18.15 As shown in the figure, given $\angle A=\angle B, A A_{1}, P P_{1}, B B_{1}$ are all perpendicular to $A_{1} B_{1}$, and $A A_{1}=17, P P_{1}=16, B B_{1}=20, A_{1} B_{1}=12$. Then the value of $A P+P B$ is
(A) 12.
(B) 13.
(C) 14.
(D) 15.
[Solution] As shown in the figure on the next page, extend $BP$ to intersect $AA_1$ at $A_1'$, draw $PP' \parallel A_1B_1$ intersecting $AA_1$ at point $P'$, and draw $A'D \parallel A_1B_1$ intersecting $BB_1$ at point $D$.
$$
\begin{aligned}
\because & \angle A = \angle B, \\
\therefore & AP = A'P. \\
\therefore & AP' = P'A'. \\
\because \quad AP' & = AA_1 - PP_1 \\
& = 17 - 16 = 1, \\
\therefore \quad AA' & = 2 \times 1 = 2. \\
\therefore \quad BD & = BB_1 - A'A_1 \\
& = 20 - (27 - 2) = 5.
\end{aligned}
$$

By the Pythagorean theorem,
$$
\begin{aligned}
A'B & = \sqrt{A_1B_1^2 + BD^2} \\
& = \sqrt{12^2 + 5^2} = 13.
\end{aligned}
$$

Thus, $AP + PB = 13$.
Therefore, the answer is (B).