17.93 Let the three sides of a triangle be unequal, the height on the longest side be $x$, and the distances from any point on the longest side to the other two sides be $y$ and $z$. Then
(A) $xy+z$.
(D) The above size relationship depends on the lengths of the triangle's sides.
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1991)
[Solution] Let the three sides of the triangle be $a, b, c$, and $a<b<c$.
Given $S_{\triangle A B C}=\frac{1}{2} c x$,
and $S_{\triangle A B C}=S_{\triangle B C P}+S_{\triangle A C P}$
$$
=\frac{1}{2} a y+\frac{1}{2} b z,
$$

Therefore, $\frac{1}{2} c x=\frac{1}{2} a y+\frac{1}{2} b z$, which means $c x=a y+b z$,
Since $\quad a y+b z<c y+c z=c(y+z)$
Thus, $\quad c x<c(y+z)$, which means $x<y+z$.
Hence, the answer is $(A)$.