Proving "If there exists a line containing exactly $n$ points, then any line contains exactly $n$ points" from basic axioms of incidence I have to show that the following theorem can be proven using the axioms cited below:

If there exists a line that contains exactly $n$ points, then any line contains exactly n points, and any point has exactly $n$ lines that contain it.

The axioms are the following:

*

*There exist at least one line.

*If L is a line, then there exist at least 3 points on L

*If L is a line, then there exists one point that is not on L

*Given any two points P and Q, there exists one and only one line that contains both P and Q

*If L and M are two lines, then there exists at least one point P that is contained in both L and M.

My effort. I have thought about using induction to prove the statement, but I don't know how to start.
Can you please provide me an overview of a proof of this theorem? Thank you so much for your time!
UPDATE
I managed to show that every point is contained in at least $n$ lines, but I have no idea on how to show that it is contained in exactly $n$ lines, which is what the theorem states.
 A: Okay, let's go for it. I will use the axiom numbering present in the question. lower case letters represent points, UPPER CASE letters represent lines, because... $42$.
We use the notation $p \in L$ for a point $p$ and a line $L$ if $p$ is contained in $L$, and $p \notin L$ if it's not contained in $L$. For two lines $L$ and $M$, $L \cap M$ denotes the collection of points contained in both $L$ and $M$ i.e. where $L$ and $M$ intersect.

We first improve axiom $5$.

If $L \neq M$ are two distinct lines, then $|L \cap M| = 1$. (Note : axiom $5$ would say $|L \cap M| \geq 1$).

Proof : Suppose $p \neq q \in L \cap M$. By axiom $4$, there exists a unique line $L'$ such that $p,q \in L'$, but then $L,M$ are two distinct candidates which fulfill this criteria. This contradiction proves that $|L \cap M| \leq 1$, which combined with axiom $5$ proves that $|L \cap M| = 1$. $\blacksquare$

Next, we devise the following "projective" strategy : let's imagine looking at two different lines , from a point not on either of them. Then, each point on one of the lines "corresponds" to a point on the other line, as the two points that are in the "same direction" when viewed from that exterior point. This correspondence can be proved to be a bijection, so that all lines have the same number of points! I would suggest drawing a diagram , probably I'll attach one later on if I get the time.

So we begin with the first idea , of proving the existence of the exterior point.

For two lines $L \neq M$ there is a point $r \notin L , r \notin M$.

Proof : This one is nice! We know that $|L \cap M| = 1$ so let $q$ be that intersection point. From axiom $2$, Pick a $p \in L$ and a $p' \in M$ such that $p \neq q, p' \neq q$. From axiom $4$, consider the unique line $N$ that passes through $p$ and $p'$. By axiom $2$, there is a point $r \neq p , r \neq p', r \in N$. Can you prove that $r \notin L , r \notin M$?$\blacksquare$

 If $r \in L$, then $r \in N \cap L$ and $p \in N \cap L$ and $r \neq p$, which means $|N \cap L| > 2$, a contradiction as $N \neq L$ , $p'$ being contained in the former but not in the latter. Similarly we can contradict $r \in M$.


Now we are ready to "view $L$ and $M$" from $r$. Since $L$ and $M$ are "collections" of points, I'll treat them as sets and use functional notation. Let $p \in L$ and let $L_p$ be the unique line connecting $p$ and $r$. Since $L_p \neq M$ (as $r \notin M$) we know that $|L_p \cap M| = 1$. Let that point of intersection be $p'$. We define the function $f: L \to M$ by $f(p) = p'$. Clearly $f$ is well defined by what we've already written.

$f : L \to M$ as described above is an injective map, and hence $M$ contains at least as many points as $L$.

Proof : Suppose $f(p_1) = f(p_2)$. Let $L_{p_1}$ be the unique line joining $p_1$ and $r$, and $L_{p_2}$ be the unique line joining $p_2$ and $r$.
Since $f(p_1)=f(p_2)$, this point lies on both $L_{p_1}$ and $L_{p_2}$, but is distinct from $r$ which also lies on $L_{p_1}$ and $L_{p_2}$. Thus $|L_{p_1}\cap L_{p_2}| > 1$, which is a contradiction unless $L_{p_1} = L_{p_2} = L'$. So $L'$ contains both $p_1$ and $p_2$, but these points are also contained in $L$! If $p_1\neq p_2$ then $|L'\cap L| > 1$, a contradiction since $L'$ contains $r$ but $L$ does not. Hence, $p_1=p_2$ and injectivity is proved! $\blacksquare$
Note that the intersection point of $L$ and $M$ actually maps to itself under $f$.

But then, we can flip the roles of $L$ and $M$ above ,retain the point $r$ and obtain a function $f' : M \to L$ which is injective by the same logic above. Thus, we get that the map $f: L \to M$ is in fact bijective, and :

Corollary : All lines have $n$ points.

Proof : Suppose $T$ is the line which has $n$ points and $T'$ is any other line. Then $f : T \to T'$ as described earlier is bijective, hence $T'$ has $n$ points as well.$\blacksquare$

We now switch to the second part of the story.

The question is, how do we plan to do this? Well, think about this way : earlier, we "saw" two different lines through a point not on them. Now, we are going to see the lines through a point, from a line that doesn't pass through that point.
For this, we need that line , of course!

For a point $p$, there is a line $L$ with $p \notin L$.

Proof : If there is no line passing through $p$, then we are done by axiom $1$. So let $L$ be a line passing through $p$. By axiom $3$, pick a $q \notin L$. By axiom $2$ , let $r \in L, r \neq p$ . By axiom $4$, let $L'$ be the unique line passing through $r$ and $q$. Then $p \notin L'$, because if $p \in L'$ then $q \in L', q \notin L$, but $r,p \in L \cap L'$, contradiction. So $L'$ is the desired line. $\blacksquare$

Let's make every line that passes through $p$ now intersect $L'$ at various points. This is how $L'$ "sees" $p$, and it turns out this is magnificently good!
To make this more precise, let $\mathcal F$ be the collection of lines $L$ with $p \in L$. We will define a function $F : \mathcal F \to L'$ as follows : for $L \in \mathcal F$, we know that $L \neq L'$ since $p \in L, p \notin L'$. Thus, $|L \cap L'| = 1$, and that intersection point is $f(L)$.

$F : \mathcal F \to L'$ is a bijection.

Proof: We need to show injectivity and surjectivity.
Injectivity
Suppose that $F(L_1) = F(L_2)$. Then, $L_1$ and $L_2$ intersect $L'$ at the same point, but also pass through $p$. Hence, by axiom $4$ applied to these points, we know that $L_1=L_2$.
Surjectivity
This is again following from axiom $4$, since if $r \in L'$ then $r \neq p$ so there is a line $L$ that passes through $r$ and $p$, which means that $L \in \mathcal F$ and $F(L) = r$, as desired.
Hence, $F$ is bijective!$\blacksquare$.

Corollary : As $F$ is bijective, for every point $p$ we have that $\mathcal F$ contains $n$ lines, since $L'$ contains $n$ points.
