1.19. Find the ratio of the radius of the circle inscribed in an isosceles right triangle to the height dropped to the hypotenuse.
1.19. Since $\triangle ABC$ is an isosceles right triangle, the height $CD$ is also the bisector, i.e., $\angle DCA = \angle A = 45^\circ$ (Fig. 1.17); therefore, $AD = DC$ and $AC = \sqrt{2} DC$. But $AC = AK + KC = DC + r$ (since $AK = AD$ as tangents drawn from the same point),

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-422.jpg?height=477&width=448&top_left_y=1000&top_left_x=705)

Fig. 1.17 hence $r = \sqrt{2} DC - DC$, i.e., $\frac{r}{DC} = \sqrt{2} - 1$.

Answer: $(\sqrt{2} - 1): 1$.