$25 \cdot 22$ Given three distinct points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, and $R\left(x_{1}+x_{2}\right.$, $y_{1}+y_{2}$ ), now connect these three points with the origin to form line segments, then the figure $O P R Q$, determined by the positions of points $P, Q$, and $R$, can be one of the following three possible situations:
(1) Parallelogram. (2) Line. (3) Trapezoid. The possible situations are:
(A) Only (1).
(B) Only (2).
(C) Only (3).
(D) Only (1) or (2).
(E) None of the three situations mentioned exist.
(15th American High School Mathematics Examination, 1964)
[Solution] From the problem, we know that $x_{1}, y_{1}$ or $x_{2}, y_{2}$ cannot both be 0, otherwise they would be indistinguishable and contradict the problem statement.

If the equations of $O P, O Q$ are $y_{1}=k_{1} x_{1}, y_{2}=k_{2} x_{2}$, when $k_{1}=k_{2}$, $P$, $Q, R$ are all on a line passing through $O$.

If $k_{1} \neq k_{2}$, it is easy to calculate that the slope of $R Q$ is $k_{1}$; and the slope of $P R$ is $k_{2}$.
That is, $R Q \parallel O P, P R \parallel O Q$.
Thus, $O P R Q$ is either a line or a parallelogram.
Therefore, the answer is $(D)$.