Selection and Training Session

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 $\angle C{B}_1{A}_1=180^{\circ }-\angle A{B}_1{A}_1=\angle ABC$, so $△A_1{B}_{1}C\sim △ABC$ and the ratio of similitude is equal to $k=C{B}_1/CB=\cos \angle ACB$. Since $CH$ is the diameter of circumcircle of the triangle $C{B}_1{A}_1$ ($\angle C{B}_{1}H=90^{\circ }$), we have $CH=2R\cos \angle ACB$. Since $△AOB$ 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 $C{B}_1{A}_1$ and $ABC$, we have $PO\perp CN$. Then $MH\perp CN$. Since the angle $CNH$ subtends the diameter of the circumcircle of the triangle $C{B}_1{A}_1$, we have $NH\perp CN$, so $N$, $H$, $M$ lie on the same line. Moreover, $MN$ is the altitude of $△KMC$, so $H$ is the orthocenter of $△KMC$ ($CH\perp AB$). Therefore $LS=CH/2=PH=OM$, hence the lines $LO$ and $KB$ are parallel, as required.