1. In a convex quadrilateral $A B C D$, $A D=B C, \angle A$ $=\angle C=100^{\circ}$. Then the shape of quadrilateral $A B C D$ is $\qquad$
2. Parallelogram.
As shown in Figure 3, draw $D E \perp$
$A B$ at $E$, and $B F \perp C D$ at
$F$. Then
$\angle E A D=\angle F C B$.
Also, $A D=B C$, so
$\triangle E A D \cong \triangle F C B$.
Therefore, $D E=B F, A E=C F$.
Connect $B D$.
Then Rt $\triangle B E D \cong$ Rt $\triangle D F B$.
Thus, $B E=D F$.
(2) - (1) gives $A B=C D$.
Therefore, quadrilateral $A B C D$ is a parallelogram.