smc

First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $△ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $△ACB$ is similar to $△ADE$ 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 $\boxed{\frac{15}{8}}$.