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 $g$.
Existence: I try to construct the line perpendicular to $g$ that goes through $P$. For this let $A, B$ be two distinct points on $g$. Then $d(A,P)=r$ and $d(B,P)=r’.$
By a previous result, we can find a unique point $P’$ in the half-plane that does not contain $P$ such that $d(A,P’)=r$ and $d(B,P’)=r’$.
Now consider the line going through $P$ and $P’$. This will be the line that we’re looking for, but how do I prove that it is perpendicular to $g$ now?
Let $M$ be the intersection of the two lines. Then $\triangle ABP=\triangle ABP’$.
Therefore $\angle BAP=\angle BAP’$.
Then $\triangle P’MA=\triangle PMA$. So, $\angle P’MA=\angle PMA$ and $\angle P’MA+\angle PMA$ equals two right angles.