14. As shown in Figure 5, given $\square A B C D, \angle A C B=30^{\circ}$, fold $\triangle A B C$ along $A C$ so that point $B$ lands at point $E$ on the plane of $\square A B C D$, and connect $D E$. Then the value of $\frac{A C+D E}{A D}$ is $\qquad$
14. $\sqrt{3}$.

Point $B$ rotates $180^{\circ}$ around $A C$ and lands on point $E$. Points $A, B, C, D, E$ are on the same plane. $E C = B C = A D, A E = A B = C D$, so quadrilateral $A C D E$ is an isosceles trapezoid, and $\angle D A C = 30^{\circ}$. Extend $A C$ to point $F$ such that $C F = E D$, then $\triangle A D F$ is an isosceles triangle with base angles of $30^{\circ}$.
$$
\text{Therefore, } \frac{A C + D E}{A D} = \frac{A F}{A D} = \sqrt{3} \text{.}
$$