For reference:
The triangle $ABC$ is inscribed in a circle; the perpendicular bisector of $AB$ intersects $BC$ at $M$ and the extension of $AC$ at $N$. Calculate $\angle OCB$ if $O$ is the center of the circle; and $OH= OM = MN$ ($H$ is the midpoint of $AB$). (Answer: $18,5^\circ$)
My progress:
Draw $BN$.
$NO = 2OH \implies O$ is centroid $\triangle ABN$
$\angle HBY = \angle DYB\\ \angle DYN = \angle A$
$\triangle NYH $ is isosceles $\implies \angle MNY = \angle YHN$
$YM$ is a midsegment of $\triangle AON \implies YM \parallel AO\\ \therefore \angle OAY - \angle MYN$
...
Applying Menelaus's theorem in $\triangle ANH$ where traversal $CB$ intersects $NA$ and $NH$ internally and $AH$ externally.
$ \displaystyle \frac{AB}{BH} \cdot \frac{HM}{MN} \cdot \frac{NC}{CA} = 1 \implies CA = 4 NC$
Say the foot of the perp from $O$ to $CA$ is $G$ then,
$ AG = GC = \frac12 CA = 2 NC, NG = 3 NC, NA = 5 NC$
Now note that $\triangle OGN \sim \triangle AHN$.
So, $ ~ \displaystyle \frac{NH}{NA} = \frac{NG}{NO} \implies \displaystyle \frac{NH}{NA} = \frac{3NA/5}{2NH/3}$
$NH^2 = \dfrac9{10} NA^2 \implies AH^2 = \dfrac1{10} NA^2$
$NH = 3 AH \implies \angle ANH \approx 18.5^\circ$
Finally, $\angle OCB = \angle OBC = 90^\circ - \angle A = \angle ANH$