17-185 A triangle with side lengths $a, b, c$ satisfies:
$$
\frac{1}{a}-\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b-c},
$$

then this triangle is
(A) an equilateral triangle.
(B) an isosceles triangle.
(C) a scalene triangle.
(D) a right triangle.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution] From $\frac{1}{a}-\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b-c}$,
multiplying both sides by $a b c(a+b-c)$ and rearranging terms, we get
$$
(b-c)\left(a^{2}+a b-c a+b c\right)=0 .
$$

Thus, $b-c=0$ or $a^{2}+a b-c a+b c=0$.
Next, we prove that $a^{2}+a b-c a+b c=0$ is impossible.
If $a=b$, then $a^{2}+a b-c a+b c \neq 0$;
If $a \neq b$, then $c=\frac{a^{2}+a b}{a-b}>0$, so $a>b$. Also, $a+b>c$, thus $a+b>\frac{a^{2}+a b}{a-b}$.
Therefore, $1>\frac{a}{a-b}$ or $a-b>a$, which is also impossible.
Hence,
$b-c=0$ or $b=c$.
Therefore, the answer is $(B)$.