1. Given the following four propositions:
(1) A non-rhombus parallelogram is divided into two pairs of congruent triangles by its diagonals, one pair being obtuse triangles and the other pair being acute triangles:
(2) If there are two points on a line that have equal (non-zero) distances to another line, then the two lines are parallel;
(3) Two figures have exactly one axis of symmetry;
(4) The line segment with endpoints $A(1,0)$ and $B(-1,0)$ has two axes of symmetry.
Among them, the number of correct propositions is ().
(A) 1
(B) 2
(C) 3
(D) 4
- 1. A่.

Line segment $AB$ has two axes of symmetry, which are the perpendicular bisector of $AB$, the $y$-axis, and the $x$-axis on which $AB$ lies. Proposition (4) is a true proposition. The rest are false propositions.

For proposition (1), there is a counterexample as shown in Figure 4, in quadrilateral $\square ABCD$ (not a rhombus), when $\angle BAC = \angle DCA \geqslant 90^{\circ}$, $\triangle AOD \cong \triangle COB$ is a pair of obtuse triangles, but $\triangle AOB \cong \triangle COD$ is not a pair of acute triangles.
For proposition (2), there is a counterexample as shown in Figure 5, points $A$ and $B$ on line $l_1$ are on opposite sides of another line $l_2$, although their distances to $l_2$ are equal $\left(AH_1 = BH_2\right)$, the two lines intersect.

For proposition (3), there is a counterexample, for $2n$ equally spaced points $A_1, B_1, A_2, B_2, \cdots, A_n, B_n$ on the circumference, they can be connected to form two regular $n$-gons $A_1A_2\cdots A_n$ and $B_1B_2\cdots B_n$, these two $n$-gons have $n$ axes of symmetry. When $n=4$, as shown in Figure 6, there are 4 axes of symmetry; when $n=2$, the two lines in Figure 5 are axisymmetric, with 2 axes of symmetry.