Proposition 20.
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.
1