4. Below are four propositions:
(1) A quadrilateral with one pair of opposite sides equal and one pair of opposite angles equal is a parallelogram;
(2) A quadrilateral with one pair of opposite sides equal and one diagonal bisecting the other diagonal is a parallelogram;
(3) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of these angles bisecting the other diagonal is a parallelogram;
(4) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of these angles being bisected by the other diagonal is a parallelogram.
Among them, the number of correct propositions is
(A) (1) ;
(B) (2) ;
(C) (3) ;
(D) (4).
4. (A)

For (1), (2), and (4), counterexamples can be provided respectively. For example:
(1) In quadrilateral $ABCD$ in Figure 1, $\triangle ABD \cong \triangle CDE$.
(2) In Figure 2, construct isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and use $O$ as the intersection of the diagonals to construct parallelogram $ABCE$. At this point, quadrilateral $ABCD$ satisfies the conditions $AD=AE=BC$ and $AO=CO$, but it is not a parallelogram.
(4) In quadrilateral $ABCD$ in Figure 3, $B$ and $D$ are any two points on the perpendicular bisector of $AC$.
The following is a proof that proposition (3) is correct.
In Figure 4, given $\angle BAD = \angle DCB$,

and $OB = OD$.
With $O$ as the center, rotate $\triangle ABD$ counterclockwise

by $180^\circ$. Since $OB = OD$, $I$
coincides with $B$, $B$ coincides with $D$, and point $A$ coincides with some point $A'$ on ray $OC$.
If $A'$ is not $C$, then $\angle BA'D > \angle BCD$ (if $A'$ is inside segment $OC$) or $\angle BA'D < \angle BCD$ (if $A'$ is on the extension of $OC$), both of which contradict $\angle BA'D = \angle BAD = \angle BCD$!

Thus, $A'$ is indeed $C$. Therefore, $OA = OA' = OC$. So quadrilateral $ABCD$ is a parallelogram.
In summary, only proposition (3) is correct.