4・14 Four distinct positive numbers $a, b, c, d$, where $a$ is the largest and $d$ is the smallest, and $a: b = c: d$. The relationship between $a+d$ and $b+c$ is
(A) $b+c>a+d$.
(B) $a+d>b+c$.
(C) $a+d=b+c$.
(D) Cannot be determined.
(China Guangzhou, Wuhan, Fuzhou, and other five cities Junior High School Mathematics League, 1986)
[Solution] $\because a: b=c: d, \quad \therefore \quad a=\frac{b c}{d}$.
$$
\therefore \quad(a+d)-(b+c)=\frac{b c-b d-c d+d^{2}}{d}=\frac{(b-d)(c-d)}{d} \text {, }
$$
$\because d$ is the smallest, $\therefore b>d, c>d$, and $d>0$,
$$
\text { then } \begin{array}{c}
\quad \frac{(b-d)(c-d)}{d}>0, \\
\therefore \quad a+d>b+c .
\end{array}
$$

Therefore, the answer is $(B)$.