10. (14 points) As shown in Figure 1, in the right triangular prism $A B C-A_{1} B_{1} C_{1}$, it is known that $\angle B A C=90^{\circ}, A B=a, A C=2, A A_{1}=1$, and point $D$ is on edge $B_{1} C_{1}$, with $B_{1} D: D C_{1}=1: 3$.
(1) Prove: $B D \perp A_{1} C$;
(2) For what value of $a$ is the dihedral angle $B-A_{1} D-B_{1}$ equal to $60^{\circ}$?
10. Draw $D E / / A_{1} B_{1}$ intersecting $A_{1} C_{1}$ at point $E$. Then $D E \perp A_{1} C_{1}$.
From the right triangular prism $A B C-A_{1} B_{1} C_{1}$, we know that plane $A_{1} B_{1} C_{1} \perp$ plane $A_{1} C$.
Therefore, $D E \perp$ plane $A_{1} C$.
Connect $A E$. Then $A E$ is the projection of $B D$ on plane $A_{1} C$. In the rectangle $A_{1} C_{1} C A$, we calculate that $A E \perp A_{1} C$.
By the three perpendiculars theorem, we get $B D \perp A_{1} C$.
(2) Draw $B_{1} F \perp A_{1} D$, with the foot of the perpendicular at $F$, and connect $B F$. Then $B F \perp A_{1} D$,
$\angle B F B_{1}$ is the plane angle of the dihedral angle $B-A_{1} D-B_{1}$.
Thus, $\angle B F B_{1}=60^{\circ}, B_{1} F=\frac{\sqrt{3}}{3}$.
Therefore, $A_{1} D=\sqrt{A_{1} E^{2}+D E^{2}}=\frac{\sqrt{9 a^{2}+4}}{4}$.
Also, $S_{\triangle A_{1}, B_{1} D}=\frac{1}{4} S_{\triangle A_{1} B_{1} C_{1}}=\frac{a}{4}$, so
$\frac{1}{2} A_{1} D \cdot B_{1} F=\frac{1}{2} \times \frac{\sqrt{9 a^{2}+4}}{4} \times \frac{\sqrt{3}}{3}=\frac{a}{4}$.
Solving for $a$, we get $a=\frac{2 \sqrt{3}}{3}$.
Thus, when $a=\frac{2 \sqrt{3}}{3}$, the dihedral angle $B-A_{1} D-B_{1}$ is $60^{\circ}$.