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
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
let the parallelogram GM equal to the trangle 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
the angle HKF is also equal to the angle GHM. [C.N. 1]
Let the angle KHG be added
therefore the angles KFH, KHG are equal to the angles KHG, GHM.
But the angles FKH,
KHG are equal to two right angles;
thereore 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
not lying on the same side make the adjacent angles
equal to two right angles;
therefore HK is in a straight line with HM [I. 14]
And, since the straight line HG falls upon the parallels KM, FG, the alternate angles MHG, HGF are equal to one another. [I. 29]
Let the angle HGL be added
therefore the angles MHG, HGL are equal to the angles HGF, HGL. [C.N. 2]
And, since FK is equal and
parallel to HG,
and HG to ML also,
KF is also equal and parallel to ML; [C.N. 1; ] [I. 30]
and the straight lines KM, FL join them (at their extremities); therefore KM, FL are also equal and parallel. [I. 33]
Therefore KFLM is a parallelogram.
And, since the triangle ABD is equal
to the parallelogram
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.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
Next: Proposition 46
Previous: Proposition 44
This proposition in other editions: