1.53. Given a triangle $A B C$ such that $A B=15 \text{ cm}, B C=12 \text{ cm}$ and $A C=18 \text{ cm}$. In what ratio does the center of the inscribed circle divide the angle bisector of angle $C$?
1.53. Since $CD$ is the bisector

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-437.jpg?height=424&width=480&top_left_y=964&top_left_x=60)
(Fig. 1.41), then $\frac{AD}{DB} = \frac{AC}{BC} = \frac{18}{12} = \frac{3}{2}$, from which $AD = 9 \text{ cm}, DB = 6 \text{ cm}$. Next, we find the length of the bisector using formula (1.37):

$$
CD = \sqrt{12 \cdot 18 - 9 \cdot 6} = 9 \sqrt{2} \text{ cm}
$$

We draw the radii to the points of tangency and let $CF = x$. Then, using the equality of tangents drawn from the same point, we get $18 - x + 12 - x = 15$, from which $x = \frac{15}{2}$ cm. Using Heron's formula, we find

$$
S = \sqrt{\frac{45}{2} \cdot \frac{21}{2} \cdot \frac{15}{2} \cdot \frac{9}{2}} = \frac{135 \sqrt{7}}{4} \text{ cm}^2
$$

But $S = pr$, from which $r = \frac{135 \sqrt{7}}{4} : \frac{45}{2} = \frac{3 \sqrt{7}}{2}$ cm. Therefore,
$OC = \sqrt{x^2 + r^2} = \sqrt{\frac{225}{4} + \frac{63}{4}} = 6 \sqrt{2}$ cm. Since $CD = 9 \sqrt{2}$ cm, point $O$ divides the bisector in the ratio $2:1$.

Answer: $2:1$.