$17 \cdot 116 P$ is a point inside an equilateral triangle of height $h$. Let the distances from $P$ to each side be $x, y, z$. If the three segments of lengths $x, y, z$ can form a triangle, then the conditions that $x, y, z$ must satisfy are
(A) $x<h, y<h, z<h$.
(B) $x<\frac{h}{2}, y<\frac{h}{2}, z<\frac{h}{2}$.
(C) $x \leqslant \frac{h}{2}, y \leqslant \frac{h}{2}, z \leqslant \frac{h}{2}$.
(D) $x<\frac{h}{3}, y<\frac{h}{3}, z<\frac{h}{3}$.
(E) $x \leqslant \frac{h}{3}, y \leqslant \frac{h}{3}, z \leqslant \frac{h}{3}$.
(Jilin Province, China, Junior High School Mathematics Competition, 1986)
［Solution］Using the relationship of the areas of triangles, it is easy to prove that
$$
h=x+y+z \text {. }
$$

Since $x, y, z$ can form a triangle, then
$$
\left\{\begin{array}{l}
x+y>z, \\
y+z>x, \\
z+x>y .
\end{array}\right.
$$

Substituting (1) into (2), (3), and (4) respectively, we get $x<\frac{h}{2}, y<\frac{h}{2}, \quad z<\frac{h}{2}$. Therefore, the answer is $(B)$.