$25 \cdot 23$ In the Cartesian coordinate system, the equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$
$(a, b$ are two different positive numbers) represents the curve
(A) Triangle.
(B) Square.
(C) Non-square rectangle.
(D) Non-square rhombus.
(China High School Mathematics League, 1994)
[Solution 1] The lines $l_{1}: x-y=0$ and $l_{2}: x+y=0$ divide the plane $x O y$ into 4 regions:
$\mathrm{I}\left\{\begin{array}{l}x+y \geqslant 0 \\ x-y \geqslant 0\end{array}\right.$
II $\left\{\begin{array}{l}x+y \geqslant 0 \\ x-y \leqslant 0\end{array}\right.$
III $\left\{\begin{array}{l}x+y \leqslant 0 \\ x-y \leqslant 0\end{array}\right.$
$\mathrm{IV}\left\{\begin{array}{l}x+y \leqslant 0 \\ x-y \geqslant 0\end{array}\right.$
In region (I), the equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$

changes to
$$
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1
$$

It represents the line $l_{3}$. Let $l_{3}$ intersect $l_{1}$ at point $A$. Then by
$$
\left\{\begin{array}{l}
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1, \\
x-y=0
\end{array}\right.
$$

we get the coordinates of point $A$ as $(a, a)$.
Let $l_{3}$ intersect $l_{2}$ at point $B$, by
$$
\left\{\begin{array}{l}
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1, \\
x+y=0
\end{array}\right. \text {. }
$$

we get the coordinates of point $B$ as $(b, -b)$.
Therefore, the curve represented by (1) in region (I) is the line segment $A B$.
Similarly, in region (II), equation (1) becomes $\frac{x+y}{2 a}+\frac{x-y}{2 b}=1$, which represents the line $l_{4}$. And $l_{4}$ intersects $l_{1}$ at point $A$, and $l_{4}$ intersects $l_{2}$ at point $D(-b, b)$, the curve represented by equation (1) in region (III) is the line segment $A D$.

Similarly, we get point $C(-a, a)$ on the curve represented by (1), and the curve represented by (1) in region III is the line segment $C D$, and the curve represented by (1) in region (IV) is the line segment $C B$.
Since $A O=C O, B O=D O, A C \neq B D, A C \perp B D$,
the quadrilateral $A B C D$ represented by (1) is a rhombus that is not a square.
Therefore, the answer is $(D)$.
[Solution 2] Rotate the Cartesian coordinate system $x O y$ counterclockwise by $45^{\circ}$ to get the new coordinate system $x^{\prime} O y^{\prime}$. The coordinates of point $P$ in the coordinate system $x^{\prime} O y^{\prime}$ are $\left(x^{\prime} y^{\prime}\right)$, and in the coordinate system $x O y$ are $(x, y)$. Then
$$
\left\{\begin{array}{l}
x^{\prime}=\frac{1}{\sqrt{2}}(x+y), \\
y^{\prime}=\frac{1}{\sqrt{2}}(-x+y) .
\end{array}\right.
$$

The given equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$

transforms into
$$
\frac{\left|x^{\prime}\right|}{a}+\frac{\left|y^{\prime}\right|}{b}=\sqrt{2} \text {. }
$$

Clearly, (2) represents a curve that is symmetric about the $x^{\prime}$-axis and $y^{\prime}$-axis. In the first quadrant of $x^{\prime} O y^{\prime}$, (2) becomes $\frac{x^{\prime}}{a}+\frac{y^{\prime}}{b}=\sqrt{2}$, which is the line segment $A B$, where $A(\sqrt{2} a, 0), B(0, \sqrt{2} b)$.
By symmetry, in the second quadrant, equation (2) is the line segment $B C$, where $C(-\sqrt{2} a, 0)$; in the third quadrant, equation (2) is the line segment $C D$, where $D(0, -\sqrt{2} b)$;
in the fourth quadrant, equation (2) is the line segment $A D$.
By symmetry, we know $A B=B C=C D=A D$.
Since $a \neq b$, we have $A C \neq B D$. Therefore, $A B C D$ is a rhombus that is not a square.
Therefore, the answer is $(D)$.