$1 \cdot 25$ Given real numbers $m, n, p, M$ satisfy:
$$
M=\frac{m}{n+p}=\frac{n}{p+m}=\frac{p}{m+n},
$$

then the value of $M$ is
(A) -1 .
(B) $\frac{1}{2}$.
(C) 1 .
(D) $\frac{1}{2}$ or -1 .
(China Beijing Junior High School Mathematics Competition, 1983;
2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
［Solution 1］From the given conditions, we have
$$
m=(n+p) M, \quad n=(p+m) M, \quad p=(m+n) M,
$$

Adding the above three equations gives $(m+n+p)=2(m+n+p) M$,
$\therefore$ When $m+n+p \neq 0$, $M=\frac{1}{2}$.
Also, from $\frac{m}{n+p}=\frac{n}{p+m}$, we get $\frac{m}{n}=\frac{n+p}{p+m}$,

Thus,
$$
\begin{aligned}
& \frac{m-n}{n}=\frac{(m+p)-(p+m)}{p+m}=\frac{n-m}{p+m}, \\
\therefore \quad & \frac{n}{p+m}=\frac{m-n}{n-m}=-1,
\end{aligned}
$$

which means $M=-1$ (at this time, $m+n+p=0$).
In summary, the answer is $(D)$.
[Solution 2] From the given conditions and the properties of proportional segments, we have $M=\frac{m+n+p}{2(m+n+p)}$, i.e., $\square$
$$
(2 M-1)(m+n+p)=0 .
$$

If $m+n+p \neq 0$, then $2 M-1=0$, i.e., $M=\frac{1}{2}$; if $m+n+p=0$, substituting $m=-(n+p)$ into the original equation gives $M=-1$. Therefore, the answer is $(D)$.