In any triangle two sides taken together in any manner are greater than the remaining one.
For let ABC be
a triangle;
I say that in the triangle ABC two sides
taken together in any manner are greater than the remaining
one, namely
BA, AC greater than BC,
AB, BC greater than AC,
BC, CA greater than AB.
For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined.
Then, since DA is
equal to AC,
the angle ADC is also equal to the
angle ACD;
[I. 5]
therefore the angle BCD is greater than the
angle ADC.
[C.N. 5]
And, since DCB is a triangle
having the angle BCD greater
than the angle BDC,
and the greater angle is subtended by the greater side,
[I. 19]
therefore DB is greater
than BC.
But DA is equal
to AC;
therefore BA,
AC are greater
than BC.
Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.
Therefore etc. Q.E.D.
Book I: Euclid, Elements, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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