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 $ACD$; [Prop. 1.5]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, [Prop. 1.19]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.