17-105 Let the lengths of the sides of $\triangle ABC$ be $a, b, c$, and $c \leqslant b \leqslant a$. Through an interior point $P$, the lines connecting $P$ to $A, B, C$ intersect the opposite sides at $A^{\prime}, B^{\prime}, C^{\prime}$, respectively. Let
$$
s=A A^{\prime}+B B^{\prime}+C C^{\prime},
$$

then for all positions of $P$, $s$ is less than
(A) $2 a+b$.
(B) $2 a+c$.
(C) $2 b+c$.
(D) $a+2 b$.
(E) $a+b+c$.
[Solution] It is easy to prove: $A A^{\prime}<\max \{A B, A C\}=\max \{b, c\}=b$ (using the proposition that in the same triangle, the larger angle faces the longer side).
Similarly,
$$
\begin{array}{l}
B B^{\prime}<\max \{a, c\}=a, \\
C C^{\prime}<\max \{a, b\}=a .
\end{array}
$$

Therefore,
$$
A A^{\prime}+B B^{\prime}+O C^{\prime}<2 a+b \text{. }
$$

Hence, the answer is $(A)$.