Proof: By Euclid
(related to Proposition: 1.14: Combining Rays to Straight Lines)
- 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.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"