$25 \cdot 9$ Choose a point $(x, y)$ on the coordinate plane such that its distances to the $x$-axis, $y$-axis, and the line $x+y=2$ are the same, then the value of $x$ is
(A) $\sqrt{2}-1$.
(B) $\frac{1}{2}$.
(C) $2-\sqrt{2}$.
(D) 1 .
(E) Cannot be uniquely determined.
(35th American High School Mathematics Examination, 1984)
[Solution 1] In the Cartesian coordinate system, a point $(x, y)$ that is equidistant from the $x$-axis and the $y$-axis must lie on the lines $y=x$ or $y=-x$ (i.e., the angle bisectors of the $x$-axis and $y$-axis), as shown in the figure.

A point $(x, y)$ that is equidistant from the $x$-axis and the line $x+y=2$ must lie on the two angle bisectors passing through $(2,0)$ (their intersection point), as shown in the figure.

Thus, we find their intersection points, labeled as $A, B, C, D$ in the figure. The x-coordinates of these four points are
$$
-\sqrt{2}, 2-\sqrt{2}, 2+\sqrt{2} \text { and } \sqrt{2} \text {. }
$$

Therefore, the answer is $(E)$.
[Solution 2] The line $x+y=2$ and the two coordinate axes form a triangle, which has an incircle and three excircles.

The centers of these circles are equidistant from the three sides, thus satisfying the given condition (and only these four points satisfy the condition), so there are four points in total.
From the figure, it is easy to see that their x-coordinates are different.
Therefore, the answer is $(E)$.