10. A.

As shown in Figure 5, establish a coordinate system. Let
$D A=D D_{1}=a, D C=b$, then
$$\begin{array}{l} D_{1}(0,0, a), C(0, b, 0), \\ D_{1} C=(0, b,-a) ; A(a, 0, \end{array}$$
$0), M\left(0, \frac{b}{2}, 0\right), A M=$
$\left(-a, \frac{b}{2}, 0\right)$. Then
$$\begin{array}{l} \cos \theta=\frac{\boldsymbol{A M} \cdot \boldsymbol{D}_{1} \boldsymbol{C}}{|\boldsymbol{A M}| \cdot\left|\boldsymbol{D}_{1} \boldsymbol{C}\right|}=\frac{\frac{b^{2}}{2}}{\sqrt{a^{2}+\frac{b^{2}}{4}} \cdot \sqrt{a^{2}+b^{2}}} \\ =\frac{b^{2}}{\sqrt{4 a^{2}+b^{2}} \cdot \sqrt{a^{2}+b^{2}}} . \end{array}$$

From the given information,
$$1-\left(\frac{b^{2}}{\sqrt{4 a^{2}+b^{2}} \cdot \sqrt{a^{2}+b^{2}}}\right)^{2}=\left(\frac{\sqrt{78}}{9}\right)^{2} \text {. }$$

Simplifying, we get $\frac{a}{b}=\sqrt{2}$, which means $\frac{A A_{1}}{A B}=\sqrt{2}$.