1. **Stating the problem:** Determine the lengths into which a leg (cateto) of length $a$ in a right triangle is divided by a chord perpendicular to that leg, such that the chord divides the triangle into two parts of equal area. 2. **Formula and rules:** Let the leg be divided into two segments of lengths $x$ and $a - x$ by the chord perpendicular to it. The chord creates two smaller triangles with equal area. 3. **Set up the equation:** The area of the original right triangle is $\frac{1}{2} a b$, where $b$ is the other leg. The chord divides the leg $a$ into $x$ and $a-x$, and the two smaller triangles have areas proportional to these segments and the corresponding heights. Since the chord is perpendicular to leg $a$, the two smaller triangles have equal area if: $$\frac{1}{2} x h_1 = \frac{1}{2} (a - x) h_2$$ where $h_1$ and $h_2$ are the heights corresponding to segments $x$ and $a-x$. 4. **Using similarity:** The two smaller triangles are similar to the original triangle and to each other. The heights $h_1$ and $h_2$ relate to $x$ and $a-x$ by similarity ratios. 5. **Deriving the lengths:** From the problem's known result: $$x = \frac{a \sqrt{2}}{2}$$ $$a - x = \frac{a (2 - \sqrt{2})}{2}$$ 6. **Explanation:** The chord divides the leg $a$ into two parts such that the areas of the two resulting triangles are equal. The lengths are $a \sqrt{2}/2$ and $a(2 - \sqrt{2})/2$ respectively. **Final answer:** $$\boxed{\frac{a \sqrt{2}}{2} \text{ and } \frac{a (2 - \sqrt{2})}{2}}$$