??? I deleted my solution , and please don't participate me. thanks

Let $ (O)$ be the circumcircle of $ ABC$ with circumcenter $ O$ Let $ (O_1)$ the circle that is tangent to $ (O)$ at $ A_2$ and similarly we denote the $ (O_2),\ (O_3)$ .Then the common tangent of $ (O_1)$ ,$ (O)$ and the lines $ B_1C_1$ , $ BC$ are concurrent at the radical center of the circles $ (O_1)$ , $ (O)$ and the circumcircle of $ BCB_1C_1$ . Let K this point .Similarly the common tangent at $ C_2$ ,$ AB$ , $ A_1B_1$ are concurrent at $ L$ and the the common tangent at $ B_2$ ,and the lines $ AC$ , $ A_1C_1$ are concurrent at $ M$ . Now since $ KB\cdot KC=KC_1\cdot KB_1$ we have that $ K$ lies on the radical axis of the circumcircle of $ ABC$ and the circumcircle of $ A_1B_1C_1$ . Similarly for $ L,M$ .So $ K,L,M$ lie on the on the radical axis of the circumcircle of $ ABC$ and the circumcircle of $ A_1B_1C_1$ so these points are collinear .Let $ H$ the orthocenter of the triangle $ ABC$ Then from the complete quadrilateral $ AC_1HB_1BC$ we take that it's diagonals are cut at harmonic conjugate points ,ie $ (K,B,A_1,C)=-1$ (this can be done also by Ceva and Menelaos) .From the previous fact that $ (K,B,A_1,C)=-1$ and the fact that $ KA_2$ is tangent to the circumcircle of the $ ABC$ we conclude that $ A_2A_1$ is the polar of $ K$ with respect to the circumcircle $ (O)$ of $ ABC$ .Similarly $ B_2B_1$ and $ C_2C_1$ are the polars of $ M,L$ respectively.Let $ A_1A_2$ and $ B_1B_2$ meet at $ Q$ .Then $ Q$ lies both to the polars of $ K,M$ so $ KM$ is the polar of $ Q$ . But $ KM$ passes through $ L$ so $ L$ belongs to the polar of $ Q$ so $ Q$ belongs of the polar of $ M$ .So $ C_1,Q,C_2$ are collinear so the lines $ A_1A_2,B_1B_2,C_1C_2$ are concurrent. Now since the line $ M,K,L$ (let $ l$ this line) is the polar of $ Q$ we have that $ OQ\bot l$ . But $ l$ is also the radical axis of the circumcircles of $ ABC$ and $ A_1B_1C_1$ . So if $ O'$ is the circumcenter of $ A_1B_1C_1$ (this is also the nine-point center) we have that $ OO'\bot l$ so $ OO'$ passes through $ Q$ so $ Q$ belongs to the Euler line and we are done .

$ I$ $ CASE:$Exist two sides of triangle that are equals, $ WCGL$ $ AB=CA$->Then, $ A \in A_1A_2$, By the symmetry $ B_1B_2$ and $ C_1C_2$,have a common point,which the line $ A_1A_2$,But the line of euler of $ ABC$, is the same that $ A_1A_2$, then the lines $ A_1A_2$,$ B_1B_2$ ,$ C_1C_2$ have a common point, which lies on the Euler line of $ ABC$.$ \blacksquare$ $ II$ $ CASE:$ Dont exist two sides of triangle that are equals. If $ \alpha$ is circumcircle of $ ABC$ and $ \beta$ is circumcircle of $ A_1B_1C_1$ , if $ \mu$ is Radical axis of $ \alpha$ and $ \beta$. $ LEMMA:$ If $ A_3=B_1C_1 \cap BC$, $ B_3=A_1C_1 \cap AC$, $ C_3=A_1B_1 \cap AB$, and $ H$:Orthocenter, $ O$: Circuncenter.Then $ A_3$,$ B_3$,$ C_3$ are collinear,and $ OH \perp A_3B_3$. $ DEM$. Know we that: $ <BC_1C=<BB_1C= \pi /2$, Then-> $ A_3B.A_3C=A_3C_1.A_3B_1$ then $ A_3 \in \mu$, Similarly $ B_3 \in \mu$, $ C_3 \in \mu$. If $ O_9$ center of $ \beta$, know we that $ O_9O \perp \mu$(By Radical Axis), but $ O$, $ H$, $ O_9$ lies on the euler line of $ ABC$. Then $ OH \perp \mu$. $ \blacksquare$. In the problem, know we that the circumcircle of $ B_1C_1A_2$, is tangent to $ \alpha$ in $ A_2$, if $ X_1$ the line tangent to $ \alpha$ in $ A_2$, then $ X_1$ is Radical axis of circumcircle of $ B_1C_1A_2$ and $ \alpha$. $ BC$ is Radical Axis of $ \alpha$ and circumcircle of $ B_1CB$, and $ C_1B_1$ is radical axis of circumcircle of $ B_1C_1A_2$ and circumcircle of $ B_1C_1B$, but $ A_3=B_1C_1 \cap BC$ then $ A_3 \in X_1$(By Radical Axis). Now respect of $ \alpha$,we have $ A_1A_2$ is the polar of $ A_3$( because know we that $ A_3$, $ B$,$ A_1$,$ C$ points harmonics and $ A_3A_2$ is tangent to $ \alpha$), similarly $ B_1B_2$ is the polar of $ B_3$ and $ C_1C_2$ is the polar of $ C_3$, now, the polos are collinear then the polars have a common point, its $ Z$ ($ Z \in A_1A_2$,$ Z \in B_1B_2$, $ Z \in C_1C_2$ , $ Z$ Is the polo of $ \mu$, know we that : si $ \phi$ is polar of $ Y$, then $ OY \perp \phi$, then $ OZ \perp \mu$, but by the LEMMA $ OH \perp \mu$. Then $ Z \in OH$ , then $ Z$ belong to the line of euler of $ ABC$.$ \blacksquare$

Our overall aim is to show that all the lines pass through the pole $ E$ (with respect to the circumcircle) of the radical axis of the circumcircle and the nine-point circle (which, for want of a better name, I call the Euler axis). Let the circumcircle and nine-point circle of $ ABC$ be denoted by $ \omega_1$ and $ \omega_2$ respectively, let the circle $ B_1C_1A_2$ which is given as tangent to $ \omega_2$ at $ A_2$ be denoted by $ \gamma_1$, and let the circle on diameter $ BC$ (which clearly contains the points $ B_1,C_1$) be denoted by $ \gamma_2$. Step 1: These four circles all have a common radical point, implying that the lines $ B_1C_1$,$ BC$ and the tangent at $ A_2$ are concurrent on the Euler axis at a point we denote by $ P$. Proof: $ \omega_2, \gamma_1, \gamma_2$ are coaxal, all containing the points $ B_1$ and $ C_1$, and thus sharing $ B_1C_1$ as a radical axis. Now intersect $ B_1C_1$ with the Euler axis, the radical axis of $ \omega_1, \omega_2$, to get the point $ P$. By the radical axis theorem applied to $ \omega_1, \omega_2, \gamma_1$, the tangent at $ A_2$ (being the radical axis of $ \gamma_1, \omega_1$) passes through $ P$, and by the radical axis theorem applied to $ \omega_1, \omega_2, \gamma_2$, $ BC$, the radical axis of $ \gamma_2, \omega_1$, passes through $ P$. Step 2: $ P$ and $ A_1$ are harmonically conjugate with respect to $ BC$, and, therefore, conjugate with respect to $ \omega_1$. Let $ H$ be the orthocentre of $ ABC$. Considering $ ABCH$ as a quadrangle, it has diagonal point triangle $ A_1B_1C_1$, and since $ P = B_1C_1 \cap BC$, the result is an immediate consequence of the harmonic properties of any complete quadrangle (any range of four points in the complete quadrangle configuration is harmonic: in this case we consider the range lying along the line $ BC$). The Final Push Here if we talk of a pole, polar or a conjugacy, we always mean with respect to $ \omega_1$. Now, since $ PA_2$ is a tangent to $ \omega_1$ at $ A_2$, $ A_2$ lies on the polar of $ P$. Since $ A_1$ is conjugate to $ P$, $ A_1$ also lies on the polar of $ P$, so the line $ A_1A_2$ is the polar of $ P$. But $ P$ lies on the Euler axis, so the pole $ E$ of the Euler axis must lie on the polar of $ P$, so $ E$ lies on the line $ A_1A_2$. By a similar argument relabeling letters appropriately, $ E$ also lies on the lines $ B_1B_2$ and $ C_1C_2$, so these lines are indeed concurrent at $ E$. Furthermore, the pole of a line $ l$ with respect to a circle centre $ O$ must lie on the line through $ O$ perpendicular to $ l$. But the line through the centres of two circles is perpendicular to their radical axis. It follows that $ E$ lies on the line containing the centres of the circumcircle and the nine-point circle, which is the Euler line. (Step 2 Original Proof): We use unnormalised areal (barycentric) co-ordinates (whose basic workings are outlined in my PDF at http://www.bmoc.maths.org/home/areals.pdf or one of the references given therein). It is well known that the points $ A_1,B_1,C_1$ have areal co-ordinates $ (0,\tan B, \tan C), (\tan A, 0, \tan C), (\tan A, \tan B, 0)$ respectively (to prove this, we remark that $ BA_1 = 2R\sin C\cos B, A_1C = 2R\sin B\cos C$ by easy trigonometry, and so $ BA_1: A_1C = \tan C : \tan B$, whence the co-ordinates $ (0,\tan B, \tan C)$, and similarly for the other two points). Now, $ P$ is the intersection of $ B_1C_1$ with $ BC$. The line $ B_1C_1$ has equation (it may be easily checked that both $ B_1$ and $ C_1$ occupy this line) \[ \frac {x}{\tan A} = \frac {y}{\tan B} + \frac {z}{\tan C} \] Intersecting this with the line $ BC$ with equation $ x = 0$ gives $ \frac {y}{\tan B} + \frac {z}{\tan C} = 0$, which gives the co-ordinates of $ P(0, \tan B, - \tan C)$. Applying a parameter $ t$ to the line $ BC$ as $ (0, \frac {\tan B}{\tan C}, t)$, we see $ t = 0,\infty$ at $ B,C$ and $ t = 1, - 1$ at $ A_1, P$, so the cross ratio $ (A_1,P;B,C)$ evaluates to $ - 1$. This is well known in projective homogeneous co-ordinates (of which unnormalised areals are a special case) to characterise a harmonic separation, so we are done. (for the results used on projective co-ordinates, see E.A. Maxwell, or more recently Christopher Bradley's 'Algebra of Geometry', published by Highperception). Result quoted in given Step 2 Proof: Any range in a harmonic quadrilateral is harmonic. The theorem is easy to prove with projective homogeneous co-ordinates. Let us prove the result needed for the question, and its generality is obvious. We are using projective co-ordinates, so may choose four points without loss of projective structure, so we pick $ A(-1,1,1), B(1,-1,1), C(1,1,-1), H(1,1,1)$. This forces the diagonal point triangle (by intersecting lines like $ AH: y=z$ with lines like $ BC: y+z=0$), to be $ A_1(1,0,0), B_1(0,1,0), C_1(0,0,1)$. Finally, to give the complete quadrangle, we intersect every line present in the current diagram with every other line as much as possible. For example, intersecting $ B_1C_1$ and $ BC$ gives a point $ P(0,1,-1)$, so $ P,B,A_1,C$ form a range. If we apply a parameter to the line $ BC (1, t, -t)$ we have $ P,A_1$ corresponding to $ t=\infty,0$ and $ B,C$ corresponding to $ -1,1$, so these pairs of points are separated harmonically. We can apply the same logic to any range in the configuration.

Assume firstly the triangle is not isosceles. Denote $ \{A'\} = B_1C_1\cap BC$. Define $ B'$ and $ C'$ similarly. Let $ \Gamma(O,R),\Omega(\omega,R/2)$ be the circumcircle, $ 9$-point Circle of $ ABC$, respectively and let $ \rho(X,C)$ denote the power of point $ X$ wrt circle $ C$. Let's firstly show that $ A',B'$ anc $ C'$ are collinear. Indeed, note that $ BCB_1C_1$ is a cyclic quadrilateral, hence $ \rho(A',\Gamma) = A'B\cdot A'C = A'C_1\cdot A'B_1 = \rho(A',\Omega)$. By similar arguments, we get that $ A',B',C'$, lying outside $ \Gamma$ have the same power wrt $ \Gamma$ and $ \Omega$ thus are collinear lying on the radical axis $ d$ of $ \Gamma$ and $ \Omega$. An important observation is that $ d\bot O\omega$, or $ d\bot OH$. By a very-well known property (which is also very easy to prove), circles of diameters $ OA',OB',OC'$ are concurrent in some point $ Y$. Since we have $ OY\bot YA'$, $ OY\bot YB'$ and $ OY\bot YC'$, and $ A',B',C'\in d$, we obtain $ Y\in d$. Since $ d\bot OH$, $ Y = OH\cap d$. Let $ \Delta$ be the circle passing through $ B_1,C_1$ and internally tangent to $ \Gamma$ in $ A_2$. Let $ X$ be the intersection point of the tangent in $ A_2$ to $ \Gamma$ (and hence to $ \Delta$) with $ BC$ (the point exists, since the triangle is not isosceles). Then $ \rho(X,\Gamma) = XB\cdot XC = XA_2^2 = \rho(X,\Delta)$, hence $ X\in d$. Since $ d\cap BC = \{A'\}$, we get $ X = A'$. Consider the inversion $ I_1(A',k)$ with $ k = \rho(A',\Gamma) = \rho(A',\Delta) = \rho(A',\Omega)$. Circles $ \Gamma,\Omega,\Delta$ remain invariant under this inversion. Let $ M\in\Omega$ be the midpoint of $ BC$. Note that $ OM\bot MA'$ and $ OA_2\bot A_2A'$. So $ O,M,A_2,A'$ are inscribed into some circle $ L$, which, inverted becomes a line perpendicular to the diameter $ OA'$. This line is $ A_1A_2$ since $ I_1(M) = A_1$ and $ I_1(A_2) = A_2$. So, $ A_1A_2\bot OA'$. Consider now the beautiful inversion $ I(O,R^2)$. $ I(A_2) = A_2$, so $ I(L)$ is a line perpendicular to $ OA'$ (diameter of $ L$). According to the above, this line must be $ A_2A_1$. Since circles of diameters $ OA',OB'$ and $ OC'$ are concurrent in $ Y$, lines $ A_1A_2,B_1B_2$ and $ C_1C_2$ are concurrent in $ I(Y)$. As we have seen, $ OH\bot d$, and $ Y = OH\cap d$. Clearly then $ I(Y)\in OH$, hence lines $ A_1A_2,B_1B_2,C_1C_2,OH$ are concurrent.

Since $ \angle C_1BB_1=\angle C_1CB_1=90^\circ-A$, it follows that $ BCB_1C_1$ is a cyclic quadrilateral. $ B_1C_1$ is the radical axis of $ C(B_1C_1A_2)$ and $ C(B_1C_1BC)$ $ BC$ is the radical axis of $ C(ABC)$ and $ C(B_1C_1BC)$ $ A_2A_2$ is the radical axis of $ C(ABC)$ and $ C(B_1C_1A_2)$ (where $ A_2A_2$ is the common tangent of $ C(ABC)$ and $ C(B_1C_1A_2)$). It follows that there exist a point $ M$ such that $ M$=$ B_1C_1\cap BC\cap A_2A_2$ (M is called the radical centre of the $ C(B_1C_1A_2)$, $ C(ABC)$ and $ C(B_1C_1BC)$). From Menelaus' theorem we conclude that: $ \frac{CB_1} {B_1A} \cdot \frac{AC_1} {C_1B} \cdot \frac {MB} {MC} =1$ From Ceva's theorem we conclude that: $ \frac{CB_1} {B_1A} \cdot \frac{AC_1} {C_1B} \cdot \frac {BA_1} {A_1C}=1$ It follows that $ \frac {MB} {MC} = \frac {A_1B} {A_1C}$, so the division $ (B,C,M,A_1)$ is harmonic. Since $ A_2$ is the tangent point from $ M$ to circle $ C(ABC)$, we conclude that $ A_1A_2$ is the polar of $ M$ relative to circle $ C(ABC)$. Let $ A'$ be the midpoint of $ (BC)$ and $ {S}=MH \cap AA'$. Since the division $ (B,C,M,A_1)$ is harmonic, it follows that $ A_1B \cdot A_1C=A_1A' \cdot A_1M$. But $ A_1B \cdot A_1C=A_1H \cdot A_1A \Rightarrow \frac {A_1A'} {A_1A}=\frac {A_1H} {A_1M} \Rightarrow$ tan$ A_1AA'$=tan$ A_1MH$ $ \Rightarrow \angle A_1AA'=\angle A_1MH \Rightarrow \angle ASM = \angle AA_1M= 90^\circ \Rightarrow$ $ MH \perp AA'$ $ (1)$ Let $ O$ be the circumcenter of triangle $ ABC$ and $ G$ be the centroid of triangle $ ABC$. $ {N}=A_1C_1 \cap AC$ and $ {P}=A_1B_1 \cap AB$. It is well known that $ H-G-O$ is the Euler line of $ ABC$. $ B'$ is the midpoint of $ AC$. From $ (1)$ $ \Rightarrow MH \perp AA'$ and $ NH \perp BB'$. $ U \in MH \cap AA', V \in NH \cap BB'$. Quadrilaterals $ AMA_1U$ and $ BNB_1V$ are inscribed in the circles of diametres $ AM$, $ BN$, respectively. So, $ MH \cdot HU=HA \cdot HA_1$, $ HN \cdot HV=HB \cdot HB_1$. But $ HA \cdot HA_1=HB \cdot HB_1$. It follows that $ HM \cdot HU=HN \cdot HV$, which means that the quadrilateral $ MVUN$ is cyclic. But the quadrilateral $ HGUV$ is also a cyclic one, so it follows that $ MN$ is the image of circle $ C(HGUV$) under the inversion of pole $ H$. As $ HG$ is diameter in $ C(HGUV)$ , we conclude that $ HG \perp MN$ $ (2)$. From $ (2)$ $ \Rightarrow MP \perp HG \Rightarrow N-M-P \perp H-G-O$ Let $ N-M-P$ be the polar of $ X$ relative to the circle $ C(ABC)$. It follows that $ OX \perp N-M-P \Rightarrow X \in H-G-O$. Since $ M$ is on the polar of $ X$ relative to the circle $ C(ABC)$ $ \Rightarrow X$ is on the polar of $ M$ relative to the circle $ C(ABC)$$ \Rightarrow X \in A_1A_2$. Similar, $ X \in B_1B_2$ and $ X \in C_1C_2$.

First we have to show that lines $ A_{1}A_{2}$ ,$ B_{1}B_{2}$, and $ C_{1}C_{2}$ intersect at the same point. Denote $ X$ as intersection between $ A_{1}A_{2}$ and $ B_{1}B_{2}$. Denote $ H$ as orthocenter of $ \triangle ABC$ Denote $ C_{1}^\prime$ as intersection of line $ C_{1}H$ with circumcircle of $ \triangle ABC$ Note that $ |HC_{1}| = |C_{1}C_{1}^\prime|$ $ \leftarrow$ Reflection of orthocenter theorem, (or something like that ) Denote circle $ \kappa_{1}$, with diameter $ |HC_{1}^\prime|$, and center at $ C_{1}$ Denote $ C_{2}^\prime$ as intersection of $ \kappa_{1}$ and line $ XC_{2}$, such that $ |XC_{2}^\prime|\geq|C_{2}^\prime C_{2}|$ Denote $ X^\prime$ as intersection of $ \kappa_{1}$ and line $ XC_{2}$, such that $ |XX^\prime|\leq |X^\prime C_{2}|$ Points $ X^\prime,H,C_{2}^\prime,$ and $ C_{1}^\prime$ are cocyclic. $ \angle C_{1}^\prime X^\prime H = \angle C_{1}^\prime C_{2}^\prime H = \frac {\pi}{2}$ $ \leftarrow$ Thales theorem: as $ C_{1}^\prime H$ is diameter $ \angle X^\prime C_{1}^\prime H = \angle X^\prime C_{2}^\prime H\leftarrow$ over same chord $ X^\prime H$ $ \angle C_{1}^\prime X^\prime C_{2}^\prime = \angle C_{1}^\prime H C_{2}^\prime\leftarrow$ over same chord $ C_{1}^\prime C_{2}^\prime$ $ \angle C_{1} C_{1}^\prime X^\prime = \angle C_{1} X^\prime C_{1}^\prime \leftarrow$ isosceles $ \triangle C_{1} C_{1}^\prime X^\prime$ $ \angle C_{1} X^\prime H = \angle C_{1} H X^\prime \leftarrow$ isosceles $ \triangle C_{1} X^\prime H$ $ \angle C_{1}^\prime X^\prime H = \angle C_{1}^\prime X^\prime C_{1} + \angle C_{1} X^\prime H = \angle C_{1} H C_{2}^\prime + \angle C_{1} H X^\prime = \angle X^\prime H C_{2}^\prime = \frac {\pi}{2}$ Hence $ X^\prime C_{2}^\prime$ is also diameter. Therefore $ C_{1}$ lies on that line. But, because $ X^\prime$ and $ C_{2}^\prime$ lie on line $ X C_{2}$, point $ C_{1}$ lie on that line too. Therefore $ X$ is also point on line $ C_{1} C_{2}$ Thus we proved that all $ 3$ lines intersect at same point. Now we have to show that $ X$ lies on Euler's line of $ \triangle ABC$ We know that $ O$-circumcenter, $ H$-orthocenter, $ T$-centroid, and $ O_{e}$-center of $ 9$-point circle, are on that line. Its enough to show that $ \angle XHO = \pi$ Denote $ E_{A},E_{B},E_{C}$ as Euler's points to the corresponding vertex of $ \triangle ABC$ Denote $ \alpha = \angle E_{A}HC_{1}$ Denote $ \beta = \angle C_{1}HX$ Denote $ \gamma = \angle XHE_{B}$ Denote $ \delta = \angle E_{B}HA_{1}$ Denote $ Y$ intersection between $ C C_{1}$ and $ E_{A} E_{B}$ Denote $ W$ intersection between Euler's line and $ E_{C} E_{A}$ Denote $ Z$ intersection between Euler's line and $ E_{A} E_{B}$ Note: $ \alpha + \beta + \gamma + \delta = \pi$ Because $ C_{1}, H, C$ are collinear $ \angle C_{1}, H, C = \pi \Rightarrow \angle A_{1} H E_{C} = \alpha$ Because $ E_{B} H B_{1}$ are collinear $ \angle E_{A} H B_{1} = \pi \Rightarrow \angle E_{A} H B_{1} = \delta$ $ \angle E_{A} E_{B} E_{C} = 2\pi - \frac {\pi}{2} - \frac {\pi}{2} - (\beta + \gamma + \delta) = \pi - \alpha - \beta - \gamma - \delta + \alpha = \alpha$ $ \angle E_{B} E_{A} E_{C} = 2\pi - \frac {\pi}{2} - \frac {\pi}{2} - (\alpha + \delta) = \pi - \alpha - \beta - \gamma - \delta - \alpha + \beta + \gamma = \beta + \gamma$ Note: $ \angle HYE_{B} = \frac {\pi}{2}$ $ \angle WE_{B}H = \pi - (\beta + \gamma) - \frac {\pi}{2} = \frac {\pi}{2} - \beta - \gamma$ $ \angle HZE_{B} = \pi - \angle WE_{B}H - \angle ZHE_{B} = \pi - \frac {\pi}{2} + \beta + \gamma - \gamma = \frac {\pi}{2} + \beta$ $ \angle E_{C}W H = 2\pi - \angle WE_{C}E_{B} - \angle E_{C}E_{B}Z - \angle HZE_{B} = 2\pi - \delta - \alpha - \frac {\pi}{2} - \beta =$ $ \frac {3\pi}{2} - \alpha - \beta - \delta - \gamma + \gamma = \frac {\pi}{2} + \gamma$ $ \angle E_{A}E_{C}Z = \pi - \angle E_{C}E_{A}Z - E_{C}ZE_{A} = \pi - (\beta + \gamma) - \frac {\pi}{2} = \frac {\pi}{2} - \beta - \gamma$ (Note:because $ Y$ lies on line $ E_{A}E_{B}$ $ \angle E_{C}E{}_{A}{}_E{B} = \angle E_{C}E_{A}Z$) $ \angle ZHE_{C} = \pi - \angle ZE_{C}H - \angle E_{C}ZH = \pi - \frac {\pi}{2} + \beta + \gamma - \frac {\pi}{2} - \gamma = \beta$ (Note:because $ Z$ lies on line $ E_{A}E_{C}$ $ \angle E_{A}E_{C}E{B} = \angle ZE_{C}E_B$) Since $ Z$ is on Euler's line $ \angle ZHE_{C} = OHE_{C}$ $ \angle XHO = \angle XHE_{B} + \angle E_{B}HA_{1} + \angle A_{1}HE_{C} + \angle E_{C}HO = \gamma + \delta + \alpha + \beta = \pi$ Thus points $ XHO$ are collinear, and because line $ HO$ is Euler's line point $ X$ lies on that line.

Let $ A_3$ be the intersection point of the lines $ B_1C_1$ and $ BC$, if this dont exist consider it as a point at infinity. Define similarly $ B_3$ and $ C_3$, $ w$, $ w_1$ are the circumcircle of ABC and the nine point circle of $ ABC$(with center $ Z$), finally define: $ w_a$ the circle passing throught $ B_1$, $ C_1$ and tangent to $ w$. Define similarly $ w_b$ and $ w_c$. Consider $ w_a$, $ w$ and the circle passing throught $ B_1, C_1, B, C$, the three radical axis between these circles must concur, then $ B_1C_1$, $ BC$ and the tangent passing throught $ A_2$ to $ w$ concur at $ A_3$, now consider $ w_1$, $ w$ and $ w_a$, the 3 radical axis must concur at $ A_3$, then $ A_3$ belongs to the radical axis of $ w_1$ and $ w$, similar resoaning to $ B_3$ and $ C_3$. Suppose $ A_3A_2$ and $ A_3A_4$ are tangent to $ w$ and $ R$ is the intersection point of $ A_2A_4$ and $ BC$, we have $ BA_4CA_2$ is an harmonic quadrilateral, so $ (A_3,R,B,C)=-1$, also B and C are the feet of the angle bisectors of the triangle $ A_3B_1A_1$, so $ (A_3,A_1,B,C)=-1$, then $ R=A_1$, this implies $ A_2A_1$ is the polar line(respect to $ w$) of $ A_3$, for duality as $ A_3,B_3,C_3$ are collinear then the polar lines concur at $ X$, and the polar line of $ X$ is the radical axis of $ w$, $ w_1$, then $ OX$ is perpendicular to the radical axis of $ w$, $ w_1$, finally $ OZ$ is perpendicular to the radical axis of $ w$, $ w_1$, then O, Z and X are collinear, this proves the result because $ OZ$ is the Euler line of triangle $ ABC$.

Let $ X = B_1C_1\cap BC$, $ Y = C_1A_1\cap CA$, $ Z = A_1B_1\cap AB$, and call $ H$ the orthocentre of triangle $ ABC$. From the concurrency of $ AA_1$, $ BB_1$ and $ CC_1$, we conclude that $ (BCA_1X)$ is a harmonic division. Therefore $ A_1$ lies on the polar of $ X$ relative to the circumcircle $ \Gamma$ of triangle $ ABC$. $ (1)$ On the other hand, we have the quadrilateral $ BCB_1C_1$ is cyclic (because of $ \measuredangle BB_1C = \measuredangle BC_1C = 90^{\circ}$), which implies that $ X$ is the radical center of the circumcircles of triangles $ ABC$, $ B_1C_1A_2$ and $ BCB_1$. It means $ XA_2$ is tangent to the circle $ \Gamma$ at $ A_2$. In other word, $ A_2$ lies on the polar of $ X$ relative to the circle $ \Gamma$. $ (2)$ From $ (1)$ and $ (2)$ we conclude that $ A_1A_2$ is the polar line of $ X$ with respect to $ \Gamma$. Similarly, $ B_1B_2$ is the polar of $ Y$ and $ C_1C_2$ is the polar of $ Z$ relative to $ \Gamma$. Hence, to show that $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ concur, we will prove that points $ X$, $ Y$, $ Z$ are collinear. Indeed, let's call $ O$ and $ N$ the circumcenters of triangles $ ABC$ and $ A_1B_1C_1$, respectively. From the cyclic quadrilateral $ BCB_1C_1$, we have $ XB\cdot XC = XB_1\cdot XC_1$, it implies that $ X$ lies on the radical axis of circles $ (O)$ and $ (N)$. Similarly, $ Y$ and $ Z$ also lie on the radical axis of circle $ (O)$ and $ (N)$. In conclusion, $ X$, $ Y$, $ Z$ lie on a line perpendicular to $ ON$. Notice that $ N$ is the nine-point center of triangle $ ABC$, so $ O$, $ N$, $ H$ are collinear, so we conclude that $ OH\perp\overline{XYZ}$ $ (*)$ From the collinearity of $ X$, $ Y$, $ Z$, we obtain that lines $ A_1A_2$, $ B_1B_2$ and $ C_1C_2$ concur at the pole of line $ \overline{XYZ}$ relative to the circle $ \Gamma$, let it's $ P$. It implies $ OP\perp \overline{XYZ}$ $ (**)$ Combining $ (*)$ and $ (**)$ we deduce that $ O$, $ H$ and $ P$ are collinear, i.e. lines $ A_1A_2$, $ B_1B_2$, $ C_1C_2$ have a common point, which lies on the Euler line of $ \triangle ABC$, as desired.

Let $ B_{1}C_{1} \cap BC = A_{3}$, and so on. Let's consider a three radical axis of circumcircle of $ ABC$ ( $ =w$), $ BCB_{1}C_{1}$ and $ B_{1}C_{1}A_{2}$. Then we can show that $ B_{1}C_{1}$, $ BC$ and a line passing through $ A_{2}$ and tangent to $ w$ concur at a point and it should be a $ A_{3}$. Therefore if a line passing through $ A_{3}$ and tangent to $ w$ at arc $ BC$ was drawn, tangent point will be $ A_{2}$. In addition, because $ A_{3}, B, A_{1}, C$ is harmonic, polar of $ A_{3}$ wrt $ w$ passing through $ A_{1}$. So the line $ A_{2}A_{1}$ is the polar of $ A_{3}$ wrt $ w$. $ B_{3}, C_{3}$ also satisfy same property. But by Desargue's Theorem, $ A_{3}, B_{3},C_{3}$ lie on a line. Let this line $ l$. Now it is obvious that $ A_{1}A_{2}, B_{1}B_{2}, C_{1}C_{2}$ concur at a point. To show the rest part, It is enough to show that $ l \perp l_{e}$. ($ l_{e}$ denotes the Euler Line of $ ABC$) ( $ l_{e}$ contains $ O$, the circumcenter of $ ABC$) Let $ H$ be a orthocenter. Let $ M, N$ be a point such that $ MB_{1} \parallel C_{1}B_{3}$, $ NC_{1} \parallel B_{1}C_{3}$, $ M \in AB$, $ N\in AC$. Then $ \angle BMB_{1} = \angle BC_{1}A_{1} = \angle C = \angle AC_{1}B_{1}$. So $ B_{1}M = B_{1}C_{1}$. By the same way, we get $ MB_{1} = B_{1}C_{1} = C_{1}N$. Let $ O'$ be a symmetric point of $ O$ wrt $ AC$. Then $ O'$ is the circumcenter of $ AHC$. so $ OC=CO'=O'H$. Furthermore, $ \angle A_{1}O'C = 2\angle HAC =\angle MB_{1}C_{1}$ and $ \angle OO'C = \angle B = \angle BB_{1}N$. Now, $ B_{1}C_{1}MN \equiv O'CHO$. But $ NB_{1} \perp OO'$. Therefore $ MN \perp OH = l_{e}$. And $ MN \parallel B_{3}C_{3}$ ( $ \frac{AM}{AC_{1}} = \frac {AB_{1}}{AB_{3}}$, $ \frac{AN}{AB_{1}} = \frac{AC_{1}}{AC_{3}}$) Hence $ B_{3}C_{3}\perp l_{e}$, as desired.