2. In the convex quadrilateral $ABCD$,
$$
\angle BAD=\angle BCD=120^{\circ}, BC=CD=10.
$$

Then $AC=$ . $\qquad$
2. 10 .

From the problem and the sum of the interior angles of a convex quadrilateral being $360^{\circ}$, we know
$$
\begin{array}{l}
\angle A B C+\angle A D C=120^{\circ} . \\
\text { If } A C>B C=C D=10 \text {, then } \\
\angle A B C+\angle A D C \\
>\angle B A C+\angle D A C=\angle B A D=120^{\circ} \text {, }
\end{array}
$$

This contradicts equation (1).
If $A C<B C=C D=10$, similarly we get a contradiction.
Therefore, $A C=10$.