$22 \cdot 7$ Given a dihedral angle $\alpha-P Q-\beta . A, B$ are two points on the half-plane $\alpha$, and point $C$ is on the half-plane $\beta$ ($A, B, C$ are not on $P Q$, and the distances from $A, B$ to $P Q$ are not equal). Using $A, B, C$ as three vertices to construct a trapezoid, such that the fourth vertex of the trapezoid is also on the half-plane of the dihedral angle, then the number of trapezoids that meet the above conditions is
(A) does not exist.
(B) there is only one.
(C) there are two.
(D) there are infinitely many.
(China Beijing High School Mathematics Competition, 1986)
[Solution] As shown in the figure, connect $\mathrm{B}, \mathrm{A}$ to intersect $\mathrm{PQ}$ at $\mathrm{R}$, and connect $R C$. This draws the intersection line $R B$ of the plane through points $A, B, C$ with plane $\alpha$, and the intersection line $R C$ with plane $\beta$.

Through $B$, draw a line parallel to $A C$ intersecting $R C$ at $D_{2}$, completing the trapezoid $A B D_{2} C$; through $A$, draw a line parallel to $B C$ intersecting $R C$ at $D_{1}$, completing the trapezoid $A B C D_{1}$. If a line parallel to $A B$ is drawn through $C$, it will not intersect plane $\beta$ at any point other than $C$, so no trapezoid can be formed. Therefore, there are two trapezoids that meet the conditions.
Hence, the answer is $(C)$.