[asy][asy] import olympiad; size(6cm); defaultpen(fontsize(10pt)); defaultpen(linewidth(0.4)); dotfactor *= 1.5; pair B = dir(110), A = dir(200), C = dir(340), O = origin, M = (A+C)/2, A1 = extension(O,M,B,C), C1 = extension(O,M,B,A), O1 = circumcenter(B,C1,A1); draw(A--B--C--A); draw(O--C1--B); draw(A1--O1--C1--A1^^A--O--C--A, linewidth(1)); draw(C1--foot(C1,A,O)--O^^O--M, dashed); dot("$B$", B, dir(150)); dot("$A$", A, dir(210)); dot("$C$", C, dir(330)); dot("$O$", O, dir(140)); dot("$A_1$", A1, dir(210)); dot("$C_1$", C1, dir(90)); dot("$O_1$", O1, dir(15)); [/asy][/asy] We proceed by casework on $\angle B$. If $\angle B = 90^\circ$, then $O \in \overline{AC}$ and $O_1 \in \overline{AC}$ and the conclusion is obvious since $\overline{A_1C_1} \perp \overline{AC}$. If $\angle B < 90^\circ$ (as shown in diagram above), then $O$ lies on the same side of $\overline{AC}$ as $B$, and $O_1$ lies on the opposite side of $\overline{A_1C_1}$ as $B$. If $\angle B > 90^\circ$, then $O$ lies on the opposite side of $\overline{AC}$ as $B$, and $O_1$ lies on the same side of $\overline{A_1C_1}$ as $B$. Either way, we are done if $\angle O_1C_1A_1 = \angle OAC$ since $\overline{A_1C_1} \perp \overline{AC}$ (consider rotating this pair of perpendicular lines). Since $\angle C_1O_1A_1 = \angle COA = 2\angle B$ and both $\triangle C_1OA_1$ and $\triangle COA$ are isosceles, they must be similar. This implies the conclusion.

let A1O meets AC at point M. <AOM=<ABC so ABA1O is cyclic...<BAO=<BA1C1=<BTC1 (T-C101∩circumcircle of C1BA1)...<BC1T=90-<BTC1....<BC1T+<BAO=90

Assume $\angle B< 90$, the other case is probably similar. Then, note that \[\angle C_1AO = 90 - \angle C\]and if we let $D=C_1O\cap AC$, then \[\angle AC_1O_1 = \angle AC_1D + \angle A_1C_1O_1 = (90-\angle A) + \frac{180-\angle C_1O_1A_1}{2} = (90-\angle A) + \frac{180-(2*(180-\angle C_1BA_1))}{2} \]\[=(90-\angle A) +(90-\angle B)\]Thus, \[\angle O_1C_1A+\angle C_1AO = (180-\angle A - \angle B) + (90-\angle C) = 90\]from which it clearly follows that $C_1O\perp AO$.

Note $\angle BOA=90-\angle C=\angle BA_1A$ so $A_1=\overline{BC}\cap(BAO)$ and similarly $C_1=\overline{AB}\cap(BCO).$ Hence, the inversion about $(ABC)$ sends $A_1$ to $C_1$ and thus sends $(A_1BC_1)$ to itself. Hence, $(ABC)$ and $(A_1BC_1)$ are orthogonal and $$\angle AC_1O_1+\angle OAC_1=\angle O_1BC_1+\angle ABO=90.$$$\square$