[Theorem on the lengths of a tangent and a secant; the product of the entire secant and its external part [Tangent lines and touching circles

A circle passing through vertex $A$ of triangle $A B C$ touches side $B C$ at point $M$ and intersects sides $A C$ and $A B$ at points $L$ and $K$, respectively, different from vertex $A$. Find the ratio $A C: A B$, if it is known that the length of segment $L C$ is twice the length of segment $K B$, and the ratio $C M: B M=3: 2$.

#
Apply the Tangent-Secant Theorem.

## Solution

By the Tangent-Secant Theorem,

$$
C M^{2}=C L \cdot A C \text { and } B M^{2}=B K \cdot A B,
$$

therefore,

$$
\left(\frac{3}{2}\right)^{2}=\frac{C M^{2}}{B M^{2}}=\frac{C L \cdot A C}{B K \cdot A B}=\frac{C L}{B K} \cdot \frac{A C}{A B}=2 \cdot \frac{A C}{A B}
$$

Thus, $\frac{A C}{A B}=\frac{9}{8}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-20.jpg?height=421&width=640&top_left_y=104&top_left_x=709)

## Answer

$9: 8$.