Out of three straight lines, which are equal to three given straight lines, to construct a triangle: thus it is necessary that two of the straight lines taken together in any manner should be greater than the remaining one.
Let the three given straight lines
be A, B,
C,
and of these let two taken together in any manner be
greater than the remaining one,
namely
A, B greater than C,thus it is required to construct a triangle out of straight lines equal to A, B, C.
A, C greater than B,
and
B, C greater than A;
Let there be set out a straight line DE, terminated at D but of infinite length in the direction of E, and let DF be made equal to A, FG equal to B, and GH equal to C.
With centre F and
distance FD let the
circle DKL be described;
again, with centre G and
distance GH let the
circle KLH be
described;
and let KF,
FG be joined;
I say that the triangle KFG has
been constructed out of three straight lines equal to
A, B,
C.
For, since the point F
is the centre of the circle
DKL,
FD is equal to
FK.
But FD is equal to
A;
therefore KF is also
equal to A.
Again, since the point G
is the centre fo the circle LKH,
GH is equal
to GK.
But GH is
equal to C;
therefore GK is
equal to C.
And FG is also
equal to B;
therefore the three straight lines
KF, FG,
GK are equal to
the three straight lines A,
B, C.
Therefore out of the three straight lines KF, FG, GK, which are equal to the three given straight lines A, B, C, the triangle KFG has been constucted. Q.E.F.
Book I: Euclid, Elements, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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