18.31 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 this pair of 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 this pair of angles being bisected by the other diagonal is a parallelogram.

Among these, the number of correct propositions is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(Chinese Junior High School Mathematics League, 1988)
[Solution] For (1), (2), and (4), counterexamples can be provided respectively. For example:
(1) In quadrilateral $A B C D$ as shown in Figure (1), $\triangle A B D \cong \triangle C D E$.
(2) As shown in Figure (2), construct isosceles $\triangle A D P$, extend the base $P D$ to any point $O$, and use $O$ as the intersection of the diagonals to construct parallelogram $A B C P$.

At this point, quadrilateral $A B C D$ satisfies the condition $A D=(A P=) B C$, and $A O=C O$, but it is not a parallelogram.
(4) In quadrilateral $A B C D$ as shown in Figure (3), $B, D$ are any two points on the perpendicular bisector of $A C$.
The following is a proof that proposition (3) is correct.
As shown in Figure (4), given $\angle B A D=\angle D C B$, and $O B=O D$.
With $O$ as the center, rotate $\triangle A B D$ counterclockwise by $180^{\circ}$. Since $O B=O D$, $D$ coincides with $B$, and $B$ coincides with $D$, point $A$ coincides with some point $A^{\prime}$ on ray $O C$.

If $A^{\prime}$ is not $C$, then $\angle B A^{\prime} D>\angle B C D\left(A^{\prime}\right.$ is inside segment $O C$) or $\angle B A^{\prime} D<\angle B C D\left(A^{\prime}\right.$ is on the extension of $O C$), both of which contradict $\angle B A^{\prime} D=\angle B A D=\angle B C D$.

Thus, $A^{\prime}$ is indeed $C$, i.e., $O A=O A^{\prime}=O C$, so quadrilateral $A B C D$ is a parallelogram.
In summary, only proposition (3) is correct.
Therefore, the answer is $(A)$.