Proof: By Euclid

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.
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