4. Among the following 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 this pair of opposite angles being bisected by the other diagonal is a parallelogram;
(4) A quadrilateral with one pair of opposite angles equal and the diagonal connecting the vertices of this pair of opposite angles bisecting the other diagonal is a parallelogram.
Among them, the correct proposition numbers are $\qquad$
4. (4).

Propositions (1), (2), and (3) can be refuted with the following counterexamples:
Proposition (1): The quadrilateral $ABCD$ in Figure 5(a), where $\triangle ABD \cong \triangle CDE$.

Proposition (2): As shown in Figure 5(b), construct an isosceles $\triangle ADE$, extend the base $ED$ to any point $O$, and use $O$ as the intersection of the diagonals to construct $\square ABCE$. In this case, the quadrilateral $ABCD$ satisfies the conditions $AD = (AE =) BC$ and $AO = CO$, but it is not a parallelogram.

Proposition (3): The quadrilateral $ABCD$ in Figure 5(c), where $A$ and $C$ are any two points on the perpendicular bisector of $BD$.
The following is a proof that Proposition (4) is correct.
As shown in Figure 5(d), given $\angle BAD = \angle DCB$ and $OB = OD$.

With point $O$ as the center, rotate $\triangle ABD$ counterclockwise by $180^\circ$. Since $OB = OD$, point $D$ coincides with $B$, and point $B$ coincides with $D$, while point $A$ coincides with some point $A_1$ on ray $OC$. If $A_1$ is not $C$, then $\angle BA_1D > \angle BCD$ (if $A_1$ is inside segment $OC$) or $\angle BA_1D < \angle BCD$ (if $A_1$ is on the extension of $OC$), both of which contradict $\angle BA_1D = \angle BAD = \angle BCD$. Therefore, $A_1$ must be $C$, i.e., $OA = OA_1 = OC$. Thus, quadrilateral $ABCD$ is a parallelogram.