And because at the point H in the straight line GH, the two straight lines KH, HM, on the opposite sides of it, make the adjacent angles together equal to two right angles, KH is in the same straight line with HM. [I. 14.
And because the straight line HG meets the parallels
KM, FG, the alternate angles MHG, HGF are equal. [I. 29.
Add to each of these equals the angle HGL ;
therefore the angles MHG, HGL, are equal to the angles HGF, HGL. [Axiom 2.
But MHG,HGL are together equal to two right angles; [I. 29.
therefore HGF, HGL are together equal to two right angles.
Therefore FG is in the same straight line with GL. [I. 14.
And because KF is parallel to HG, and HG to ML,[Constr.
KF is parallel to ML ; [I. 30.
and KM, FL are parallels ; [Construction.
therefore KFLM a parallelogram. [Definition.
And because the triangle ABD is equal to the parallelo-
gram HF, [Construction.
and the triangle DBC to the parallelogram GM ; [Constr.
the whole rectilineal figure ABCD is equal to the whole
parallelogram KFLM. [Axiom 2.
Wherefore, the parallelogram KFLM has been de- scribed equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E. q.e.f.
Corollary. From this it is manifest, how to a given
straight line, to apply a parallelogram, which shall have an
angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure ; namely, by applying to the
given straight line a parallelogram equal to the first tri-
angle ABD, and having an angle equal to the given angle;
and so on. [I. 44.
