10. In a regular $\triangle A B C$, $D$ and $E$ are the midpoints of sides $A B$ and $A C$ respectively. Fold $\triangle A B C$ along $D E$ to form a dihedral angle $A-D E-C B=60^{\circ}$. If $B C=10 \sqrt{13}$, then the distance between the skew lines $A E$ and $B D$ is $\qquad$
10. Let $M$ be the midpoint of $BC$, and the median $AM$ of $\triangle ABC$ intersects $DE$ at $N$. After folding into a dihedral angle, $\angle ANM$ is the plane angle of the dihedral angle $A-DE-CB$, i.e., $\angle ANM = 60^{\circ}$.

Connect $EM, AM, DM. \because EM \parallel DB$, $\therefore$ the distance $d$ between $AE$ and $DB$ is the distance from $DB$ to the plane $AEM$, which is also the distance from $D$ to the plane $AEM$.

Prove $AF = a$, then $EM = a, AN = NM = \frac{\sqrt{3}}{2} a, \triangle ANM$ is an equilateral triangle,
$$
\begin{array}{c}
AM = \frac{\sqrt{3}}{2} a, S_{\triangle AM} = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} a \\
\cdot \sqrt{a^{2} - \left(\frac{\sqrt{3}}{4} a\right)^{2}} = \frac{\sqrt{39}}{16} a^{2} .
\end{array}
$$

Let $O$ be the midpoint of $MN$, connect $AO$, then it is easy to see that $AO \perp$ base $DECB$, and $AO = AN \sin 60^{\circ} = \frac{3}{4} a$, thus
$$
\begin{array}{l}
d \cdot 3_{\triangle AEM} = d \cdot \frac{\sqrt{39}}{10} a = V_{A-DEM} \\
= 4O \cdot S_{\triangle DEM} = \frac{3}{4} a \cdot \frac{\sqrt{3}}{4} a^{2} . \\
d = \frac{3}{\sqrt{13}} a = \sqrt{13} \cdot 5 \sqrt{13} = 15 .
\end{array}
$$