$23 \cdot 3$ Tetrahedron $A B C D$ has all edge lengths equal to $1, P$ is the midpoint of edge $A B$, and $Q$ is the midpoint of edge $C D$. Then the length of line segment $P Q$ is
(A) $\frac{1}{2}$.
(B) $\frac{\sqrt{3}}{2}$.
(C) $\frac{3}{4}$.
(D) $\frac{\sqrt{2}}{2}$.
(Beijing High School Mathematics Competition, 1987)
[Solution] Connect $P C, P D$, then $P C=P D$. (The medians of congruent equilateral triangles are equal).

Therefore, $P Q$ is the median on the base $C D$ of the isosceles triangle $P C D$, it is also the altitude on the base, i.e., $P Q \perp C D$ (as shown in the figure).
It is easy to calculate $C P=\frac{\sqrt{3}}{2}, C Q=\frac{1}{2}$,
$$
\therefore \quad P Q=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{2}}{2} \text {. }
$$

Therefore, the answer is $(D)$.