5. Two circles touch each other internally at point K. A chord $AB$ of the larger circle touches the smaller circle at point $L$, and $AL=10$. Find $BL$, if $AK: BK=2: 5$.
Answer: $B L=25$.

Solution. Draw the chord $P Q$ and the common tangent $M N$ (see fig.), then

$$
\angle P Q K=\frac{1}{2} \smile P K=\angle P K M=\frac{1}{2} \smile A K=\angle A B K
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_87fbe4949747c6e4495eg-2.jpg?height=411&width=431&top_left_y=291&top_left_x=1498)

therefore $P Q \| A B$ and $\smile P L=\smile Q L$. Consequently, $\angle P K L=\angle Q K L$, i.e., $K L$ is the bisector of triangle $A K B$, from which we obtain

$$
\frac{A L}{B L}=\frac{A K}{B K}=\frac{2}{5} \Longrightarrow B L=10 \cdot \frac{5}{2}=25
$$