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;
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.
Then, since the angle E is equal to each of the angles HKF, GHM,
the angle HKF is also equal to the angle GHM.
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;
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 in straight line with HM.
And, since the straight line HG falls upon the parallels KM, FG, the alternate
angles MHG, HGF are equal to one another.
Let the angle HGL be added to each;
therefore the angles MHG, HGL are equal to the angles HGF, HGL.
But the angles MGH, HGL are equal to two right angles;
therefore the angles HGF, HGL are also equal to two right angles.
Therefore FG is in a straight line with GL.
And since it is equal and parallel to HG, and HG to ML also,
KF is also equal and parallel to ML;
and the straight lines KM, FL join them (at their extremities); therefore KM,
FL are also equal and parallel.
Therefore KLFM 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 which is equal to the given angle E.
Q.E.F