If two internal straight-lines are constructed on one of the sides of a triangle, from its ends,  the constructed (straight-lines) will be less than the two remaining  sides of the triangle, but will encompass a greater angle. For let the two internal straight-lines $BD$ and $DC$ have been constructed on one of the sides $BC$ of the triangle $ABC$, from its ends $B$ and $C$ (respectively). I say that  $BD$ and $DC$ are less than the (sum of the) two  remaining sides of the triangle $BA$ and $AC$, but encompass an angle $BDC$ greater than $BAC$.  For let $BD$ have been drawn through to $E$. And since in any triangle (the sum of any) two sides is greater than the remaining (side) [Prop.~1.20],  in triangle $ABE$ the  (sum of the) two sides $AB$ and $AE$ is thus  greater than $BE$. Let $EC$ have been added to both. Thus, (the sum of) $BA$ and $AC$ is greater than (the sum of) $BE$ and $EC$.  Again, since in triangle $CED$ the (sum of the) two sides $CE$ and $ED$ is  greater than $CD$, let $DB$ have been added to both. Thus,  (the sum of) $CE$ and $EB$ is greater than  (the sum of) $CD$ and $DB$. But, (the sum of) $BA$ and $AC$ was shown (to be) greater than (the sum of) $BE$  and $EC$. Thus, (the sum of) $BA$ and $AC$ is much greater than (the sum of) $BD$ and $DC$.  Again, since in any  triangle the external angle  is greater than the internal and opposite (angles) [Prop. 1.16],  in triangle $CDE$ the external angle $BDC$ is thus greater than $CED$.  Accordingly, for the same (reason),  the external angle $CEB$ of the triangle $ABE$ is also greater than $BAC$. But, $BDC$ was shown (to be)  greater than $CEB$. Thus, $BDC$ is much greater than $BAC$.  Thus, if two internal straight-lines are constructed on one of the sides of a triangle, from its ends,  the constructed (straight-lines) are less than the two remaining  sides of the triangle, but  encompass a greater angle. (Which is) the very thing it was required to show.