In the figure, $ABCD$ is a rectangle with $AB=4$ and $BC=3$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $BE$?

Let $x=BE$ and $\alpha=\angle AJG$. Since $JG=BC=3$ and $JK=BE=x$, the vertical and horizontal constraints in the figure give
$$3\sin\alpha+x\cos\alpha=3,\qquad 3\cos\alpha+x(1+\sin\alpha)=4.$$
The first equation implies $x(1+\sin\alpha)=3\cos\alpha$. Substituting into the second gives $6\cos\alpha=4$, so $\cos\alpha=\frac23$ and $\sin\alpha=\frac{\sqrt5}{3}$. Therefore
$$x=\frac{3(1-\sin\alpha)}{\cos\alpha}=\boxed{\frac{3(3-\sqrt5)}2}.$$