Answer: $\frac{10}{3}$ cm and $\frac{8}{3}$.

Solution. Points $C$ and $C_{1}$ are symmetric with respect to the fold line. Therefore, $C C_{1} \perp K M$. From the equality of angles $A K M$ and $K M B$, it follows that the adjacent angles are also equal. Therefore, triangle $C K M$ is isosceles. Its height, lying on $C C_{1}$, will also be the bisector.

![](https://cdn.mathpix.com/cropped/2024_05_06_84fc2054d2075e5d2c9fg-02.jpg?height=505&width=784&top_left_y=1258&top_left_x=633)

Thus, $C C_{1}$ is the bisector of triangle $A C B$. By the Angle Bisector Theorem,

$$
\frac{A C_{1}}{C_{1} B}=\frac{A C}{C B}=\frac{5}{4}
$$

Therefore, $A C_{1}=\frac{5}{9} A B=\frac{10}{3}$ cm, $C_{1} B=\frac{4}{9} A B=\frac{8}{3}$ cm.

Evaluation. 12 points for a correct solution.
3. A triangle $ABC$ with side lengths $AB=6 \text{~cm}, BC=5 \text{~cm}, CA=4 \text{~cm}$ has been cut out of paper. It was folded along a line so that vertex $C$ ended up at point $C_{1}$ on side $AB$. Moreover, in the resulting quadrilateral $AKMB$, two angles adjacent to the fold line $KM$ turned out to be equal. Find $AC_{1}$ and $C_{1}B$.