2. In quadrilateral $A B C D$, it is known that $A B=B C=$ $C D$, and $\angle A=80^{\circ}, \angle D=40^{\circ}$. Then the degree measure of $\angle B$ is $\qquad$.
2. $80^{\circ}$.

As shown in Figure 4, draw $BO \parallel CD$, and connect $AO, DO$.
Then quadrilateral $BCDO$ 

is a rhombus. Therefore,
$$
\begin{array}{l}
OB=OD \\
=BC=AB. \\
\text{By } \angle ABC + \angle C \\
=360^{\circ} - \angle BAD - \angle ADC = 240^{\circ}, \\
\angle OBD + \angle C = 180^{\circ}, \\
\angle ABO = 240^{\circ} - 180^{\circ} = 60^{\circ}.
\end{array}
$$

Thus, $\angle ABO = 240^{\circ} - 180^{\circ} = 60^{\circ}$.
Therefore, $\triangle ABO$ is an equilateral triangle.
Then $\angle ODA = \angle OAD = \angle BAD - \angle BAO = 20^{\circ}$, $\angle OBC = \angle ODC = \angle ADC - \angle ODA = 20^{\circ}$. Hence $\angle ABC = \angle ABO + \angle OBC = 80^{\circ}$.