that $O_AAOH$ is a parallelogram! Let $O'$ be the midpoint of $AH$. So, we would just need that the midpoint of $O_A$ is the midpoint of $AH$. But, consider the spiral similarity that takes $BF$ to $CE$. Let this center of spiral similarlity be $S$.Then, $SO_AO\sim SB_1B$ and $SO_AO'\sim SB_1F$ and $SO'O\sim SFB$. Thus, $O'$ is the midpoint of $OO_A$ and the result follows. Now, for part $b)$, my angle chase is very similar to khanhnx's above angle chase so I will omit that.

khanhnx wrote: Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$ Can you explain me how $180^{\circ} - \angle{BAC}=\angle{O'_bNO'_c}?$

Jack125 wrote: khanhnx wrote: Here is my solution for this problem Solution First, we prove the following lemma Lemma. Given $\triangle ABC;$ $B', C'$ be reflections of $B, C$ in $CA, AB$. Prove that the line through $A$ and perpendicular to $B'C'$ is isogonal conjugate of the line through $A$ and NPC center of $\triangle ABC$ Proof. Let $O$ be center of $(ABC);$ $O_a$ be center of $(OBC)$. The problem is equivalent to proving $AO_a$ $\perp$ $B'C'$. Let $S$ $\equiv$ $BC'$ $\cap$ $CB'$. We have $\angle{SBC} + \angle{SCB} = 360^{\circ} - \angle{CBC'} - \angle{BCB'} = \angle{BOC}$. Let $AD$ $\perp$ $BC$ at $D$. Note that $A$ is $S$ - excenter of $\triangle SBC,$ we have $O_aB'^2 - O_aC'^2 = B'C \cdot B'S - C'B \cdot C'S = BC (CS - BS) = BC (BD - CD) = AB^2 - AC^2 = AB'^2 - AC'^2$ or $AO_a$ $\perp$ $B'C'$ Back to the main problem. If $N$ is orthocenter of $\triangle ABC,$ relabel $N$ as $H$. Let $O$ be center of $(ABC)$. From the above lemma, we have $m_a, HO$ are isogonal conjugate in $\angle{BHC}$. Similarly, we have $m_b, HO$ are isogonal conjugate in $\angle{AHC};$ $m_c, HO$ are isogonal conjugate in $\angle{AHC}$ If $N$ is NPC center of $\triangle ABC,$ let $D, E, F, m'_a, m'_b, m'_c$ be reflections of $N, m_a, m_b, m_c$ in $BC, CA, AB;$ $O_b, O_c$ be center of $(ANC), (ANB);$ $S$ $\equiv$ $m'_b$ $\cap$ $m'_c;$ $O'_b, O'_c$ be reflections of $O_b, O_c$ in $CA, AB$. Then, from the above lemma, we have $EO_b, m'_b$ are isogonal conjugate in $\angle{CNA};$ $FO_c, m'_c$ isogonal conjugate in $\angle{ANB}$. We have $\angle{ESF} = 360^{\circ} - \angle{EAF} - \angle{AES} - \angle{AFS} = 180^{\circ} - 2\angle{BAC} + \angle{CEO_b} - \angle{BFO_c}$ $= 180^{\circ} - 2\angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c} = \angle{O'_bNO'_c} - \angle{BAC} + \angle{CNO'_b} - \angle{BNO'_c}$ $= \angle{BNC} - \angle{BAC} = \angle{ABN} + \angle{ACN} = \angle{EDF}$ or $S$ $\in$ $(DEF)$. Similarly, if we let $S'$ $\equiv$ $m'_c$ $\cap$ $m'_a$ then $S'$ $in$ $(DEF),$ so $S'$ $\equiv$ $S$ or $m'_a, m'_b, m'_c$ concur at a point $S$ which lies on $(DEF)$ Can you explain me how $180^{\circ} - \angle{BAC}=\angle{O'_bNO'_c}?$ I think you can see here