Proving two lines intersect without using Euclid's parallel axiom

Question

I am a beginner of Euclidean geometry, and now I am stuck on the following problem.

Let $L$ be a line and $A,B$ be two different points to the same side of $L$ (not on $L$). Now I would like to find a point $Q$ on $L$ so that a beam of light goes from $A$ will bounce back and pass $B$.

The construction is clear to me. I need to just find $A'$ which is symmetric to $A$ with $L$ as a mirror, and then link $A'$ to $B$. The intersection of $A'B$ with $L$ is just $Q$.

Now the problem is, I could not use the parallel axiom. How could I use Euclid's propositions 1.1-1.28 to conclude that $Q$ exists?

Answer 1

I am not too sure what Euclid's propositions 1.1-1.28 are. But I think in Euclid's geometry proof by construction is good. And your construction does not make any use of the parallel Axiom. So you're good. Here's a very quick try to generalize your construction

1. There is a line $a$: $A \in a$ and $a\perp l$

2. $a\cap l=\{M\}$

3. There is $A' \in a$ so that $|AM|$ and $|A'M|$ are congruent

4. $A'$ is in a different half plane (parted by $l$) then $A$ and $B$

5. $A'B \cap l \neq \emptyset $. There is $Q$