(Edited by Sir Thomas L. Heath, 1908)

[Euclid, ed. Heath, 1908, on

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 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
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 KFH,
KHG are equal to the angles
KHG, GHM.

But the angles FKH,
KHG are equal to two right angles;
[I. 29]

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
KH, HM
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
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;
[I. 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
[I. 14]

And, since FK is equal and
parallel to HG,
[I. 34]

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
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.

Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)

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