$\triangle ABC$ is given, in which $AC > BC$ and the incircle $k(O)$ touches $BC$ and $AC$ in $M$ and $N$, respectively. Point $B_1$ is the image of B with respect to the line $CO$. Show that $M$ and $N$ are symmetric with a respect to $CO$ and $B_1M = BN$ and $AM > BN$.
As it is in the drawing, $B_1$ lies on $AC$, but I do not know how to show it. For the first part of the problem, I should show that $CO$ is the perpendicular bisector of $MN$. This is because $NC=MC$ (tangent segments) and $ON=OM=r$, thus points $C$ and $O$ determine the perpendicular bisector of $MN$. Let us denote $CO$ with $a$.
$\sigma_a:B \to B_1$
$\sigma_a:M \to N$ (here I am not sure if there's a difference with $N \to M$, and if there is I would be very grateful if someone could explain to me)
I don't know how to show the next things. Thank you in advance!
Better to denote the incenter by $I$, and leave $O$ for the circumcenter. Anyway:
There is no difference using $\sigma_a\colon M\mapsto N$ or $N\mapsto M$, since $\sigma_a$ is an involution (i.e., do it twice will give you back the original point). However, in the name of consistency, it is better to use $N\mapsto M$ since you used $B\mapsto B_1$ before, for the next line:
Finally, to show $AM>BN$, note that $CB_1B$ is isosceles, so $\angle AB_1B$ is obtuse. Thus $\angle AB_1M$ is also obtuse, and $$AM^2>B_1M^2+AB_1^2>B_1M^2=BN^2.$$