# Solution.

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Let $MN$ be the diameter of the given semicircle. First, note that the quadrilateral $C_1AB_1O$ is cyclic, since the sum of its opposite angles is $180^\circ$. Let $\Omega$ be the circle circumscribed around the quadrilateral $C_1AB_1O$, obviously, $AO$ is its diameter. Since the segment $AO$ is seen at a right angle from points $B_1$ and $H$, the points $A, B_1, H$, and $O$ lie on the same circle, namely on the same circle $\Omega$. We introduce the following notations: $AC_1 = AB_1 = t, BC_1 = x, BM = b, OH = h$, $NC = c, CB_1 = y$, and $R$ is the radius of the semicircle with diameter $MN$. Since the semicircle with diameter $MN$ is inscribed in the angle $BAC$, $AO$ is the angle bisector of this angle. By the angle bisector theorem of a triangle, we get:

$$\frac{x+t}{y+t}=\frac{b+R}{c+R}$$

Next, consider the circle $\Omega$ and the pairs of secants $CA, CO$ and $BA, BH$ drawn from points $C$ and $B$ respectively. By the property of secants drawn from the same point, we have: $y(y+t)=(c+R-h)(c+R)$ and $x(x+t)=(b+R)(b+R+h)$, from which $c+R-h=\frac{y(y+t)}{c+R}$ and $b+R+h=\frac{x(x+t)}{b+R}$.

Thus,

$$\begin{gathered} \frac{AC_1}{C_1B} \cdot \frac{BH}{HC} \cdot \frac{CB_1}{B_1A}=\frac{t}{x} \cdot \frac{b+R+h}{c+R-h} \cdot \frac{y}{t}=\frac{y}{x} \cdot \frac{b+R+h}{c+R-h}=\frac{y}{x} \cdot \frac{x(x+t)}{b+R} \cdot \frac{c+R}{y(y+t)}= \\ =\frac{x+t}{y+t} \cdot \frac{c+R}{b+R}=\mid \text{ by equality }(1) \left\lvert\,=\frac{x+t}{y+t} \cdot \frac{y+t}{x+t}=1\right. \end{gathered}$$

Comment. It is proven that the points $H, O, C_1, A$, and $B_1$ lie on the same circle - 3 points.