17・106 Through any point $P$ inside $\triangle ABC$, draw $IF \parallel BC, EH \parallel AC, DG \parallel AB$. Then the value of $\frac{IF}{BC} + \frac{EH}{AC} + \frac{DG}{AB}$ is
(A) $\frac{3}{2}$.
(B) 2.
(C) $\frac{4}{3}$.
(D) $\frac{5}{3}$.
(Chinese Jiangsu Province Suzhou Junior High School Mathematics Competition, 1985; Chinese Jiangxi Province Nanchang Junior High School Mathematics Competition, 1989)
[Solution 1] Let the three sides of $\triangle ABC$ be $a, b, c$, and let $DE=a^{\prime}, \quad GF=b^{\prime}$, $HI=c^{\prime}$. From the conclusion of the previous problem, we have
$$
\frac{IF}{BC}=\frac{a-a^{\prime}}{a}=1-\frac{a^{\prime}}{a}=1-\frac{S_{\triangle PBC}}{S_{\triangle ABC}},
$$

Similarly,
$$
\begin{array}{l}
\frac{EH}{AC}=1-\frac{S_{\triangle PAC}}{S_{\triangle ABC}}, \\
\frac{DG}{AB}=1-\frac{S_{\triangle PAB}}{S_{\triangle ABC}},
\end{array}
$$

Thus, $\frac{IF}{BC}+\frac{EH}{AC}+\frac{DG}{AB}=3-\frac{S_{\triangle PBC}+S_{\triangle PAC}+S_{\triangle PAB}}{S_{\triangle ABC}}=2$.
Therefore, the answer is $(B)$.
[Solution 2] Consider the case where $P$ is the centroid of $\triangle ABC$. In this case, by the properties of the centroid of a triangle, we have
$$
\frac{IF}{BC}=\frac{EH}{AC}=\frac{DG}{AB}=\frac{2}{3},
$$

Then, $\frac{IF}{BC}+\frac{EH}{AC}+\frac{DG}{AB}=2$.
Therefore, the answer is $(B)$.