9.22. (GDR, 64). On the side $BC$ of triangle $ABC$, a point $P$ is taken such that $PC = 2BP$. Find $\angle ACB$ if $\angle ABC = 45^\circ$ and $\angle APC = 60^\circ$.
9.22. If point $C_{1}$ is symmetric to point $C$ with respect to line $A P$ (Fig. 18), then $C_{1} P = C P = 2 B P$ and $\angle C_{1} P B = 180^{\circ} - \angle A P C - \angle A P C_{1} = 180^{\circ} - 60^{\circ} - 60^{\circ} = 60^{\circ}$. Therefore, $\angle C_{1} B P = 90^{\circ}$ (since triangle $C_{1} P B$ is similar to a right triangle with hypotenuse 2 and leg 1), which means that $B A$ is the bisector of angle $C_{1} B P$. Thus, point $A$, which is equidistant from the lines $C_{1} P$, $P C$, and $C_{1} B$, lies on the bisector of angle $P C_{1} D$ (where point $D$ lies on the extension of segment $B C_{1}$ beyond point $C_{1}$). Therefore,

$$
\angle A C B = \angle A C_{1} P = \frac{1}{2} \left(180^{\circ} - \angle B C_{1} P \right) = \frac{1}{2} \left(180^{\circ} - 30^{\circ} \right) = 75^{\circ}
$$

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