In any triangle two sides taken together in any manner are greater than the remaining one.
For let ABC be
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.
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
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, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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