To construct a rectilinear angle equal to a given rectilinear angle at a (given) point on a given straight-line. Let $AB$ be the given straight-line,  $A$ the (given) point on it, and  $DCE$ the given rectilinear angle. So it is required to construct a rectilinear angle equal to the given rectilinear angle $DCE$ at the (given) point $A$ on the given straight-line $AB$.  Let the points $D$ and $E$ have been taken at random on each of the (straight-lines) $CD$ and $CE$ (respectively), and let $DE$ have been joined. And let the triangle $AFG$ have been constructed from three straight-lines which are equal to $CD$, $DE$, and $CE$, such that $CD$ is equal to $AF$, $CE$ to $AG$, and further $DE$ to $FG$ [Prop.~1.22].  Therefore, since the two (straight-lines) $DC$, $CE$ are equal to the two (straight-lines) $FA$, $AG$, respectively, and the base $DE$ is equal to the base $FG$, the angle $DCE$ is thus equal to the angle $FAG$ [Prop.~1.8].  Thus, the rectilinear angle $FAG$,  equal to the  given rectilinear angle $DCE$, has been constructed at the (given) point $A$ on the given straight-line $AB$. (Which is) the very thing it was required to do.