17-179 The three external angle bisectors of $\triangle A B C$ intersect to form a $\triangle L M N$, then $\triangle L M N$
(A) must be a right triangle.
(B) must be an obtuse triangle.
(C) must be an acute triangle.
(D) is not necessarily an acute triangle.
(E) must not be an acute triangle.
(China Tianjin Junior High School Mathematics Competition; China Junior High School Mathematics League, 1984)
[Solution] As shown in the figure, $L$ is the intersection point of one of the external angle bisectors of $\triangle ABC$, then
$$
\begin{array}{l}
\angle L A B=\frac{1}{2}(\angle B+\angle C), \\
\angle L B A=\frac{1}{2}(\angle A+\angle C),
\end{array}
$$

Therefore,
$$
\begin{aligned}
& \angle L A B+\angle L B A \\
= & \frac{1}{2}(\angle A+\angle B+2 \angle C) \\
= & 90^{\circ}+\frac{1}{2} \angle C>90^{\circ}
\end{aligned}
$$

Thus, $\angle L<90^{\circ}$, which means it is an acute angle. Similarly, $\angle M$ and $\angle N$ are also acute angles. Therefore, the answer is $(C)$.