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; [I. 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. [I. 44] Then, since the angle E is equal to each of the angles HKF, GHM,
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 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; [I. 29] therefore the angles HGF, HGL are also equal to two right angles. [C.N. 1]
therefore KM, FL are also equal and parallel. [I. 33]
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 which is equal to the given angle E.
Q. E. F. 1