1.6. The legs of a right triangle are 9 and 12 cm. Find the distance between the point of intersection of its angle bisectors and the point of intersection of the medians.
1.6. In $\triangle A B C$ (Fig. 1.5), we have $A B=\sqrt{9^{2}+12^{2}}=15$ (cm); the median $C D$ is half the hypotenuse, i.e., $\frac{15}{2}$ cm. Let $E$ be the point of intersection of the medians; then $C E=\frac{2}{3} C D=\frac{2}{3} \cdot \frac{15}{2}=5$ (cm) (according to the additional relations, point $1^{\circ}$). Draw $D F \perp A C$; since $D F$ is the midline in $\triangle A B C$, we have $C F=6$ cm, $D F=\frac{9}{2}$ cm. Next, draw $E K \perp A C$; then $E K \| D F$ and $\triangle C E K \sim \triangle C D F$, from which

$$
E K=\frac{2}{3} D F=\frac{2}{3} \cdot \frac{9}{2}=3(\mathrm{~cm}), C K=\frac{2}{3} C F=\frac{2}{3} \cdot 6=4(\mathrm{~cm})
$$

Let $M$ be the point of intersection of the angle bisectors of $\angle C B A$ and $\angle B C A$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-416.jpg?height=454&width=588&top_left_y=1400&top_left_x=304)

Fig. 1.5
i.e., the center of the inscribed circle in $\triangle A B C$; then $M N \perp A C$, where $r=M N$ is the radius of this circle. By formula (1.4), we have $r=\frac{S}{p}$, where $S=\frac{1}{2} \cdot 9 \cdot 12=54\left(\mathrm{~cm}^{2}\right), p=\frac{1}{2}(9+12+15)=18$ (cm); thus, $r=\frac{54}{18}=3$ (cm). Finally, considering that $\angle M C N=45^{\circ}$, i.e., $C N=M N$, we get $N K=M E=C K-C N=4-3=1$ (cm).

Answer: 1 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-417.jpg?height=352&width=585&top_left_y=678&top_left_x=46)

Fig. 1.6