If a straight-line stood on a(nother)  straight-line makes angles, it will certainly either make two right-angles, or (angles whose sum is) equal to two right-angles. For  let some straight-line $AB$ stood on the straight-line $CD$ make the angles $CBA$ and $ABD$. I say that the angles $CBA$ and $ABD$ are certainly either two right-angles, or (have a sum) equal to two right-angles.  In fact, if $CBA$ is equal to $ABD$ then they are two right-angles [Def.~1.10]. But, if not, let $BE$ have been drawn from the point $B$ at right-angles to [the straight-line] $CD$ [Prop.~1.11]. Thus, $CBE$ and $EBD$ are two right-angles. And since $CBE$ is equal to the two (angles) $CBA$ and $ABE$, let $EBD$ have been added to both. Thus, the (sum of the angles) $CBE$ and $EBD$ is equal to the  (sum of the) three (angles) $CBA$, $ABE$, and $EBD$ [C.N.~2]. Again, since $DBA$ is equal to the two (angles) $DBE$ and $EBA$, let $ABC$ have been added to both. Thus, the (sum of the angles) $DBA$ and $ABC$ is equal to the (sum of the) three (angles) $DBE$, $EBA$, and $ABC$ [C.N.~2]. But (the sum of) $CBE$ and $EBD$ was also shown (to be) equal to the (sum of the) same three (angles). And things equal to the same thing are also equal to one another [C.N.~1]. Therefore,  (the sum of) $CBE$ and $EBD$ is also equal to (the sum of) $DBA$ and $ABC$. But, (the sum of) $CBE$ and $EBD$ is two right-angles. Thus, (the sum of) $ABD$ and $ABC$ is also equal to two right-angles.  Thus, if a straight-line stood on a(nother)  straight-line makes angles, it will certainly either make two right-angles, or (angles whose sum is) equal to two right-angles. (Which is) the very thing it was required to show.