Unique perpendicular line

Question

Consider an absolute plane (i.e. it satisfies all axioms except the parallel axiom). Let g be a straight line and P a point not in g. Then there is a unique straight line going through P which is perpendicular on g.

I asked a question regarding the existence in the same problem here. Now I have a question regarding Uniqueness.

Suppose $A$ and $B$ are points on $g$ such that the lines $PA$ and $PB$ are both perpendicular to $g$. Suppose toward contradiction that $A\neq B$.

Now consider the unique point $A'$ on $AP^+$ such that d(A',P)=d(A,P). Similarly consider the unique point $B'$ on $BP^+$ such that $d(B',P) = d(B,P)$.

Then the triangles $\triangle PA'B'$ and $\triangle PAB$ are congruent.

How do I generate a contradiction from this? Am I on the right track?

Answer 1

There are two ways to prove this. Use one of these theorems:

1. If alternate angles are congruent, then the lines are parallel.

2. An exterior angle of a triangle is greater than interior angle not adjacent to it.

Answer 2

Saccheri-Legendre Theorem (https://www.math.unl.edu/~bharbourne1/M812TSummer2014/Day6/Saccheri-LegendreTheorem.pdf) states that "If one assumes Euclid's postulates other than the parallel postulate, then the sum pf the interior angles of any triangle is at most 180 degrees." This Theorem is true in Your case. Now, look at the triangle PAB that appears in Your case. If A and B are different points, then the interior angle APB is greater than 0, interior angles PAB and PBA are both 90 degrees. Hence the sum of the interior angles of the triangle PAB is greater than 180 degrees. A contradiction. Hence, there could be at most one perpendicular to g from P.