An expansion of @Robert's way of drawing this out. First, a note: your question is a special case of a Medial Triangle where the outer triangle is specified to be right. That wiki link (and the Wolfram links in the bibliography) give a more complete explanation of the general case.
Anyway, draw your right triangle in such a way that the side lengths sum to 18. It doesn't matter which sides have what lengths so long as they sum to the specified 18 (If you're actually sketching this out on paper, don't even worry about the lengths. This is just to convince you that we're talking about your problem).
Let the vertices be $A$, $B$ and $C$ with $AB$ the hypotenuse. Mark the midpoints of each, with $A'$ the midpoint of the hypotenuse, and $B',C'$ the midpoints of the bases. Note then that $A'C'$ is parallel to $B'C$ and thus $A'B'$ is half the length of $AC$. By a similar argument note that $A'C'$ is half the length of $BC$. Finally, note that $B'C'$ (the hypotenuse of the inscribed triangle) is the hypotenuse of $\triangle C'B'C$ which is similar to $\triangle ABC$ and has bases half the length. Thus, $B'C'$ must be half of $AB$.
Hopefully this helps. It probably works best if you draw it out on graph paper, and are as precise as possible. However, be sure to convince yourself with the argument, not careful drawing and measuring. It may be helpful to think about it as proving the side lengths of the medial triangle are half of the outer triangle, rather than the worry about the specific lengths at first.