[proofplan] We construct an incidence structure on the point set $X \times X$ by using rows, columns, and the level sets of the Latin squares. The Latin property shows that each family of parallel candidates partitions the point set into $n$ lines of size $n$. Orthogonality shows that level sets coming from different Latin squares meet in exactly one point. These intersection facts prove that any two distinct points lie on exactly one line, and the displayed families give exactly $n+1$ parallel classes, so the incidence structure is an affine plane of order $n$. [/proofplan]

[step:Construct the points and the three types of lines] Let \begin{align*} I := \{1,\dots,n-1\} \end{align*} be the index set for the given Latin squares. Define the point set \begin{align*} \mathcal{P} := X \times X. \end{align*} For each $a \in X$, define the vertical line \begin{align*} V_a := \{a\} \times X \subset \mathcal{P}. \end{align*} For each $b \in X$, define the horizontal line \begin{align*} H_b := X \times \{b\} \subset \mathcal{P}. \end{align*} For each $i \in I$ and each $s \in X$, define the symbol line \begin{align*} S_{i,s} := \{(x,y) \in X \times X : L_i(x,y)=s\} \subset \mathcal{P}. \end{align*} Let $\mathcal{L}$ be the set consisting of all lines $V_a$, $H_b$, and $S_{i,s}$ as $a,b,s$ vary over $X$ and $i$ varies over $I$. We regard $(\mathcal{P},\mathcal{L})$ as an incidence structure with incidence given by set membership. [/step]

[step:Show that each displayed family partitions the point set into lines of size $n$]For every $a \in X$, the map $y \mapsto (a,y)$ from $X$ to $V_a$ is a bijection, so $|V_a|=n$. The family $\{V_a : a \in X\}$ partitions $\mathcal{P}$ because every point $(x,y) \in X \times X$ lies in exactly one vertical line, namely $V_x$. For every $b \in X$, the map $x \mapsto (x,b)$ from $X$ to $H_b$ is a bijection, so $|H_b|=n$. The family $\{H_b : b \in X\}$ partitions $\mathcal{P}$ because every point $(x,y) \in X \times X$ lies in exactly one horizontal line, namely $H_y$. Fix $i \in I$ and $s \in X$. Since $L_i$ is a Latin square, for each row coordinate $x \in X$ there exists a unique column coordinate $y \in X$ such that $L_i(x,y)=s$. Therefore the map $(x,y) \mapsto x$ from $S_{i,s}$ to $X$ is a bijection, and hence $|S_{i,s}|=n$. For fixed $i \in I$, the family $\{S_{i,s}:s \in X\}$ partitions $\mathcal{P}$ because every point $(x,y) \in \mathcal{P}$ has exactly one symbol value $L_i(x,y) \in X$.[/step]

[guided]We must verify two things about each proposed parallel class: every line has exactly $n$ points, and the lines in the class cover $\mathcal{P}$ without overlap. For vertical lines, fix $a \in X$. The function $y \mapsto (a,y)$ from $X$ to $V_a$ is injective because $(a,y_1)=(a,y_2)$ implies $y_1=y_2$, and it is surjective by the definition of $V_a$. Hence $|V_a|=|X|=n$. Also, every point $(x,y) \in X \times X$ belongs to $V_x$, and it cannot belong to $V_a$ with $a \neq x$. Thus $\{V_a:a \in X\}$ partitions $\mathcal{P}$. The horizontal case is the same argument with the two coordinates interchanged. For fixed $b \in X$, the function $x \mapsto (x,b)$ from $X$ to $H_b$ is a bijection, so $|H_b|=n$. Every point $(x,y) \in X \times X$ belongs to exactly one horizontal line, namely $H_y$. Now fix a Latin square index $i \in I$ and a symbol $s \in X$. The defining Latin property says that in each row $x \in X$, every symbol of $X$ appears exactly once. Applying this to the symbol $s$, for each $x \in X$ there is a unique $y \in X$ such that $L_i(x,y)=s$. Therefore projection to the first coordinate gives a bijection $(x,y) \mapsto x$ from $S_{i,s}$ to $X$. So $|S_{i,s}|=n$. Finally, for fixed $i$, every point $(x,y)$ has one symbol value $L_i(x,y)$, so it lies in $S_{i,L_i(x,y)}$; and it cannot lie in two different symbol lines $S_{i,s}$ and $S_{i,t}$ unless $s=t$. Hence $\{S_{i,s}:s \in X\}$ partitions $\mathcal{P}$.[/guided]

[step:Compute all intersections between lines from different families] For $a,b \in X$, \begin{align*} V_a \cap H_b = \{(a,b)\}. \end{align*} Fix $a,s \in X$ and $i \in I$. Since $L_i$ is Latin, in the row $a$ there is a unique $y \in X$ such that $L_i(a,y)=s$. Hence $V_a \cap S_{i,s}$ consists of exactly one point. Similarly, for fixed $b,s \in X$ and $i \in I$, the column Latin property gives a unique $x \in X$ such that $L_i(x,b)=s$, so $H_b \cap S_{i,s}$ consists of exactly one point. Finally, let $i,j \in I$ with $i \neq j$, and let $s,t \in X$. Orthogonality of $L_i$ and $L_j$ means that the map $(x,y) \mapsto (L_i(x,y),L_j(x,y))$ from $X \times X$ to $X \times X$ is a bijection. Therefore there is a unique point $(x,y) \in X \times X$ such that $L_i(x,y)=s$ and $L_j(x,y)=t$. Hence \begin{align*} |S_{i,s} \cap S_{j,t}|=1. \end{align*} [/step]

[step:Prove that any two distinct points lie on exactly one line]Let $P=(x_1,y_1) \in \mathcal{P}$ and $Q=(x_2,y_2) \in \mathcal{P}$ be distinct points. If $x_1=x_2$, then both points lie on $V_{x_1}$. No horizontal line contains both points because $P \neq Q$ implies $y_1 \neq y_2$. No symbol line $S_{i,s}$ contains both points, because for fixed $i \in I$ and fixed row $x_1$, the Latin property gives each symbol exactly once in that row. Thus $V_{x_1}$ is the unique line through $P$ and $Q$. If $y_1=y_2$, the same argument with rows and columns interchanged shows that $H_{y_1}$ is the unique line through $P$ and $Q$. It remains to assume $x_1 \neq x_2$ and $y_1 \neq y_2$. Define the collection of lines through $P$ \begin{align*} \mathcal{M}_P := \{V_{x_1},H_{y_1}\} \cup \{S_{i,L_i(P)} : i \in I\}. \end{align*} This collection has $n+1$ lines. By the intersection computations above, any two distinct lines in $\mathcal{M}_P$ meet exactly at $P$: the vertical and horizontal lines meet at $P$; each symbol line through $P$ meets the vertical and horizontal lines only at $P$; and two symbol lines through $P$ from distinct Latin squares meet only at $P$ by orthogonality. Each line in $\mathcal{M}_P$ contains $n$ points, and the only common point among any two of them is $P$. Hence their union has cardinality \begin{align*} 1 + (n+1)(n-1) = n^2. \end{align*} Since $|\mathcal{P}|=|X \times X|=n^2$, the union of the lines in $\mathcal{M}_P$ is all of $\mathcal{P}$. Because $Q$ has neither the same first coordinate nor the same second coordinate as $P$, it lies on neither $V_{x_1}$ nor $H_{y_1}$. Therefore there is a unique $i \in I$ such that $Q \in S_{i,L_i(P)}$. Equivalently, \begin{align*} L_i(x_1,y_1)=L_i(x_2,y_2), \end{align*} and the unique line through $P$ and $Q$ is $S_{i,L_i(P)}$.[/step]

[guided]The difficult case is when the two points share neither coordinate. Let \begin{align*} P=(x_1,y_1), \qquad Q=(x_2,y_2), \end{align*} with $x_1 \neq x_2$ and $y_1 \neq y_2$. The vertical and horizontal lines through $P$ cannot contain $Q$, so if a line through both points exists, it must be one of the symbol lines through $P$. For each $i \in I$, the symbol line through $P$ coming from $L_i$ is \begin{align*} S_{i,L_i(P)}=\{(x,y)\in X\times X:L_i(x,y)=L_i(P)\}. \end{align*} Together with the vertical and horizontal lines through $P$, define \begin{align*} \mathcal{M}_P := \{V_{x_1},H_{y_1}\} \cup \{S_{i,L_i(P)} : i \in I\}. \end{align*} There are $2+(n-1)=n+1$ lines in this collection. We now check that these lines overlap only at $P$. The line $V_{x_1}$ and the line $H_{y_1}$ meet at exactly $(x_1,y_1)=P$. A symbol line $S_{i,L_i(P)}$ meets $V_{x_1}$ only at $P$ because, in the row $x_1$, the symbol $L_i(P)$ appears exactly once, namely at the column $y_1$. Similarly, $S_{i,L_i(P)}$ meets $H_{y_1}$ only at $P$ because, in the column $y_1$, the symbol $L_i(P)$ appears exactly once, namely at the row $x_1$. Finally, take two distinct indices $i,j \in I$. The point $P$ lies in both $S_{i,L_i(P)}$ and $S_{j,L_j(P)}$. Orthogonality says that the ordered pair of symbols \begin{align*} (L_i(P),L_j(P)) \end{align*} occurs at exactly one cell of $X \times X$. Since it occurs at $P$, it occurs nowhere else. Hence \begin{align*} S_{i,L_i(P)} \cap S_{j,L_j(P)}=\{P\}. \end{align*} Each line in $\mathcal{M}_P$ has $n$ points. Since all $n+1$ lines share the point $P$ and otherwise are pairwise disjoint, their union contains $P$ plus $n-1$ additional points from each of the $n+1$ lines. Thus \begin{align*} \left|\bigcup_{M \in \mathcal{M}_P} M\right| = 1+(n+1)(n-1) = n^2. \end{align*} But $\mathcal{P}=X\times X$ also has $n^2$ points. Therefore these $n+1$ lines through $P$ cover all points of $\mathcal{P}$. Since $Q$ does not lie on $V_{x_1}$ or $H_{y_1}$, it must lie on at least one symbol line $S_{i,L_i(P)}$. It lies on at most one such symbol line because two different symbol lines through $P$ meet only at $P$, and $Q \neq P$. Hence there is a unique $i \in I$ such that \begin{align*} Q \in S_{i,L_i(P)}. \end{align*} This is exactly the statement that \begin{align*} L_i(x_1,y_1)=L_i(x_2,y_2), \end{align*} and the unique line through $P$ and $Q$ is $S_{i,L_i(P)}$.[/guided]

[step:Identify the parallel classes and count the lines] Define the vertical class $\mathcal{V} := \{V_a:a \in X\}$, the horizontal class $\mathcal{H} := \{H_b:b \in X\}$, and, for each $i \in I$, the symbol class $\mathcal{S}_i := \{S_{i,s}:s \in X\}$. Each of these classes partitions $\mathcal{P}$ into $n$ disjoint lines of size $n$. Lines from different classes meet in exactly one point by the intersection computations above. Therefore these classes are precisely the parallel classes of the incidence structure. The total number of lines is \begin{align*} |\mathcal{L}| = n+n+(n-1)n = n^2+n. \end{align*} We have shown that $\mathcal{P}$ has $n^2$ points, every line has $n$ points, every two distinct points lie on exactly one line, and the lines split into $n+1$ parallel classes with $n$ lines in each class. Hence $(\mathcal{P},\mathcal{L})$ is an affine plane of order $n$. [/step]