A question in textbook says that in a circle, $BN$ is a diameter, $BONC$ and $AMC$ are straight lines. If $MC = OB$, prove that $\angle ABC + \frac{3}{2}\angle ACB = 90^\circ$ .
Drawn diagram which might not be correct:
It is apparent that $\angle ABC + \angle ANB = 90^\circ$, so this leaves the question of why $\angle ANB = \frac{3}{2}\angle ACB$.
Let $\angle NOM=\alpha$. Then $\angle NAM=\alpha/2$, because $NAM$ is inscribed angle based on an arc $NM$. From triangle $CAN$ we have that $\angle CNA=180^{\circ}-3\alpha/2$ and hence $\angle ANB=3\alpha/2$.