Prove that through every point in space, not lying on a given line, there exists a unique line parallel to the given one

Question

Prove that through every point in space, not lying on a given line, there exists a unique line parallel to the given one.

let's name the point $A$, the given line $a$, and the searched line $b$.

I thought to create a a plane $(c, A)$, then the intersection between $(a,A)$ and $(c, A)$ will be a line passing through point $A$, and will be parallel to line $a$ and $c$. However firstly, I don't know how to prove that this line is unique. Also I have the sense that something is wrong with my proof.

Answer 1

This question is about Euclid's fifth postulate (parallel postulate). A postulate (or axiom) is something that we assume to be true as an initial premise.

You mentioned that this postulate doesn't come natural to you, which is undesireable for postulates. The Wikipedia atricle contains this line:

Many other statements equivalent to the parallel postulate have been suggested, some of them appearing at first to be unrelated to parallelism, and some seeming so self-evident that they were unconsciously assumed by people who claimed to have proven the parallel postulate from Euclid's other postulates.

I find that a thorough reading of the Wikipedia page may shed some light on the "comming natural to you" part of it all.

Answer 2

First proof that the Point outside the line, and the line itself are in a plane.

You are given a point P and a a line. By postulates, we know that the said line contains two points, call them points A and B, respectively. Moreover, we are also guaranteed that there is exactly one plane through all these three points and that this planes contains point P and the line through points A and B. Recall that the notion of parallel lines only makes sense if the said lines are co planar. Hence this part is essential.

Now proof that we can draw a line through P parallel to line through A and B that we call line k.

Draw a line called m through point P, assuming that there is no line through P that is parallel to line k , there be a point of intersection called D shared by both line m and k.

Clearly, there is an angle formed by the rays DP and ray DA ( assuming point A is the left of D, otherwise just pick any point to the left on D on line k. )

Call this angle formed ALPHA.Assuming that our angle Alpha is restricted between 0 and 180

0 < ALPHA < 180 , note it can't be either 0 or 180 as this would mean that point P is actually on the line m.

Taking a look at our line m, we see that we have point D to the right of point P, now choose a point to the left of point P call it E.

By another postulate, the protractor postulate to be exact, We can pair all the angles based at point P with the numbers between 0 and 180, which reflect the degree measure starting from ray PD. All we have to do now is draw the angle that has measure value ALPHA, and extend its terminal rays into a line, call this line j.

It follows that line m is a transversal line through lines k and j. Since alternate interior angles are congruent / equal in degree measurement, we can conclude from yet another theorem/postulate, that lines k and j are parallel.

QED.

I'm using Geometry by as Jurgenson reference.This is a proof by contradiction.

I used from the book

Postulate 3 on pg 18 Postulate 5 on pg 23 Postulate 8, 9 on pg 23 ThEOREM 1-1 ON pg 23. Theorem 3-5 on pg 83.

My proof only shows the existence part. I think in these types of proofs the existence is the hardest part.