64. In $\triangle A B C$, the angle bisectors of $\angle C$ and $\angle A$ intersect at point $O$. Given that $A C+A O=B C, \angle B=25^{\circ}$, then $\angle A C B=$ $\qquad$.
Answer: $105^{\circ}$
Solution: On $\mathrm{BC}$, take point $\mathrm{D}$ such that $\mathrm{CD}=\mathrm{AC}$, and connect $\mathrm{OB}, \mathrm{OD}$,
It is easy to see that
$$
\triangle A O C \cong \triangle D O C,
$$

which gives us
$$
\mathrm{AO}=\mathrm{DO} \text { . }
$$

Also, since
$$
A C+A O=B C, \quad C D+D B=B C,
$$

we have
$$
\mathrm{DB}=\mathrm{AO} \text {, }
$$

which leads to
$$
\mathrm{DO}=\mathrm{DB} \text {, }
$$

thus
$$
\angle D B O=\angle D O B=\frac{1}{2} \angle A B C \text {, }
$$

so
$$
\angle C A O=\angle C D O=\angle A B C=25^{\circ} \text {, }
$$

therefore
$$
\angle A C B=180^{\circ}-25^{\circ}-50^{\circ}=105^{\circ} \text { . }
$$