Let $ABC$ be a right triangle with a right angle at $C.$ Two lines, one parallel to $AC$ and the other parallel to $BC,$ intersect on the hypotenuse $AB.$ The lines split the triangle into two triangles and a rectangle. The two triangles have areas $512$ and $32.$ What is the area of the rectangle?

[i]Author: Ray Li[/i]
1. Let $ABC$ be a right triangle with a right angle at $C$. Let $D$ lie on $AC$, $E$ on $AB$, and $F$ on $BC$. The lines $DE$ and $EF$ are parallel to $BC$ and $AC$ respectively, forming a rectangle $DEFG$ inside the triangle.

2. Given that the areas of the two triangles formed are $512$ and $32$, we need to find the area of the rectangle $DEFG$.

3. Since $\triangle ADE \sim \triangle EFB$, the ratio of their areas is the square of the ratio of their corresponding sides. Let $AC = a$ and $BC = b$. Let $AD = a - l$ and $DF = l$, where $l$ is the length of the rectangle.

4. By the property of similar triangles, we have:
   \[
   \left(\frac{a - l}{l}\right)^2 = \frac{32}{512} = \frac{1}{16}
   \]
   Solving for $\frac{a - l}{l}$, we get:
   \[
   \frac{a - l}{l} = \frac{1}{4}
   \]
   Therefore:
   \[
   a - l = \frac{l}{4}
   \]
   Solving for $a$, we get:
   \[
   a = \frac{5l}{4}
   \]

5. Now, we need to find the area of the rectangle. The area of $\triangle ADE$ is given by:
   \[
   \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times DE = 32
   \]
   Since $AD = \frac{l}{4}$ and $DE = l$, we have:
   \[
   \frac{1}{2} \times \frac{l}{4} \times l = 32
   \]
   Simplifying, we get:
   \[
   \frac{l^2}{8} = 32
   \]
   Solving for $l^2$, we get:
   \[
   l^2 = 256
   \]

6. The area of the rectangle $DEFG$ is:
   \[
   \text{Area of } DEFG = l \times w = l \times l = l^2 = 256
   \]

The final answer is $\boxed{256}$