A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?

$\textbf{(A) }   1+\frac12 \sqrt2   \qquad        \textbf{(B) }   \sqrt3   \qquad    \textbf{(C) }   \frac74   \qquad   \textbf{(D) }  \frac{15}{8} \qquad  \textbf{(E) }   2$
First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that
\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\]
Thus, our answer is $\boxed{\textbf{(D) } \frac{15}{8}}$.

Note
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of $AB$, because $A$ must be reflected onto $B$.