17-108 If the three sides of $\triangle ABC$ are $BC=a, CA=b, AB=c$, and a point $P$ is taken arbitrarily inside $\triangle ABC$, and lines parallel to the three sides are drawn through $P$ to intersect the sides (as shown in the figure), if $DE=a', FG=b', HI=c'$, then the value of $\frac{a'}{a}+\frac{b'}{b}+\frac{c'}{c}$ is
(A) $\frac{1}{3}$.
(B) $\frac{1}{2}$.
(C) 1 .
(D) Cannot be determined, it depends on the position of $P$.


[Solution 1] From $I F // B C, H E // A C$, we have $\triangle A B C \sim \triangle H I P$, thus
$$
\frac{c^{\prime}}{c}=\frac{I P}{a}=\frac{B D}{a} .
$$

Also, $\triangle A B C \backsim \triangle G P F$, so $\frac{b^{\prime}}{b}=\frac{P F}{a}=\frac{E C}{a}$.

Therefore, the answer is $(C)$.
[Solution 2] Let the altitude from $B C$ be $h_{a}$, and the distance from $P$ to $B C$ be $h_{a}^{\prime}$. Connect $P B, P C$.

Then
Similarly, $\frac{b^{\prime}}{b}=\frac{S_{\triangle P A C}}{S_{\triangle A B C}}, \frac{c^{\prime}}{c}=\frac{S_{\triangle P A B}}{S_{\triangle A B C}}$,
Thus,
$$
\begin{aligned}
& \frac{a^{\prime}}{a}+\frac{b^{\prime}}{b}+\frac{c^{\prime}}{c} \\
= & \frac{S_{\triangle P B C}+S_{\triangle P A C}+S_{\triangle P A B}}{S_{\triangle A B C}}=1 .
\end{aligned}
$$

Therefore, the answer is $(C)$.