8. The triangle formed by the intersection of the three external angle bisectors of a triangle ( ).
(A) must be an acute triangle
(B) must be an obtuse triangle
(C) must be a right triangle
(D) is similar to the original triangle
8. A.

According to the theorem that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles (as shown in Figure 5), we have
$$
\begin{array}{l} 
\angle C B A^{\prime}+\angle A^{\prime} C B \\
= \frac{1}{2}(\angle A B C+ \\
\angle B C A+2 \angle C A B) \\
= \frac{1}{2} \times 180^{\circ}+\frac{1}{2} \angle C A B>90^{\circ} .
\end{array}
$$

Therefore, $\angle A^{\prime}<90^{\circ}$.
Similarly, $\angle B^{\prime}<90^{\circ}, \angle C^{\prime}<90^{\circ}$.
Thus, the triangle formed by the lines of the three exterior angle bisectors is definitely an acute triangle.