In a right triangle the square of the hypotenuse is equal to twice the product of the legs.  One of the acute angles of the triangle is:
$ \textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ} \qquad\textbf{(E)}\ 75^{\circ} $
1. Let's denote the legs of the right triangle as \(a\) and \(b\), and the hypotenuse as \(c\). According to the problem, the square of the hypotenuse is equal to twice the product of the legs:
   \[
   c^2 = 2ab
   \]

2. By the Pythagorean theorem, we know that in a right triangle:
   \[
   c^2 = a^2 + b^2
   \]

3. Equating the two expressions for \(c^2\), we get:
   \[
   a^2 + b^2 = 2ab
   \]

4. Rearrange the equation to form a standard quadratic equation:
   \[
   a^2 + b^2 - 2ab = 0
   \]

5. Notice that the left-hand side can be factored as a perfect square:
   \[
   (a - b)^2 = 0
   \]

6. Solving for \(a\) and \(b\), we find:
   \[
   a = b
   \]

7. Since \(a = b\), the right triangle is an isosceles right triangle, where the two legs are equal. In an isosceles right triangle, the angles opposite the equal legs are equal, and the sum of the angles in a triangle is \(180^\circ\). Therefore, each of the acute angles is:
   \[
   \frac{90^\circ}{2} = 45^\circ
   \]

8. Thus, one of the acute angles of the triangle is:
   \[
   \boxed{45^\circ}
   \]