2. In $\triangle A B C$, $\angle A=60^{\circ}, \angle B=20^{\circ}$, extend $B C$ to $D$ such that $C D=A B$. Then $\angle C A D=$ $\qquad$
2. $80^{\circ}$.

As shown in Figure 4,
Take point $E$ on side $CD$ such that $BE = AB$, and connect $AE$. Then
$$
\begin{array}{l}
\angle AEC = \angle BAE = \frac{180^{\circ} - \angle B}{2} = 80^{\circ} \\
= \angle B + \angle BAC = \angle ACE.
\end{array}
$$

Therefore, $AC = AE$.
It is easy to see that $\triangle ABE \cong \triangle ADC$.
Thus, $\angle CAD = \angle EAB = 80^{\circ}$.