17・184 The lengths of the three sides of a triangle $a, b, c$, satisfy the relation
$$
\frac{1}{a}-\frac{1}{b}+\frac{1}{c}=\frac{1}{a-b+c}
$$

The triangle must be
(A) an equilateral triangle.
(B) an isosceles triangle with $a$ as the base.
(C) an isosceles triangle with $c$ as the base.
(D) a scalene triangle.
(E) none of the above conclusions is correct.
(China Beijing Junior High School Mathematics Competition, 1982)
[Solution] Given $\frac{b c-a c+a b}{a b c}=\frac{1}{a-b+c}$, that is,
$$
(a-b+c)(b c-a c+a b)=a b c,
$$

then $(a-b)(b c-a c+a b)+b c^{2}-a c^{2}+a b c=a b c$,
or $(a-b)(b c-a c+a b)-c^{2}(a-b)=0$.
That is, $(a-b)\left(b c-a c+a b-c^{2}\right)=0$.
Or $(a-b)(b-c)(a+c)=0$.
By the given, $a+c \neq 0$, so $a-b=0$ or $b-c=0$, hence $a=b$ or $b=c$.
Therefore, the triangle is isosceles.
Thus, the answer is $(E)$.