1.2. A point on the hypotenuse, equidistant from both legs, divides the hypotenuse into segments of 30 and 40 cm. Find the legs of the triangle.
1.2. By the condition, $\angle C=90^{\circ}, A D=30 \text{ cm}$, $D B=40$ cm (Fig. 1.1). Let $A C=x, B C=y$. Since the point equidistant from the sides of the angle lies on its bisector, then $\frac{x}{y}=\frac{30}{40}$, i.e., $y=\frac{4 x}{3}$. But $x^{2}+y^{2}=A B^{2}$ and, therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-414.jpg?height=398&width=315&top_left_y=1447&top_left_x=825)

Fig. 1.1
$x^{2}+\frac{16 x^{2}}{9}=70^{2}$, from which $x^{2}=1764$. Thus, $x=42$ (cm), then $y=\frac{4}{3} \cdot 42=56(\text{ cm})$

Answer: 42 and 56 cm.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-415.jpg?height=295&width=298&top_left_y=414&top_left_x=59)

Fig. 1.2