$17 \cdot 40$ In a regular $\triangle A B C$, point $P$ is on side $B C$. A new triangle is formed with $A P, B P, C P$ as its sides, and the largest interior angle of this new triangle is $\theta$, then
(A) $\theta \geqslant 90^{\circ}$.
(B) $\theta \leqslant 120^{\circ}$.
(C) $\theta=120^{\circ}$.
(D) $\theta=135^{\circ}$.
(Anhui Province, Anqing and other counties and cities Junior High School Mathematics Competition, 1991)
[Solution 1] This problem is a special case of the previous one, where point $P$ is on side $BC$, thus the angles it forms with the three vertices are considered as $\alpha, \beta, 180^{\circ}$.

In this way, the three interior angles of the triangle formed by $AP, BP, CP$ are $\alpha-60^{\circ}, \beta-60^{\circ}, 180^{\circ}-60^{\circ}$, with the largest being $120^{\circ}$.
Therefore, the answer is $(C)$.
[Solution 2] With $A$ as the center, rotate $\triangle ABP$ counterclockwise to the position of $\triangle ACP^{\prime}$, and connect $PP^{\prime}$.
Then $CP^{\prime}=BP$, and it is easy to prove that $\triangle APP^{\prime}$ is an equilateral triangle.
In $\triangle PP^{\prime}C$, $PP^{\prime}=AP$, and $P^{\prime}C=BP$, so $\triangle PCP^{\prime}$ is the new triangle sought. Its largest angle is $120^{\circ}$.
Therefore, the answer is $(C)$.