If two straight-lines, not lying on the same side,  make adjacent angles (whose sum is) equal to two right-angles  with some straight-line, at a point on it, then the two straight-lines will be straight-on (with respect) to one another. For let two straight-lines $BC$ and $BD$, not lying on the same side,  make adjacent angles $ABC$ and $ABD$ (whose sum is) equal to two right-angles with some straight-line $AB$, at the point $B$ on it. I say that $BD$ is  straight-on with respect to $CB$.  For if $BD$ is not straight-on to $BC$ then let $BE$ be straight-on to $CB$.  Therefore, since the straight-line $AB$ stands on the straight-line $CBE$, the (sum of the) angles $ABC$ and $ABE$ is thus equal to two right-angles [Prop.~1.13]. But (the sum of) $ABC$ and $ABD$ is also equal to two right-angles. Thus,  (the sum of angles) $CBA$ and $ABE$ is equal to (the sum of angles) $CBA$ and $ABD$ [C.N.~1]. Let (angle) $CBA$ have been subtracted from both. Thus, the remainder $ABE$ is equal to the remainder $ABD$ [C.N.~3], the lesser to the greater. The very thing is impossible. Thus, $BE$ is not straight-on with respect to $CB$. Similarly, we can show that neither (is) any other (straight-line)   than $BD$. Thus, $CB$ is straight-on with respect to $BD$.  Thus, if two straight-lines, not lying on the same side,  make adjacent angles (whose sum is) equal to two right-angles with some straight-line, at a point on it, then the two straight-lines will be straight-on (with respect) to one another. (Which is) the very thing it was required to show.