orl wrote: A triangle $ ABC$ is given. Let $ B_1$ be the reflection of $ B$ across the line $ AC$, $ C_1$ the reflection of $ C$ across the line $ AB$, and $ O_1$ the reflection of the circumcentre of $ ABC$ across the line $ BC$. Prove that the circumcentre of $ AB_1C_1$ lies on the line $ AO_1$. Let $ (O)$ be circumcircle of $ \triangle ABC$. Let parallel to $ BC$ through $ A$ cut $ (O)$ again at $ Z.$ By Ptolemy for isosceles trapezoid $ AZBC,$ $ AZ \cdot BC = AB^2 - AC^2.$ Let $ D \equiv BC_1 \cap CB_1.$ $ \triangle ABC \cong \triangle ABC_1 \cong \triangle AB_1C$ are congruent $ \Longrightarrow$ $ \angle BDC = 2\pi - (3 \angle A + \angle B + \angle C) = \pi - 2 \angle A = \pi - \angle COB$ $ \Longrightarrow$ $ OBDC$ is cyclic, let $ (P)$ be its circumcircle. Let $ CD$ cut $ (O)$ again at $ E$ $ \Longrightarrow$ $ \angle BDE = \pi - 2 \angle A,$ $ \angle DEB = \angle A$ $ \Longrightarrow$ $ \triangle BDE$ is isosceles with $ BD = ED$ and $ \angle CBE = \pi - (\angle A + 2 \angle B) = \angle C - \angle B.$ Consequently, $ AZ = CE = CD - ED = CD - DB.$ Let $ K, L$ be midpoints of $ BD, CD,$ respectively. Then $ AB_1^2 - AC_1^2 + KB_1^2 - KD^2 + LD^2 - LC_1^2 = AB^2 - AC^2 + BC_1 \cdot DB - CB_1 \cdot DC =$ $ = AZ \cdot BC - BC \cdot (CD - DB) = AZ \cdot BC - BC \cdot CE = 0$ By Carnot theorem for $ \triangle DB_1C_1,$ perpendiculars to $ B_1C_1, C_1D, DB_1$ from $ A, L, K$ are concurrent at $ P.$ Let $ N$ be 9-point circle center of $ \triangle ABC$; $ N$ is midpoint of $ AO_1.$ Kosnita point $ Ko \equiv N^*$ is isogonal conjugate of $ N$ (well known, see hidden stuff below) and $ Ko \in AP.$ Triangles $ \triangle ABC, \triangle AB_1C_1$ have common internal bisector of $ \angle A.$ Since $ AKoP$ is A-altitude of $ \triangle AB_1C_1$ passing through its orthocenter $ \Longrightarrow$ its isogonal $ ANO_1$ WRT $ \angle C_1AB_1$ passes through its circumcenter. $ \square$

Let us to present an elementary solution of this nice problem. We denote as $ M,\ N,$ the midpoints of the side-segments $ AC,\ AB$ respectively and let be the points $ D\equiv AC_{1}\cap OM$ and $ E\equiv AB_{1}\cap ON.$ It is easy to show that $ DE\parallel B_{1}C_{1}$ $ ,(1)$ because of $ \frac {AD}{AE} = \frac {AC}{AB} = \frac {AC_{1}}{AB_{1}}$ from the similar right triangles $ \bigtriangleup MAD,\ \bigtriangleup NAE$ and the similar isosceles triangles $ \bigtriangleup ABB_{1},\ \bigtriangleup ACC_{1}.$ So, from $ (1)$ we conclude that $ DE\parallel KL$ $ ,(2)$ where $ K,\ L,$ are the midpoints of the segments $ AC_{1},\ AB_{1},$ respectively. $ \bullet$ Let $ BB',\ CC'$ be, the altitudes of the given triangle $ \bigtriangleup ABC$ $ ($ where $ B'\equiv AC\cap BB_{1}$ and $ C'\equiv AB\cap CC_{1}$ $ )$ and we denote the points $ P\equiv CC'\cap OM$ and $ Q\equiv BB'\cap ON.$ Because of $ \angle CPM = 90^{o} - \angle PCM = 90^{o} - \angle OCB = 90^{o} - \angle O_{1}CB = \angle CO_{1}O,$ we conclude that the quadrilateral $ OPCO_{1}$ is cyclic and then we have that $ \angle OPO_{1} = \angle OCO_{1}$ $ ,(3)$ From $ (3)$ $ \Longrightarrow$ $ \angle OPO_{1} = 2\angle OCB = 2\angle PCA = \angle CAD$ $ \Longrightarrow$ $ \angle MAD = \angle MPK'$ $ ,(4)$ where $ K'\equiv AD\cap O_{1}P.$ From $ (4)$ and because of $ PM\perp AC,$ we conclude that $ PK'\equiv O_{1}K'\perp AD$ and by the same way we can prove that $ QL'\equiv O_{1}L'\perp AE,$ where $ L'\equiv AE\cap O_{1}Q.$ $ \bullet$ Because of the similar right triangles $ \bigtriangleup K'PD,\ \bigtriangleup L'QE$ $ ($ from $ \angle K'PD = \angle MAD = \angle NAE = \angle L'QE$ $ ),$ we have that $ \frac {K'D}{L'E} = \frac {PD}{QE}$ $ ,(5)$ Because of the similar triangles $ \bigtriangleup APD,\ \bigtriangleup AQE,$ $ ($ from $ \angle PDA = \angle QEA$ and $ \angle PAD = \angle PAC + \angle CAD = \angle PCA + \angle CAD = \angle QBA + \angle BAE$ $ \Longrightarrow$ $ \angle PAD = \angle QAB + \angle BAE = \angle QAE$ $ ),$ we have that $ \frac {AD}{AE} = \frac {PD}{QE}$ $ ,(6)$ From $ (5),\ (6)$ $ \Longrightarrow$ $ \frac {K'D}{L'E} = \frac {AD}{AE}$ $ \Longrightarrow$ $ K'L'\parallel DE$ $ ,(7)$ From $ (2),\ (7)$ $ \Longrightarrow$ $ KL\parallel K'L'$ $ \Longrightarrow$ $ \frac {AK}{AK'} = \frac {AL}{AL'}$ $ (8)$ $ \bullet$ We consider now, the points $ K,\ A,\ K',$ as the orthogonal projections of $ O',\ A,\ O_{1}$ respectively, on the line segment $ AC_{1},$ where $ O'$ is the circumcenter of the triangle $ \bigtriangleup AB_{1}C_{1}.$ Similarly, we consider the points $ L,\ A,\ L',$ as the orthogonal projections of $ O',\ A,\ O_{1}$ respectively, on the line segment $ AB_{1}.$ Hence, because of $ (8),$ we conclude that the points $ O',\ A,\ O_{1},$ are collinear and the proof is completed. $ ($ As well known, if the orthogonal ( or parallel ) projections of three given points, on two lines intersect each other, form segments with the same ratio then, these points are collinear with the same ratio $ ).$ $ \bullet$ This proof is dedicated to Jean-Louis Ayme. Kostas Vittas.

Dear Kostas, thank you for your dedication. Your attached figure is very interesting and has much more to say. Sincerely Jean-Louis

First we will prove the lemma: Let $ ABC$ be a triangle and $ H$ be its orthocenter.$ BH,CH$ meet $ AC,AB$ at $ B_1,C_1$.$ B_2,C_2$ are reflection points of $ A$ through $ B_1,C_1$.$ P$ be the reflection point of H through $ B_2C_2$.Let the reflections of $ BH,CH$ through $ AB,AC$ meet at $ Q$.Then $ A,P,Q$ are collinear. We see that $ PB_2 \| AC_3,PC_2 \| PB_3,B_2C_2 \| B_3C_3$.Then $ AP,B_2C_3,B_3C_2$ are concurrent.Now we need to prove that $ AQ,B_2C_3,B_3C_2$ are also concurrent. Let $ AB \cap CQ = B_4,AC \cap BQ = C_4$.We have $ \angle{B_4BC_4} = \angle {C_4CB_4}$ then $ B,C,B_4,C_4$ lie on a circle.Apply Reim's theorem to the tranversals $ BC_1B_4,CB_1C_4$ we conclude that $ B_4C_4 \| B_1C_1$.Let $ B_1C_1 \cap B_2C_2 = R$ in the infinity line,then $ B_4,C_4,R$ are collinear.Apply Desargue's theorem to the triangles $ AB_2C_2$ and $ QC_3B_3$ we have $ AQ,B_2C_3,B_3C_2$ are concurrent. Return to our problem,apply the previous lemma to triangle $ AB'C'$, we have completed the proof. Figure of the lemma: Image not found

Let $ H$ be the orthocenter of $ ABC$. Let $ AB_1$ and $ CH$ meet at $ P$, and $ AC_1$ and $ BH$ meet at $ Q$. Let $ K_1$ and $ K_2$ be the circumcircles of $ BHC$ and $ AB_1C_1$, respectively. We have $ \angle HBC = 90^{o} - \angle BCA$ and $ \angle BCH = 90^{o} - \angle ABC$, so $ 180^{o} - \angle CHB = \angle CAB$. Now as $ O$ is the circumcentre of $ ABC$, $ \angle COB = 2\angle CAB$. Therefore $ \angle BO_1C = 2(180^{o} - \angle CHB)$. Using that $ O_1C = O_1B$, we get that $ O_1$ must be the center of $ K_1$. Consider the quadrilateral $ B_1AHC$. We have $ \angle HCA = 90 - \angle CAB = \angle ABH = \angle HB_1A$. Therefore $ B_1AHC$ is cyclic. Similarly, $ C_1AHB$ is cyclic. Let $ K_3$ and $ K_4$ be their circumcircles, respectively. Now, the radical axis of $ K_2$ and $ K_3$ is $ aB_1$, and the radical axis of $ K_1$ and $ K_3$ is $ CH$. Then $ P$ is the radical center of $ C_1$, $ C_2$ and $ C_3$, and it must be in the radical axis of $ C_1$ and $ C_2$. Similarly we can get that $ Q$ lies on the radical axis of those circumferences. Therefore, $ PQ$ is perpendicular to $ O_1O_2$. Now, $ \angle QC_1P = \angle HCA = 90 - \angle CAB = \angle ABH = \angle QB_1P$. Then $ PQB_1C_1$ is cyclic, and $ AQP$ is similar to $ AC_1B_1$. As $ \angle C_1AO_2 = 90^{o} - \angle AB_1C_1 = 90^{o} - \angle PQA$, we get that $ O_2A$ is perpendicular to $ PQ$. But then both $ O_2A$ and $ O_2O_1$ are perpendicular to $ PQ$, and $ A$, $ O_1$ and $ O_2$ are collinear.

Solution. First, let's introduce some new points: \[ \bigoplus \ \ \ \boxed{\begin{array}{l} M - \mathrm{ midpoint \ of }\ [AC], \ N - \mathrm{midpoint \ of}\ [AB], \ P - \mathrm{midpoint \ of} \ [BC] \\ \\ M_1 - \mathrm{midpoint \ of}\ [AC_1], \ N_1 - \mathrm{midpoint \ of}\ [AB_1] \\ \\ O_2 = \mathrm{sim}_{AC}\ O, \ O_3 = \mathrm{sim}_{AB}\ O, \ O' = O_3 M_1 \ \cap\ O_2 N_1 \\ \\ S = \mathrm{pr}_{O'M_1} \ O_1, \ T = \mathrm{pr}_{O'N_1} \ O_1\end{array}}\] Now, it's obvious to observe that $ O'$ is the circumcenter of $ \triangle{AB_1 C_1}.$ We have: $ O_1\in AO' \ \Longleftrightarrow\ \frac {\mathrm{d}(A;O'O_3)}{\mathrm{d}(A;O'O_2)} = \frac {\mathrm{d}(O_1;O'O_3)}{\mathrm{d}(O_1;O'O_2)} \ \Longleftrightarrow\ \frac {AM_1}{AN_1} = \frac {OS}{OT} \ \Longleftrightarrow$ $ \Longleftrightarrow \ \frac {AC}{AB} = \frac {O_1O_3}{O_1O_2}\cdot\frac {\cos{\widehat{SO_1O_3}}}{\cos{\widehat{TO_1O_2}}}$ $ \bullet$ Note that: $ \triangle{ABC}\ \equiv\ \triangle{O_1 O_2 O_3}\ \Longrightarrow\ \frac {O_1O_3}{O_1O_2} = \frac {AC}{AB}\dots\dots (*)$ $ \bullet$ We also have: $ \angle{SO_1O_3} = \angle{(O_3O_1;SO_1)} = \angle{(AC; AC_1)} = \angle{(AB_1; AB)} =$ $ = \angle{(O_1T; O_1O_2)} = \angle{TO_1O_2}\ \Longrightarrow\ \cos{\widehat{SO_1O_3}} = \cos{\widehat{TO_1O_2}}\dots\dots (**)$ According to $ (*)$ and $ (**)$ we get the result. I denote: $ \boxed{\begin{array}{l} \mathrm{sim_{YZ}\ X} - \mathrm{the \ reflection \ of} \ X \ \mathrm{across \ line} \ YZ \\ \\ \mathrm{d(X,YZ)} - \mathrm{the\ distance\ between}\ X \ \mathrm{and} \ YZ \\ \\ \mathrm{pr_{YZ}\ X} - \mathrm{the \ projection\ of\ point}\ X\ \mathrm{onto\ line}\ YZ \end{array}}$

Quote: A triangle $ ABC$ is given. Let $ B_1$ be the reflection of $ B$ across the line $ AC$, $ C_1$ the reflection of $ C$ across the line $ AB$, and $ O_1$ the reflection of the circumcentre of $ ABC$ across the line $ BC$. Prove that the circumcentre of $ AB_1C_1$ lies on the line $ AO_1$. Proposed by A. Akopyan We note that if we let $ ABC$ be a triangle with $ H$ is its orthocenter. Denote $ H_a,H_b,H_c$ respectively by the projections of $ H$ onto $ BC,CA,AB$. Let $ P_b,P_c$ respectively be the intersections of $ H_aH_c$ with $ HB$; $ H_aH_b$ with $ HC$. Let $ K_b,K_c$ respectively be the intersections of $ (HAB)\cap ([AC])$; $ (HAC)\cap ([AB])$. Denote $ I_a,O_a,$ $ \mathcal {E}$ respectively by the circumcenters of $ (AK_bK_c)$, $ (HBC)$ and the Euler circle wrt $ \triangle ABC$. Then $ \mathcal {E}$ $ ,O_a,I_a\in d_a$. (Where $ ([PQ])$ here means the circle with diameter $ PQ$) Indeed, consider the inversion through pole $ A$, power $ k = \overline {AH}\cdot \overline {AH_a}$ $ = \overline {AH_c}\cdot \overline {AB}$ $ = \overline {AH_b}\cdot \overline {AC}$. We have $ \mathcal {I}(A,k):$ $ H_c\mapsto B,$ $ H_b\mapsto B,$ $ H\mapsto H_a$. Therefore by $ \mathcal {I}(A,k)$, $ H_cH_a\mapsto (AHB)$, $ H_aH_b\mapsto (AHC)$, $ HB\mapsto ([AC])$ and $ HC\mapsto ([AB])$. As the result, $ P_b\mapsto K_b$ and $ P_c\mapsto K_c$ $ \Longrightarrow$ $ P_bP_c\mapsto (AK_bK_c)$, which leads to the fact that $ I_a\in d_a$ $ (*)$. In the other hand, we have $ \overline {P_bK_b}\cdot \overline{P_bA} =$ $ \overline {P_bH_c}\cdot \overline {P_bH_a} =$ $ \overline {P_bH}\cdot \overline {P_bB}$ $ \Longrightarrow$ $ \mathcal {P}_{P_b/(AK_bK_c)} =$ $ \mathcal {P}_{P_b/(H_aH_bH_c)} =$ $ \mathcal {P}_{P_b/(HBC)}$. With the same argument for $ P_c$. Then it is followed that $ P_bP_c$ is the radical axes of $ (H_aH_bH_c)$, $ (HBC)$ and $ (AK_bK_c)$ $ \Longrightarrow$ $ \mathcal {E},$ $ O_a$ and $ I_a$ are collinear and $ \overline {\mathcal {E}O_aI_a}\perp P_bP_c$ $ (**)$. From $ (*)$ and $ (**)$, we obtain result. $ \square$ Back to our problem. Let $ K_b,K_c$ respectively be the intersections of $ (HAB)\cap ([AC])$; $ (HAC)\cap ([AB])$ and $ H_a,H_b,H_c$ respectively be the projections from $ H$ onto $ BC,CA,AB$.In order to prove the circumcenter $ I_a$ of $ \triangle AB_1C_1$ $ \in AO_1$ and also by the notice mentioned above, which is $ A,O_a,O_1$ are collinear- where $ O_a$ is the circumcenter of $ \triangle AK_bK_c$; we will prove that $ A,I_a,O_a$ are collinear. Indeed: Consider the inversion through pole $ A$, power $ k = \overline {AH}\cdot \overline {AH_a}$ $ = \overline {AH_c}\cdot \overline {AB}$ $ = \overline {AH_b}\cdot \overline {AC}$. We have $ \mathcal {I}(A,k):$ $ (AK_bK_c)\mapsto K_bK_c$ $ \equiv P_bP_c$ $ (*)$- where $ P_b,P_c$ respectively are the intersections of $ H_aH_c,HB$ and $ H_aH_b,HC$. In the other hand, since $ C_1$ is the reflection of $ C$ wrt $ AB$ $ \Longrightarrow$ $ (C_1A,C_1B)$ $ \equiv (CB,CA)$ $ \equiv (HA,HB)$ $ \pmod \pi$, which implies that $ C_1\in (HAB)$, also note that $ C_1\in CH_c$. Now, we have known that $ \mathcal {I}(A,k):$ $ (AHB)\mapsto H_cH_a$ and $ CH_c\mapsto ([AB])$. Therefore if $ C_1\mapsto Q_c$ then $ Q_c\equiv H_aH_c\cap ([AB])$. With the same argument, we also conclude $ Q_b\equiv H_aH_b\cap ([AC])$. $ \mathcal {I}(A,k):$ $ (AC_1B_1)\mapsto Q_cQ_b$ $ (**)$. In the other hand, we have $ (H_bA,H_bQ_c)\equiv (H_aA,H_aQ_c)$ $ \pmod \pi$ but $ (H_aA,H_aH_c)\equiv (CA,CH_c)$ $ \equiv (H_bA,CH_c)$ $ \pmod \pi$ $ \Longrightarrow$ $ CH_c\| Q_cH_b$. With the same argument, we also have $ Q_bH_c\| P_bH_b$. Therefore, by the Pappus theorem apply for $ \begin{vmatrix}H_c & Q_c & P_b \\ H_b & P_c & Q_b\end{vmatrix}$, we have $ H_cP_c\cap Q_cH_b\equiv \infty$, $ H_cQ_b\cap P_bH_b\equiv \infty$, thus $ Q_cQ_b\cap P_bP_c\equiv \infty$, which implies that $ Q_cQ_b\|P_cP_c$. From $ (*)$ and $ (**)$, we conclude that $ (AB_1C_1)$ tangents to $ (AK_bK_c)$ at $ A$, which leads to the fact that $ A,I_a,O_a$ are collinear. Our proof is completed then. $ \square$

It’s easy to see that $ AO_1$ passes through the center $ N$ of nine-point circle of triangle $ ABC.$ We want to have the center of $ (AB_1C_1)$ lies on $ AN$ therefore the A-altitude of triangle $ AB_1C_1$ must pass through the isogonal conjugate point $ K$ of $ N$ wrt triangle $ ABC$ ( Kosnita point). We know that $ AK$ passes through the circumcenter $ J$ of triangle $ BOC$. So $ AJ\perp B_1C_1 \Leftrightarrow AB^2_1 - AC^2_1 = JB^2_1 - JC^2_1$ $ \Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = (B_1C^2 + CJ^2 - 2B_1C.JC.\cos \angle B_1CJ - BC_1^2 - BJ^2 + 2BC_1.BJ.\cos \angle C_1BJ)$ $ \Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = 2BC.R_{BOC}(\cos \angle C_1BJ - \cos \angle B_1CJ)$ $ \Leftrightarrow 4R^2(\sin^2 C - \sin^2 B) = 2R.\sin A.\frac {R}{\cosh}.(\cos(2A + 2B - 90^o) - \cos (2A + 2C - 90^o))$ $ \Leftrightarrow 2(\sin C - \sin B)(\sin C + \sin B) = \tan A(\sin2B - \sin2C)$ (right!) We are done.

Sorry for digging up old posts but here's my solution to this problem. See attachment..

if $M$ is the midpoint of $AB$ and $D=BB_{1}\cap AC$ then clearly $AMO\sim ADC$ where $O$ is the circumcenter of $ABC$. so $BO*BD=BM*BC$ so $BB_{1}*BO=BA*BC$ but $\angle O_{1}BA=\angle B_{1}BC_{1}$ so $BB_{1}C_{1}\sim BO_{1}A$ and $\angle BAO_{1}=\angle BB_{1}C_{1}$ if $L$ is the circumcenter of $AB_{1}C_{1}$ then $\angle C_{1}AO_{1}=\angle C_{1}AB+\angle BAO_{1}=\angle BAC+\angle BB_{1}C_{1}=90-\angle BB_{1}A+\angle BB_{1}C_{1}=90-\angle C_{1}B_{1}A=\angle C_{1}AL$ so $L$ is on $AO_{1}$

Well, I used the same similar triangle pair of leader, but my finishing is a bit different.

This problem is the combination of two properties of nine point center $ (1) $ Let $ O' $ be the reflection of the circumcenter $ O $ of $ \triangle ABC $ in $ BC $ . Then the nine point center of $ \triangle ABC $ lie on $ AO' $ . $ (2) $ Let $ A', B', C' $ be the reflection of $ A, B, C $ in $ BC, CA, AB $ . Then $ \triangle A'B'C' $ is homothety with the pedal triangle of the nine point center of $ \triangle ABC $ .

Just consider point $F$ such that $<FO1B=<BAC$ and $<O1BF=<CBA$.Now,we have spiral similarity with center B that pictures $BFO1$ in $BCA$ so we have that $BO1A$ is similar to $BFC$ and we have that $FO1C$ is similar to $AB1C1$,so from here it is pure angle chasing,so we are finished.

An extension. Let $ABC$ be a triangle inscribed incircle $(O)$. $P$ is an any point. $PC,PB$ cut $(O)$ again at $M,N$. $E,F$ are reflections of $B,C$ through $AM,AN$. $K,L$ are circumcenters of triangles $AEF$ and $PBC$. $G$ is inversion image of $L$ through $(O)$. Prove that $\angle KAG=180^\circ-|\angle NAB-\angle MAC|$. See the figure in the case $P$ outside $(O)$. When $PB,PC$ are tangent to $(O)$ we get Tuymaada problem.

buratinogigle wrote: An extension. Let $ABC$ be a triangle inscribed incircle $(O)$. $P$ is an any point. $PC,PB$ cut $(O)$ again at $M,N$. $E,F$ are reflections of $B,C$ through $AM,AN$. $K,L$ are circumcenters of triangles $AEF$ and $PBC$. $G$ is inversion image of $L$ through $(O)$. Prove that $\angle KAG=180^\circ-|\angle NAB-\angle MAC|$. By simple angle chasing $ \Longrightarrow $ $ \measuredangle EMP $ $ = $ $ \measuredangle FNP, $ so combining $ \tfrac{EM}{PM} $ $ = $ $ \tfrac{BM}{PM} $ $ = $ $ \tfrac{CN}{PN} $ $ = $ $ \tfrac{FN}{PN} $ and $ \triangle BPL $ $ \stackrel{+}{\sim} $ $ \triangle BMO $ $ \stackrel{+}{\sim} $ $ \triangle BEA $ we get $ \triangle PME $ $ \stackrel{+}{\sim} $ $ \triangle LOA $ $ \stackrel{+}{\sim} $ $ \triangle PNF, $ hence $ \measuredangle (MN, EF) $ $ = $ $ \measuredangle (OL,AL) $ $ \Longrightarrow $ $ \measuredangle (AL, \perp EF) $ $ = $ $ \measuredangle (BC, MN) $ $ (\star) $ $.$ On the other hand, it's easy to see that the angle between the bisector of $ \angle BAC $ and $ \angle EAF $ is equal to $ \measuredangle (BC, MN), $ so from $ (\star) $ and notice $ AG, $ $ AL $ are isogonal conjugate WRT $ \angle A $ (well-known) we conclude that $ \measuredangle (AG, AK) $ $ = $ $ \measuredangle (BC, MN). $

Another proof. Let $O_a, O_b, O_c, J$ be the circumcenters of triangles $BHC, CHA, AHB$, $AB_1C_1$. By simple angle chasing, $\angle O_bOO_c=\angle OO_bJ=\angle OO_cJ=180^\circ-\angle BAC$. From lemma in my post at here, we get $J$ lies on Euler line of triangle $OO_bO_c$ or $J,A,O_a$ are collinear.

Let $B_2,C_2,C_3,B_3$ be projections of $B,C,C_1,B_1,$ onto lines $AC_1,AB_1.$ Also let $U \equiv B_2B \cap C_2C.$ Easy angle chasing shows us that $\angle O_1BC=\angle O_1BC=\angle ACC_2=\angle ABB_2=\frac{\pi}{2}-\alpha$ while $\angle B_2AB=\angle CAC_2=\alpha.$ Hence by Jacobi theorem on hexagon $AB_2BO_1CC_2$ we deduce that $B_2C,BC_2,AO_1$ concur at $V.$ By Dual of Desargues Involution theorem it follows that $AB_2 \mapsto AC_2, AB \mapsto AC, AU \mapsto AV$ is involution which then coincides with reflection over A-angle bisector. Hence $\angle BAV=\angle CAU$ so easy angle chasing gives us $AV \perp B_2C_2.$ Since $V$ lies on $AO_1$ it follows that $AO_1 \perp B_2C_2.$ Note that $\frac{AB_2}{AC_2}=\frac{AB}{AC}=\frac{AB_1}{AC_1} \implies C_1B_2C_2B_1$ is cyclic. But $C_1C_3B_3B_1$ is also cyclic, hence by Reim's theorem it follows that $B_2C_2 \parallel B_3C3 \implies AO_1 \perp B_3C_3 \implies$ center of $\odot(AB_1C_1)$ lies on $AO_1$ as desired.

Complex numbers rule!!! Let $(ABC)$ be the unit circle.Denote with $x$ the affixe of point $X$.We have: $$b_1=a+c-a\overline{b}c,c_1=a+b-ab\overline{c},o_1=b+c$$By making a translation which takes $A$ to the origin we have to show that the circumcenter $o_2$ of $0(b_1-a)(c_1-a)$ lies on the line $0(o_1-a)$. It is known that the circumcenter of $0xy$ is $o=\frac{xy(\overline{x}-\overline{y})}{\overline{x}y-x\overline{y}}$,so $o_2=\frac{(c-a\overline{b}c)(b-ab\overline{c})(\overline{c-b+a(b\overline{c}-\overline{b}c)})}{(\overline{c-a\overline{b}c})(b-ab\overline{c})-(c-a\overline{b}c)\overline{b-ab\overline{c}}}=\frac{cb(1-\frac{a}{b})(1-\frac{a}{c})\frac{b-c}{bc}(1-\frac{\overline{a(b+c)}}{\overline{bc}})}{\frac{b}{c}(1-\frac{b}{a})(1-\frac{a}{c})-\frac{c}{b}(1-\frac{c}{a})(1-\frac{a}{b})}=\frac{\frac{b-c}{bc}(-1+\frac{\frac{b+c}{bca}}{\frac{1}{bc}})}{\frac{b}{ac^2}-\frac{c}{ab^2}}=\frac{bc(b-c)(b+c-a)}{b^3-c^3}=(b+c-a)\cdot \frac{bc}{b^2+bc+c^2}$. Now,just note that $\frac{o_2}{o_1-a}=\frac{bc}{b^2+bc+c^2}=\frac{\frac{1}{bc}}{\frac{1}{b^2}+\frac{1}{bc}\frac{1}{c^2}}=\overline{\frac{bc}{b^2+bc+c^2}}\Rightarrow \frac{o_2}{o_1-a}\in\mathbb{R}\Rightarrow o_2 \text{ is on } 0(o_1-a)$ $\square$