To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.Let $ABCD$ be the given rectilinear figure, and $E$ the given rectilinear angle. So it is required to construct a parallelogram equal to the rectilinear figure $ABCD$ in the given angle $E$.  <prf> And since triangle $ABD$ is equal to parallelogram $FH$, and $DBC$ to $GM$, the whole rectilinear figure $ABCD$ is thus equal to the whole parallelogram $KFLM$. Thus, the parallelogram $KFLM$, equal to the given rectilinear figure $ABCD$, has been constructed in the angle $FKM$, which is equal to the given (angle) $E$.