[proofplan] We first fix one line and use the affine parallel axiom to count the lines through an external point. This gives a bijection between the points on a line and most of the lines through the external point. Next we show that any line through an external point has $n$ points, and then use this to compare an arbitrary line with the fixed line. Once the common line size is $n$, a fixed parallel class partitions the point set, a transversal counts the number of lines in that class, and incidence counting gives the number of points, the number of lines, and the number of parallel classes. [/proofplan] [step:Count the lines through a point outside a fixed line] For a line $\ell \in L$, define its point set \begin{align*} P_\ell := \{x \in P : (x,\ell) \in I\}. \end{align*} Choose a line $\ell \in L$ and a point $p \in P$ with $(p,\ell) \notin I$; such a point exists because the affine plane has non-collinear points. Let \begin{align*} n := |P_\ell|. \end{align*} Since every line contains at least two points, $n \ge 2$. For each $x \in P_\ell$, let $m_x \in L$ denote the unique line incident with both $p$ and $x$. Define two lines $m,m' \in L$ to be parallel when either $m=m'$ or their point sets are disjoint. For any line $m \in L$, define $P_m:=\{y \in P:(y,m)\in I\}$. Define $\mathcal L(p,\ell)$ to be the set of lines through $p$ whose point sets meet $P_\ell$: \begin{align*} \mathcal L(p,\ell):=\{m \in L : (p,m) \in I,\ P_m \cap P_\ell \neq \varnothing\}. \end{align*} Define $\Phi:P_\ell \to \mathcal L(p,\ell)$ by $\Phi(x):=m_x$. This map is bijective. Indeed, if $m_x=m_y$, then $x$ and $y$ are both incident with $\ell$ and with $m_x$. Since two distinct points determine a unique line, $x \neq y$ would force $m_x=\ell$, contradicting $(p,\ell) \notin I$. Hence $x=y$, so $\Phi$ is injective. If $m$ is a line through $p$ meeting $\ell$ at a point $x$, then the uniqueness of the line through $p$ and $x$ gives $m=m_x$, so $\Phi$ is surjective. By the affine parallel axiom, there is a unique line $m_\parallel$ through $p$ disjoint from $\ell$. Therefore the lines through $p$ are exactly the $n$ lines $m_x$ with $x \in P_\ell$, together with $m_\parallel$. Hence $p$ is incident with exactly $n+1$ lines. [guided] We fix a line $\ell$ and measure the size of the plane using the number of points on this one line. Define \begin{align*} P_\ell := \{x \in P : (x,\ell) \in I\}, \end{align*} and let $n:=|P_\ell|$. Since each line contains at least two points, $n \ge 2$. Choose a point $p \in P$ not incident with $\ell$. The non-collinearity axiom guarantees that not every point lies on $\ell$, so such a point exists. For every point $x \in P_\ell$, the joining axiom gives a unique line $m_x$ through $p$ and $x$. Define two lines $m,m' \in L$ to be parallel when either $m=m'$ or their point sets are disjoint. For any line $m \in L$, define $P_m:=\{y \in P:(y,m)\in I\}$. Define $\mathcal L(p,\ell)$ to be the set of lines through $p$ whose point sets meet $P_\ell$: \begin{align*} \mathcal L(p,\ell):=\{m \in L : (p,m) \in I,\ P_m \cap P_\ell \neq \varnothing\}. \end{align*} Define $\Phi:P_\ell \to \mathcal L(p,\ell)$ by $\Phi(x):=m_x$. We verify that $\Phi$ is injective. Suppose $\Phi(x)=\Phi(y)$, so $m_x=m_y$. If $x \neq y$, then the two distinct points $x$ and $y$ lie on both $\ell$ and $m_x$. The joining axiom says that two distinct points have only one common line, so $m_x=\ell$. But $p$ lies on $m_x$ and does not lie on $\ell$, a contradiction. Thus $x=y$. We verify that $\Phi$ is surjective. Let $m$ be any line through $p$ meeting $\ell$. Choose a point $x \in P$ incident with both $m$ and $\ell$. Then $x \in P_\ell$, and both $m$ and $m_x$ are the unique line through the two distinct points $p$ and $x$. Hence $m=m_x$. Thus the $n$ points of $\ell$ correspond exactly to the $n$ lines through $p$ that meet $\ell$. The affine parallel axiom contributes one further line through $p$, namely the unique line disjoint from $\ell$. Therefore exactly $n+1$ lines pass through $p$. [/guided] [/step] [step:Show that all lines contain the same number of points] We first prove that intersecting lines have the same number of points. Let $a,b \in L$ be distinct lines with $P_a \cap P_b\neq \varnothing$, and let $o \in P_a\cap P_b$ be their unique common point. Choose $u \in P_a\setminus\{o\}$ and $v \in P_b\setminus\{o\}$, which exist because every line contains at least two points. Let $s \in L$ be the unique line through $u$ and $v$. The line $s$ is distinct from $a$ and $b$. Let $r \in L$ be the unique line through $o$ parallel to $s$. Since $s$ meets $a$ at $u$ and meets $b$ at $v$, the line $s$ is not parallel to either $a$ or $b$. Hence $r$ is distinct from both $a$ and $b$. Choose $p \in P_r\setminus\{o\}$, which exists because every line contains at least two points. If $p\in P_a$, then the two distinct points $o$ and $p$ would determine both $a$ and $r$, forcing $a=r$, a contradiction. Similarly $p\notin P_b$. Thus $p\notin P_a\cup P_b$. Let $r_b\in L$ be the unique line through $p$ parallel to $b$. The line $r_b$ cannot be parallel to $a$: if it were, then $a$ and $b$ would be two distinct lines through $o$ parallel to $r_b$, contradicting the affine parallel axiom applied at $o$ to $r_b$. Since $r_b$ is not parallel to $a$, it meets $a$; let $a_b\in P_a$ denote the unique point in $P_a\cap P_{r_b}$. Likewise, let $r_a\in L$ be the unique line through $p$ parallel to $a$. The same argument with $a$ and $b$ interchanged shows that $r_a$ is not parallel to $b$, so it meets $b$; let $b_a\in P_b$ denote the unique point in $P_b\cap P_{r_a}$. Define \begin{align*} A_0&:=P_a\setminus\{a_b\}, & B_0&:=P_b\setminus\{b_a\}. \end{align*} For each $x\in A_0$, let $\pi(x)$ be the unique point where the line through $p$ and $x$ meets $b$. This defines a map $\pi:A_0\to B_0$. The line through $p$ and $x$ is not parallel to $b$ by the exclusion of $a_b$, so it meets $b$; it cannot meet $b$ in $b_a$, because the line through $p$ and $b_a$ is parallel to $a$ and hence disjoint from $a$. If $\pi(x)=\pi(x')$, then the two points $p$ and $\pi(x)$ determine a unique line, so the lines through $p,x$ and through $p,x'$ coincide; intersecting this line with $a$ gives $x=x'$. Conversely, if $y\in B_0$, the line through $p$ and $y$ is not parallel to $a$, so it meets $a$ in a point $x\in A_0$, and then $\pi(x)=y$. Hence $\pi$ is bijective, so $|P_a|-1=|P_b|-1$ and $|P_a|=|P_b|$. If two lines $a,b\in L$ are parallel, choose $u\in P_a$ and $v\in P_b$. The unique line $t$ through $u$ and $v$ meets both $a$ and $b$. By the intersecting-line case, $|P_a|=|P_t|=|P_b|$. Therefore any two lines have the same number of points. Since $n:=|P_\ell|$, every line contains exactly $n$ points. [guided] The goal is to replace the incorrect direct count of the line through $p$ parallel to $\ell$ with a projection argument that compares two arbitrary lines. First take two distinct intersecting lines $a$ and $b$, and let $o$ be their unique intersection point. Choose points $u\in P_a\setminus\{o\}$ and $v\in P_b\setminus\{o\}$; the line $s$ through $u$ and $v$ is neither $a$ nor $b$, because $v\notin P_a$ and $u\notin P_b$. We need a projection centre $p$ away from the two lines being compared. The auxiliary line $s$ gives one directly. Let $r$ be the unique line through $o$ parallel to $s$. Because $s$ meets $a$ at $u$ and meets $b$ at $v$, it is not parallel to either $a$ or $b$; therefore $r$ cannot equal $a$ or $b$. Choose $p\in P_r\setminus\{o\}$, which exists because every line has at least two points. If $p$ lay on $a$, then the two distinct points $o$ and $p$ would determine both $a$ and $r$, forcing $a=r$; this is impossible. The same argument shows $p\notin P_b$. Thus $p\notin P_a\cup P_b$. Now construct the two exceptional points carefully. Let $r_b$ be the unique line through $p$ parallel to $b$. This line is not parallel to $a$: otherwise $a$ and $b$ would be two distinct lines through $o$ parallel to $r_b$, contradicting the affine parallel axiom. Hence $r_b$ meets $a$ in a unique point; call this point $a_b$. Similarly, let $r_a$ be the unique line through $p$ parallel to $a$. The same argument with $a$ and $b$ interchanged shows that $r_a$ meets $b$ in a unique point; call this point $b_a$. Define \begin{align*} A_0&:=P_a\setminus\{a_b\}, & B_0&:=P_b\setminus\{b_a\}. \end{align*} For each $x\in A_0$, the line through $p$ and $x$ is not parallel to $b$, by the definition of the excluded point $a_b$. Therefore it meets $b$ in a unique point; uniqueness of the point of intersection follows because two distinct common points of two lines would force the two lines to be equal. Let $\pi(x)$ denote that intersection point. The point $\pi(x)$ is not $b_a$, because the line through $p$ and $b_a$ is parallel to $a$ and hence cannot also pass through a point of $a$. The map $\pi:A_0\to B_0$ is injective: if $\pi(x)=\pi(x')$, then both lines through $p,x$ and through $p,x'$ contain the two points $p$ and $\pi(x)$, so the line-uniqueness axiom makes them equal, and their common intersection with $a$ is a single point, giving $x=x'$. The map is surjective: given $y\in B_0$, the line through $p$ and $y$ is not parallel to $a$, so it meets $a$ in a point $x$; this point cannot be $a_b$, because the line through $p$ and $a_b$ is parallel to $b$. Hence $x\in A_0$ and $\pi(x)=y$. Thus $\pi$ is a bijection, so $|P_a|-1=|P_b|-1$ and $|P_a|=|P_b|$. Finally, if $a$ and $b$ are parallel, choose any $u\in P_a$ and $v\in P_b$. The unique line $t$ through $u$ and $v$ meets both $a$ and $b$, so the intersecting-line result gives $|P_a|=|P_t|$ and $|P_t|=|P_b|$. Hence all lines have the same number of points, and because $n$ was defined as $|P_\ell|$, every line has exactly $n$ points. [/guided] [/step] [step:Show that every point lies on exactly $n+1$ lines] Let $q \in P$ be arbitrary. Choose a line $\ell \in L$ not incident with $q$; such a line exists because the plane has non-collinear points and every line has at least two points. Since every line has $n$ points, the argument from the first step applied to the pair $(q,\ell)$ shows that the lines through $q$ consist of one line joining $q$ to each of the $n$ points of $\ell$, together with the unique line through $q$ parallel to $\ell$. Hence exactly $n+1$ lines are incident with $q$. [/step] [step:Use one parallel class to count the points] For a line $\ell \in L$, define its parallel class by \begin{align*} [\ell] := \{m \in L : m=\ell \text{ or } P_m \cap P_\ell=\varnothing\}. \end{align*} The affine parallel axiom implies that for every point $p \in P$ there is a unique line in $[\ell]$ incident with $p$: if $p \in P_\ell$, the line is $\ell$; if $p \notin P_\ell$, it is the unique line through $p$ disjoint from $\ell$. Thus the lines in $[\ell]$ partition $P$. Since each such line has $n$ points, if $c:=|[\ell]|$, then \begin{align*} |P|=cn. \end{align*} Fix a point $p \in P_\ell$. The $n+1$ lines through $p$ consist of $\ell$ and $n$ further lines. Each further line meets every line in $[\ell]\setminus\{\ell\}$ in exactly one point, and two distinct such lines through $p$ cannot meet the same parallel line in the same point, because then they would share both $p$ and that point. Choose one of the further lines through $p$ and call it $t$. The line $t$ is not parallel to $\ell$, so it meets every line in $[\ell]$ in exactly one point: it meets $\ell$ at $p$, and it meets each line in $[\ell]\setminus\{\ell\}$ because a line not parallel to another line in an affine plane is not disjoint from it. The intersection has at most one point, since two distinct common points would determine both lines and force the two lines to be equal; it has at least one point because the lines are not parallel. These intersection points are distinct, since the lines in $[\ell]$ are pairwise disjoint. Therefore $t$ contains exactly one point on each of the $c$ lines in $[\ell]$. Since every line has $n$ points, $c=n$, and hence \begin{align*} |P|=n^2. \end{align*} [/step] [step:Count incidences to obtain the number of lines and parallel classes] Let \begin{align*} \mathcal I := \{(p,m) \in P \times L : (p,m) \in I\} \end{align*} be the set of incident point-line pairs. Counting $\mathcal I$ by lines gives \begin{align*} |\mathcal I|=n|L|, \end{align*} because every line contains $n$ points. Counting $\mathcal I$ by points gives \begin{align*} |\mathcal I|=(n+1)|P|, \end{align*} because every point lies on $n+1$ lines. Since $|P|=n^2$, we obtain \begin{align*} n|L|=(n+1)n^2. \end{align*} Dividing by $n>0$ gives \begin{align*} |L|=n^2+n. \end{align*} Each parallel class contains $n$ lines by the previous step. The relation on $L$ defined by equality or disjointness is transitive: if $a$ is parallel to $b$ and $b$ is parallel to $c$, then $a$ and $c$ cannot meet, because a common point of $a$ and $c$ would have two distinct lines through it parallel to $b$, contradicting the affine parallel axiom. Hence parallel classes partition $L$, and the number of parallel classes is \begin{align*} \frac{|L|}{n}=\frac{n^2+n}{n}=n+1. \end{align*} Thus the affine plane has order $n$, every line has $n$ points, every point lies on $n+1$ lines, $|P|=n^2$, $|L|=n^2+n$, and the lines split into $n+1$ parallel classes of $n$ lines each. [/step]