I'm needing some guidance in order to prove that the following proposition:
Any point in the interior of an (not $\pi$) angle $\angle AÔB$ is in a segment whose extremes are in the rays $\vec{OA}$ and $\vec{OB}$.
is a substitute to Euclid's fifth postulate, which I can use the Playfair substitute, since that equivalence I already proved:
Given a line $r$ and a point $P \notin r$, there is exactly one line $s$ such that $P \in s$ and $r \cap s = \phi $.
Here is my attempt: Suppose that there is a point $P$ in the interior of $ \angle AÔB$. Since $P$ is not in the support line of $OA$, there is exactly one line through $P$ not intersecting sup. line $OA$. consider the perpendicular line to said line. My aim is to show that this perpendicular will intersect the ray $OB$, but I'm having trouble doing this. Can someone say if this is a good approach and how to fix it? I would also accept different solutions to this problem.
Thanks.
Let's recall first of all that an immediate consequence of Euclid's fifth postulate is the following: if two lines are intersected by a common transversal and form with it non supplementary consecutive interior angles, then these lines meet on that side of the transversal where the sum of consecutive interior angles is less than $\pi$ (as a matter of fact, this is exactly how the postulate was presented by Euclid himself).
But even without Euclid's fifth postulate we can state a weaker result:
LEMMA - If two lines are intersected by a common transversal and form with it non supplementary consecutive interior angles, then these lines cannot meet on that side of the transversal where the sum of consecutive interior angles is greater than $\pi$.
This can be proven by the exterior angle theorem: if $\alpha+\beta>\pi$ (see picture below) then $\gamma=\pi-\beta<\alpha$; if $AA'$ and $BB'$ would meet at $P$ on the right, then triangle $PAB$ would have an external angle $\gamma$ greater then an internal angle, which is impossible.
Let's now prove that Euclid's fifth postulate implies your proposition. Consider a convex angle $\angle AOB\ne\pi$ and a point $P$ in its interior. Draw from $P$ the parallel to $OB$, which meets ray $OA$ at $A$ (see picture below), and take a point $Q$ on $OA$ such that $A$ is between $O$ and $Q$. We have $\alpha>\phi$ and $\alpha=\pi-\theta$, whence $\phi+\theta<\pi$: it follows that line $PQ$ meets ray $OB$.
Let's now show the converse, in its contrapositive form: supposing that uniqueness of a parallel line is not guaranteed, I'll prove that one can construct an angle, and a point inside it, for which your proposition does not hold.
Let $a$ be a line in the plane and $P$ a point not belonging to $a$ and $O$ a point on $a$. Through $P$ we can construct a line $a'$ forming equal alternating interior angles with $OP$, so we know that $a'\parallel a$. Suppose now that uniqueness of parallel line is violated, so that there exists another line $b'$ through $P$ parallel to $a$. We can construct a line $b$ through $O$ forming equal alternating interior angles with $OP$ and thus parallel to $b'$. Of the four convex angles formed by $a$ and $b$ choose the one containing $P$: I'll call this $\angle ab$ and from now on I will use $a$ and $b$ to denote those rays of the lines which are the sides of $\angle ab$. I want to prove that there exists no segment with endpoints on $a$, $b$ and containing $P$.
To this end, take any point $A$ on $a$, and let $\angle AOP=\alpha$ (see picture below). Denote then by $\beta$ the angle formed by $b$ with $OP$ and notice that $\alpha+\beta<\pi$ (because the angle is convex). Moreover, we have $\angle APO<\beta$, because if $\angle APO=\beta$ then $A$ would belong to $b'$ by construction (impossible, as $a$ and $b'$ are parallel) and if $\angle APO>\beta$ then $b'$ would intersect $AO$ (impossible again).
It follows that line $AP$ and ray $b$ form with $OP$ consecutive interior angles greater than $\pi$: by the lemma stated above, that proves the assertion.