10.049. A point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments of length 30 and $40 \mathrm{~cm}$. Find the legs of the triangle.
Solution.

Given $\angle C=90^{\circ}, A D=30 \text{ cm}, B D=40$ cm (Fig. 10.49). Let $A C=x, B C=y$. Since the point equidistant from the sides of an angle lies on its bisector, $\frac{x}{y}=\frac{30}{40}$, i.e., $y=\frac{4 x}{3}$. But $x^{2}+y^{2}=A B^{2}$ or $x^{2}+\frac{16 x^{2}}{9}=70^{2}$, from which $x^{2}=1764$. Therefore, $x=42$ (cm), $y=56$ (cm).

Answer: 42 and 56 cm.