For each point on a line there exists a unique perpendicular line through that point

Question

I'm trying to show that in an absolute plane (only the first four axioms without the parallel axiom hold) for each Point $P\in l$ there exists exactly one perpendicular line through $P$. My idea was to construct a equilateral triangle by choosing $A,B\in l$ with $d(P,A)=d(P,B)=1$, but I fail to construct the triangle from there on. We haven't defined circles, so I can't find the third point as an intersection of two circles.

Once I found the third point of the triangle $C$, the triangles $\triangle ACP$ and $\triangle BCP$ are congruent from which it follows that the line is perpendicular. Not quite sure how I can show uniqueness.

We define right angle as an angle being congruent to its adjacent angles.

Answer 1

To prove the existence of a perpendicular you need to know that:

1. There exists a right angle.
2. An angle congruent to a right angle is a right angle.

Now you use one of the axioms of congruence, which says, that an angle can be laid off upon a given side of a given half-ray.

To prove the uniqueness, you need to know, that:

3. All right angles are congruent.

Then you use one of the axioms of congruence, which says, that an angle can be laid off upon a given side of a given half-ray in only one way.