In any triangle, (the sum of) two sides taken together in any (possible way) is greater than the remaining (side). For let $ABC$ be a triangle. I say that in triangle $ABC$ (the sum of) two sides taken together in any (possible way) is greater than the remaining (side). (So), (the sum of) $BA$ and $AC$ (is greater) than $BC$, (the sum of) $AB$ and $BC$ than $AC$, and (the sum of) $BC$ and $CA$ than $AB$.  For let $BA$ have been drawn through to point $D$, and let $AD$ be made equal to $CA$ [Prop.~1.3], and let $DC$ have been joined.  Therefore, since $DA$ is equal to $AC$, the angle $ADC$ is also equal to $ACD$ [Prop.~1.5]. Thus, $BCD$ is greater than $ADC$. And since  $DCB$ is a triangle having the angle $BCD$ greater than $BDC$, and the greater angle subtends the greater side [Prop.~1.19], $DB$ is thus greater than $BC$. But $DA$ is equal to $AC$. Thus, (the sum of) $BA$ and $AC$ is greater than $BC$. Similarly, we can show that (the sum of) $AB$ and $BC$ is also greater than $CA$, and (the sum of) $BC$ and $CA$ than $AB$.  Thus, in any triangle, (the sum of) two sides taken together in any (possible way) is greater than the remaining (side). (Which is) the very thing it was required to show.