(Edited by Sir Thomas L. Heath, 1908)

[Euclid, ed. Heath, 1908, on

To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.

Let AB be the given straight
line, C the given triangle and
D the given rectilineal
angle;

thus it is required to apply to the given straight line
AB, in an angle equal to the
angle D, a parallelogram equal
to the given triangle C.

Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42] ; let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and let AH be drawn through A parallel to either BG or EF [I. 31]

Let HB be joined.

Then, since the straight line HF
falls upon the parallels AH,
EF,

the angles AHF,
HFE are equal to two right angles.
[I. 29]

Therefore the angles BHG,
GFE are less than two
right angles;

and straight lines produced indefinitely from angles less
than two right angles meet;
[Post. 5]

therefore
HB, FE,
when produced, will meet.

Let them be produced and meet at K;
through the point K let
KL be drawn parallel to either
EA or FH,
[I. 31]

and let HA, GB
be produced to the points
L, M.

Then HLKF is a
parallelogram,

HK is its diameter,
and AG, ME
are parallelograms, and
LB, BF
the so-called complements, about
HK;

therefore LB is equal to
BF.
[I. 43]

But BF is equal to the
triangle C;

therefore LB is also
equal to C.
[C.N. 1]

And, since the angle GBE
is equal to the angle ABM,
[I. 15]

while the angle GBE is equal to
D,

the angle ABM is also equal to
the angle D.

Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM 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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