4. (4th "Hope Cup" Invitational Competition Question) Rectangle $A B C D$ is congruent to rectangle $A B E F$, $D-A B-E$ is a right dihedral angle, $M$ is the midpoint of $A B$, $F M$ and $B D$ form an angle $\theta, \sin \theta=\frac{\sqrt{78}}{9}$, find $\left.\frac{A B \mid}{\mid B C} \right\rvert\,$.
4. Let $BC = l$, $\frac{|AB|}{|BC|} = \lambda$, then $AB = \lambda l$. Given $\sin \theta = \frac{\sqrt{78}}{9}$, we know $\cos \theta = \frac{\sqrt{3}}{9}$. Since $D-AB-E$ is a right dihedral angle, we have: the angle $\theta_1$ formed by $BD$ and plane $AE$ is $\theta_1 = \angle DBA$, so $\cos \theta_1 = \frac{AB}{DB} = \frac{\lambda}{\sqrt{1+\lambda^2}}, \theta_2 = \angle AMF$, hence $\cos \theta_2 = \frac{AM}{FM} = \frac{\lambda}{\sqrt{4+\lambda^2}}, \frac{\sqrt{3}}{9} = \frac{\lambda}{\sqrt{1+\lambda^2}} \cdot \frac{\lambda}{\sqrt{4+\lambda^2}}$, which simplifies to $26 \lambda^4 - 5 \lambda^2 - 4 = 0, \lambda^2 = \frac{1}{2}$. Therefore, $\lambda = \frac{\sqrt{2}}{2}$ is the solution.