$17 \cdot 29$ Let $AD$ be the altitude from the vertex $A$ to the base $BC$ of isosceles $\triangle ABC$, and let $BM_1, BM_2, \cdots, BM_n$ be the $n+1$ angle bisectors of $\angle ABC$. Their intersections with $AD$ (from $A$ to $D$) are $M_1, M_2, \cdots, M_n$ respectively. Extend $CM_2$ to intersect $AB$ at $E$. Then the positional relationship between line $EM_1$ and line $BM_n$ is
(A) parallel.
(B) intersecting.
(C) perpendicular intersecting.
(D) dependent on $n$ and cannot be determined.
(Anhui Hefei Junior High School Mathematics Competition, 1990)
[Solution] As shown in the figure, the $n$ segments $B M_{1}$ to $B M_{n}$ divide $\angle A B C$ into $n+1$ equal parts, with each part being $\alpha$. Connect $C M_{1}$, and connect $B M_{n}$ to intersect $C E$ at $F$.
It is easy to prove: $M_{1}$ is the intersection of the angle bisectors of $\triangle A E C$, so $E M_{1}$ bisects $\angle A E C$.
$$
\begin{aligned}
\therefore \angle M_{1} E C & =\frac{\angle A E C}{2}=\frac{1}{2}(\angle B+\angle E C B) \\
& =\frac{1}{2}[(n+1) \alpha+(n-1) \alpha] \\
& =n \alpha .
\end{aligned}
$$

And $\square$
$$
\begin{aligned}
\angle E F B & =\angle F B C+\angle F C B=\alpha+(n-1) \alpha \\
& =n \alpha . \\
\therefore \quad & \angle M_{1} E C=\angle E F B, \therefore \quad E M_{1} / / B M_{n} .
\end{aligned}
$$

Therefore, the answer is $(A)$.