$17 \cdot 55$ As shown, all points on $AB$ correspond to points on $A^{\prime} B^{\prime}$, and vice versa. Let $x$ represent the distance from a point $P$ on $AB$ to $D$, and $y$ represent the distance from a corresponding point $P^{\prime}$ on $A^{\prime} B^{\prime}$ to $D^{\prime}$. For any pair of corresponding points, if $x=a$, then $x+y$ equals
(A) $13 a$.
(B) $17 a-51$.
(C) $17-3 a$.
(D) $\frac{17-3 a}{4}$.
(E) $12 a-34$.
(9th American High School Mathematics Examination, 1958)
[Solution] From the figure and the properties of similar triangles, we have $\frac{P B}{P^{\prime} B^{\prime}}=\frac{A P}{A^{\prime} P^{\prime}}$,
$$
\frac{P B}{P^{\prime} B^{\prime}}=\frac{A B}{A^{\prime} B^{\prime}},
$$

which means
$$
\frac{4-x}{y-1}=\frac{x-3}{5-y}=\frac{1}{4},
$$

Thus, $y=17-4x$, so $x+y=a+(17-4a)=17-3a$. Therefore, the answer is $(C)$.