In the solution, we use directed lengths on the line of the diameter. Let $A'$, $B'$, $C'$ be the feet of the perpendiculars dropped onto the diameter from $A$, $B$, $C$. We have
$$\frac{ZB'}{ZA'} = \frac{|SB'|}{|RA'|} = \frac{|SB'|}{|CC'|} \cdot \frac{|CC'|}{|RA'|} = \frac{MB'}{MC'} \cdot \frac{NC'}{NA'}$$
Now let $OA' = a$, $OB' = b$ and $OC' = c$, thus we have $OM = r^2/a$ and $ON = r^2/b$ where $r$ is the radius of the semicircle. In this notation, the above equality reads
$$\frac{OZ - b}{OZ - a} = \frac{r^2/a - b}{r^2/a - c} \cdot \frac{r^2/b - c}{r^2/b - a} = \frac{r^2/c - b}{r^2/c - a}$$
Now since $a \neq b$, one has $OZ = r^2/c$ which shows that $ZC$ is tangent to the semicircle.