Proof: By Euclid

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.
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