Here your solution>>>>>> Proof: Because CO ís perpendicular to A1B1 .This easy to prove it. Then C,O,C2 is colinear because that CC2 is perpendicular to A1B1. We have Angle CAO=Angle OCA( because OC=OA so triangle OAC is scalen) and Angle ACC2 =Angle CC2B1 because that B1C=C2B1 by reflection) So Angle CAO =Angle CC2B1 then OB1AC2 is cyclic. Then CO.CC2=CB1.CA. But HA1AB1 is cyclic, so CH.CA1 =CB1.CA.Then CO.CC2=CH.CA1 . So O,H,A1,C2 are concyclic

Consider the inversion through pole $ C$, power $ \overline {CA_1}\cdot \overline {CB}=\overline {CB_1}\cdot \overline {CA}=\overline {CH}\cdot \overline {CC_1}=k^2$. We have $ \mathcal {I}(C,k): A_1\mapsto B$, $ B_1\mapsto A$, hence $ A_1B_1\mapsto (O)$ Therefore $ O$ will be the image of $ C_2$, which is the reflection of $ C$ through $ A_1B_1$, through $ \mathcal {I}(C,k)$. Hence $ \overline {CC_2}\cdot\overline {CO}=k^2=\overline {CH}\cdot \overline {CC_1}\Longrightarrow C_2,O,H,C_1$ are concyclic. Our proof is conpleted then.

April wrote: Suppose $ H$ and $ O$ are the orthocenter and the circumcenter of acute triangle $ ABC$; $ AA_1$, $ BB_1$ and $ CC_1$ are the altitudes of the triangle. Point $ C_2$ is the reflection of $ C$ in $ A_1B_1$. Prove that $ H$, $ O$, $ C_1$ and $ C_2$ are concyclic. Let $ CC_2 \cap A_1B_1=H_1$. we know that $ O \in CH_1$. So, $ CO\cdot CC_2=R\cdot 2 CH_1=2R\cdot CB_1\sin B=b\cdot CB_1=CH\cdot CC_1$. So, $ C_1$ and $ C_2$ are concyclic.

Dear Mathlinkers, in order to have a proof with a minimum of calcul, we can involve the Monge theorem Sincerely Jean-Louis

Let CC2 meet A1B1 at P. ∠B1CP = 90 - ∠PB1C = 90 - ∠B and ∠B1CO = 90 - ∠B so C,O,C2 are collinear So we can prove CO.CC2 = CH.CC1 . Note that CH.CC1 = CB1.CA so we instead can prove AB1OC2 is cyclic. ∠B1AO = 90 - ∠B = ∠B1CC2 = ∠B1C2O so AB1OC2 is cyclic. we're Done.