$17 \cdot 60$ In an isosceles triangle, the angle bisector of one of the base angles divides the perimeter into two parts, 63 and 35. Then the length of the leg must be
(A) 26.4.
(B) 33.
(C) 38.5.
(D) None of the above.
(Shanghai Junior High School Mathematics Competition, 1984)
〔Solution 1] Let the length of the legs of the triangle be $x$, and the base be $y$. According to the problem, we have $2 x+y=63+35=98$.
Also, by the angle bisector property, $u=\frac{x}{x+y} x, t=\frac{y}{x+y} x$.
Then $(x+u):(y+t)$
$$
\begin{array}{l}
=\left(1+\frac{x}{x+y}\right) x:\left(1+\frac{x}{x+y}\right) y \\
=x: y,
\end{array}
$$

Therefore, $x: y=63: 35=9: 5$, which means $y=\frac{5}{9} x$. Substituting into $\circledast$ we get $\frac{23}{9} x=98$, i.e., $x=\frac{98 \cdot 9}{23} \approx 38.3$. When $y: x=9: 5$, substituting $y=\frac{9 x}{5}$ into $\circledast$ we have
$$
\frac{19}{5} x=98 \text {, i.e., } x=\frac{98 \cdot 5}{19} \approx 25.8 \text {. }
$$

Therefore, the answer is $(D)$.
[Solution 2] In fact, from the two scenarios shown in the above and right figures, it is clear that the statement that the length of the legs is certain is not true.
Therefore, the answer is (D).