Let $O$ and $L$ be the circumcenters of the triangles $ABC$ and $KMC$, respectively, $H$ the orthocenter of the triangle $ABC$. Let $S$ and $P$ be the midpoints of the segments $KM$ and $CH$, respectively. Let $R$ denote the circumradius of the triangle $ABC$. Since $A$, $B$, $A_1$, $B_1$ lie on the circle with $AB$ as its diameter, we have $\angle CB_1A_1 = 180^\circ - \angle AB_1A_1 = \angle ABC$, so $\triangle A_1B_1C \sim \triangle ABC$ and the ratio of similitude is equal to $k = CB_1/CB = \cos \angle ACB$. Since $CH$ is the diameter of circumcircle of the triangle $CB_1A_1$ ($\angle CB_1H = 90^\circ$), we have $CH = 2R \cos \angle ACB$. Since $\triangle AOB$ is isosceles and $\angle AOB = 2\angle ACB$, we have
$$OM = AO \cos \angle AOM = R \cos \angle ACB = CH/2 = CP = PH.$$

Taking into account that $PH \perp AB$ and $OM \perp AB$, we obtain $PH \parallel OM$, so $PHMO$ is a parallelogram. Since $PO$ is the line of centers of circumcircles of the triangles $CB_1A_1$ and $ABC$, we have $PO \perp CN$. Then $MH \perp CN$. Since the angle $CNH$ subtends the diameter of the circumcircle of the triangle $CB_1A_1$, we have $NH \perp CN$, so $N$, $H$, $M$ lie on the same line. Moreover, $MN$ is the altitude of $\triangle KMC$, so $H$ is the orthocenter of $\triangle KMC$ ($CH \perp AB$). Therefore $LS = CH/2 = PH = OM$, hence the lines $LO$ and $KB$ are parallel, as required.