$36 \cdot 39$ As shown, $A, B$ are fixed points, $O$ is a moving point. On the side different from point $O$, take two points $A', B'$ such that $\angle OAA' = \angle OBB' = 90^{\circ}$, and $AA' = OA$, $BB' = OB$. Let the midpoint of $A'B'$ be $O'$. When $O$ moves on one side of $AB$, then
(A) $O'$ moves along a certain straight line.
(B) $O'$ moves along a certain circle.
(C) $O'$ remains stationary.
(D) None of the above conclusions are correct.
(Chinese Jiangsu Province Suzhou Junior High School Mathematics Competition, 1987)
[Solution] As shown in the figure, draw perpendiculars from $O, A^{\prime}, O^{\prime}, B^{\prime}$ to $A B$, with the feet of the perpendiculars being $C, D$, $M, E$ respectively.
It is easy to prove that $B E=O C=A D$, and $B^{\prime} E=B C, A^{\prime} D=A C$.
$$
\begin{aligned}
\therefore \quad O^{\prime} M & =\frac{A^{\prime} D+B^{\prime} E}{2}=\frac{A C+B C}{2} \\
& =\frac{A B}{2},
\end{aligned}
$$

Also, $O^{\prime} M \perp A B$, and
$$
A M=A D+D M=B E+E M=B M \text {, }
$$

Thus, when $O$ moves on one side of $A B$, $O^{\prime}$ remains stationary. Therefore, the answer is (C).