2. In the convex quadrilateral $ABCD$, $AB=CD$, $AC$ is a diagonal, $\angle DAC>\angle BCA$, and $\angle DAC$ and $\angle BCA$ are supplementary, $\angle BAC>\angle ACD$, and $\angle BAC$ and $\angle ACD$ are complementary. Then $\angle B=$
$2.45^{\circ}$.
Extend $B C$ to $E$, such that $C E=A D$.

Connect $A E$, then $\triangle D A C \cong$ $\triangle E C A$. We have $\angle D=\angle E$, $\angle A C D=\angle E A C, A E=C D$. $\because A B=C D$,
$\therefore A B=A E, \angle B=\angle E=45^{\circ}$.