(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 triangle.

Let ABC be the given triangle,
and D the given
rectilineal angle;

thus it is required to construct in the rectilinear
angle D a parallelogram
equal to the triangle ABC.

Let BC be bisected
at E, and let
AE be joined;

on the straight line EC, and
at the point E on it, let
the angle CEF be constructed equal to the
angle D;
[I. 23]

through A let
AG be drawn
parallel to EC,
[I. 31]

and through C let
CG be drawn parallel to
EF.

Then FECG is a parallelogram.

And, since BE is equal to
EC,

the triangle ABE is also equal
to the triangle
AEC,

for they are on equal bases BE,
EC and in the same
parallels BC,
AG;

therefore the triangle ABC
is double of the triangle AEC.

But the parallelogram FECG
is also double of the triangle AEC,
for it has the same base with it and is in the same
parallels with it;
[I. 41]

therefore the parallelogram FECG
is equal to the triangle
ABC.

And it has the angle CEF
equal to the given angle D.

Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q.E.F.

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

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