10.076. Find the ratio of the radius of the circle inscribed in an isosceles right triangle to the height drawn to the hypotenuse.
## Solution.

Since $\triangle A B C$ is an isosceles right triangle, the height $C D$ is also the bisector, i.e., $\angle D C A=\angle A=45^{\circ}$ (Fig. 10.74); therefore, $A D=D C$ and $A C=\sqrt{2 D C}$. But $A C=A K+K C=D C+r(A K=A D$ as tangents drawn from the same point), hence $r=\sqrt{2} D C-D C$, i.e., $r / D C=\sqrt{2}-1$.

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