To construct, in a given rectilineal angle, a parallelogram equal to a given rectilineal figure.

Let $ABCD$ be the given rectilineal figure and $E$ the given rectilineal angle;thus it is required to construct, in the given angle $E$, a parallelogram equal to the rectilineal figure $ABCD$.Let $DB$ be joined, and let the parallelogram $FH$ be constructed equal to the triangle $ABD$, in the angle $HKF$ which is equal to $E$; [Prop. 1.42]let the parallelogram $GM$ equal to the triangle $DBC$ be applied to the straight line $GH$, in the angle $GHM$ which is equal to $E$. [Prop. 1.44]Then, since the angle $E$ is equal to each of the angles $HKF$, $GHM$,the angle $HKF$ is also equal to the angle $GHM$. [C.N. 1]Let the angle $KHG$ be added to each;therefore the angles $FKH$, $KHG$ are equal to the angles $KHG$, $GHM$.But the angles $FKH$, $KHG$ are equal to two right angles. [Prop. 1.29]therefore the angles $KHG$, $GHM$ are also equal to two right angles.Thus, with a straight line $GH$, and at the point $H$ on it, two straight lines $KH$, $HM$ not lying on the same side make the adjacent angles equal to two right angles;therefore $KH$ is in a straight line with $HM$. [Prop. 1.14]And, since the straight line $HG$ falls upon the parallels $KM$, $FG$, the alternate angles $MHG$, $HGF$ are equal to one another. [Prop. 1.29]Let the angle $HGL$ be added to each;therefore the angles $MHG$, $HGL$ are equal to the angles $HGF$, $HGL$. [C.N. 2]But the angles $MHG$, $HGL$ are equal to two right angles; [Prop. 1.29]therefore the angles $HGF$, $HGL$ are also equal to two right angles. [C.N. 1]Therefore $FG$ is in a straight line with $GL$. [Prop. 1.14]And, since $FK$ is equal and parallel to $HG$, [Prop. 1.34]and $HG$ to $ML$ also,$KF$ is also equal and parallel to $ML$; [C.N. 1] [Prop. 1.30]and the straight lines $KF$, $ML$ join them (at their extremities);therefore $KM$, $FL$ are also equal and parallel. [Prop. 1.33]Therefore $KFLM$ is a parallelogram.And, since the triangle $ABD$ is equal to the parallelogram $FH$,and $DBC$ to $GM$,the whole rectilineal figure $ABCD$ is equal to the whole parallelogram $KFLM$.Therefore the parallelogram $KFLM$ has been constructed equal to the given rectilineal figure $ABCD$, in the angle $FKM$ equal to the given angle $E$.Q.E.F.