239. On the plane, a segment $AB$ is given. Using only one compass, double this segment (i.e., find a point $C$ on the extension of $AB$ beyond point $B$ such that $AC = 2AB$).

From the solution to problem 239, it follows that with the help of one compass, any given segment $AB$ can be increased by any integer number of times $n$, i.e., find a point $C$ on the extension of $AB$ beyond point $B$ such that $AC = n \cdot AB$. For example, to triple the segment $AB$, it is sufficient to find a point $C'$ on its extension beyond point $B$ such that $AC' = 2AB$, and then construct a point $C$ on the extension of $BC'$ beyond $C'$ such that $BC = 2BC'$; in this case, $AC = 3AB$ (Fig. 161).

1) This statement requires some clarification, similar to that made earlier when considering the problem of constructions using only a straightedge (see § 5, Chapter 1, pp. 124-125). Specifically, if the construction problem requires finding a certain straight line or a figure consisting of straight lines (for example, a triangle), then, of course, we cannot do this with the help of one compass alone. However, we can find any number of points on each of the required straight lines and all points of intersection of the required straight lines and circles (for example, all vertices of the required triangle).
239. Let's draw a circle with the center at point $B$ and radius $BA$, and make marks $M$, $N$, $C$ on this circle such that $AM = MN = NC = BA$ (Fig. 414). Clearly, $AM$, $MN$, and $NC$ are the sides of an inscribed equilateral triangle; therefore, $C$ and $A$ are diametrically opposite points on the circle, i.e., $C$, $B$, and $A$ lie on the same straight line and $AB = BC$.