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 $E$.  Let $DB$ have been joined, and let the parallelogram $FH$, equal to the triangle $ABD$, have been constructed in the angle $HKF$, which is equal to $E$ [Prop.~1.42]. And let the parallelogram $GM$, equal to the triangle $DBC$, have been applied to the straight-line $GH$ in the angle $GHM$, which is equal to $E$ [Prop.~1.44]. And since angle $E$ is equal to each of (angles) $HKF$ and $GHM$,  (angle) $HKF$ is thus also equal to $GHM$.  Let $KHG$ have been added to both. Thus, (the sum of) $FKH$ and $KHG$ is equal to (the sum of) $KHG$ and $GHM$. But, (the sum of) $FKH$ and $KHG$ is equal to two right-angles [Prop.~1.29]. Thus, (the sum of) $KHG$ and $GHM$ is also equal to two right-angles. So two straight-lines, $KH$ and $HM$, not lying on the same side, make  adjacent angles with some straight-line $GH$,  at the point $H$ on it, (whose sum is) equal to two right-angles. Thus, $KH$ is straight-on to $HM$ [Prop.~1.14]. And since the straight-line $HG$ falls across the parallels $KM$ and $FG$, the alternate angles $MHG$ and $HGF$ are equal to one another [Prop.~1.29]. Let $HGL$ have been added to both. Thus, (the sum of) $MHG$ and $HGL$ is equal to  (the sum of) $HGF$ and $HGL$. But, (the sum of) $MHG$ and $HGL$ is equal to two right-angles [Prop.~1.29]. Thus, (the sum of) $HGF$ and $HGL$ is also equal to two right-angles. Thus, $FG$ is straight-on to $GL$ [Prop.~1.14]. And since $FK$ is equal and parallel to $HG$ [Prop.~1.34], but also $HG$ to $ML$ [Prop.~1.34], $KF$ is thus also equal and parallel to $ML$ [Prop.~1.30]. And the straight-lines $KM$ and $FL$ join them. Thus, $KM$ and $FL$ are equal and parallel as well [Prop.~1.33]. Thus, $KFLM$ is a parallelogram. 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) $E$. (Which is) the very thing it was required to do.