Let \(\alpha =(\mathcal P,\mathcal L,\mathcal R)\) be an affine plane and recall that two lines are called parallel if they do not meet.

Let \(\ell ,m,n\in \mathcal L\) be three lines and assume that \(\ell \parallel m\) and \(m\parallel n\). Show that

Solution:

Assume that \(\ell \parallel m\) and \(m\parallel n\). We want to show that \(\ell =n\) or \(\ell \parallel n\).

We distinguish two cases. First, if \(\ell \parallel n\), then we are done.

Now assume that \(\ell \) is not parallel to \(n\). Then \(\ell \) and \(n\) meet, so there exists a point \(x\in \mathcal P\) such that \(x\,\mathrm {I}\,\ell \) and \(x\,\mathrm {I}\,n\). Since \(\ell \parallel m\), we have \(x\mathrel {\not \! I}m\). Thus \(x\) is a point not incident with \(m\). Moreover, both \(\ell \) and \(n\) are lines through \(x\) parallel to \(m\). By the Euclidean parallel property, we conclude \(\ell =n\).

Hence, in all cases, \(\ell =n\) or \(\ell \parallel n\).

Let \(\alpha =(\mathcal P,\mathcal L,\mathcal R)\) be an affine plane and recall that, for a line \(\ell \in \mathcal L\), we defined

Solution:

We use the notation and construction from the problem statement. Recall that \(a\in R(\ell )\), \(a\notin R(m)\), \(b\in R(m)\), \(b\notin R(\ell )\), and \(n\) is the unique line through \(a\) and \(b\). We also know that \(R(n)\cap R(\ell )=\{a\}\) and \(R(n)\cap R(m)=\{b\}\).

The map \(f:R(\ell )\longrightarrow R(m)\) is defined by \(f(a)=b\), and, if \(x\in R(\ell )\) with \(x\neq a\), then \(n_x\) is the unique line through \(x\) parallel to \(n\), and \(f(x)\) is the unique point in \(R(n_x)\cap R(m)\). We prove that \(f\) is injective and surjective.

Injectivity. Let \(x_1,x_2\in R(\ell )\), and assume that \(f(x_1)=f(x_2)\). We want to prove that \(x_1=x_2\).

First assume that \(x_1=a\). Then \(f(x_1)=f(a)=b\). Since \(f(x_1)=f(x_2)\), it follows that \(f(x_2)=b\). We claim that \(x_2=a\).

Suppose, for contradiction, that \(x_2\neq a\). Then \(n_{x_2}\) is defined, and \(f(x_2)\) is the unique point in \(R(n_{x_2})\cap R(m)\). Since \(f(x_2)=b\), we have \(b\,\mathrm {I}\,n_{x_2}\). But \(b\,\mathrm {I}\,n\), and \(n_{x_2}\parallel n\). Therefore \(n_{x_2}\) and \(n\) meet at \(b\), contradicting the fact that \(n_{x_2}\parallel n\). Hence \(x_2=a\). Therefore, if \(x_1=a\), then \(x_2=a\), and so \(x_1=x_2\).

Now assume that \(x_1\neq a\). We first show that \(x_2\neq a\). Indeed, if \(x_2=a\), then by the previous case, applied with the roles of \(x_1\) and \(x_2\) reversed, we would have \(x_1=a\). This contradicts our assumption that \(x_1\neq a\). Therefore \(x_2\neq a\).

Since both \(x_1\neq a\) and \(x_2\neq a\), the lines \(n_{x_1}\) and \(n_{x_2}\) are defined. Let \(y=f(x_1)=f(x_2)\). Then \(y\,\mathrm {I}\,n_{x_1}\) and \(y\,\mathrm {I}\,n_{x_2}\). Also, \(n_{x_1}\parallel n\) and \(n_{x_2}\parallel n\). By the first problem, we have \(n_{x_1}=n_{x_2}\).

Observe that \(n_{x_1}\) is distinct from \(\ell \) (since \(a\mathrel {\not \! I} n_{x_1}\)) and not parallel to \(\ell \) (since \(x_1\,\mathrm {I}\,\ell \)). Thus \(n_{x_1}\) intersects \(\ell \) in a unique point \(x_1\). For the same reason, \(n_{x_2}\) intersects \(\ell \) in a unique point \(x_2\). Since \(n_{x_1}=n_{x_2}\), we conclude that \(x_1=x_2\).

Surjectivity. Let \(y\in R(m)\). We want to show that there exists \(x\in R(\ell )\) such that \(f(x)=y\).

First assume that \(y=b\). Then we can take \(x=a\), since \(f(a)=b=y\).

From now on, assume that \(y\neq b\). Since \(y\in R(m)\), and since \(R(n)\cap R(m)=\{b\}\), we have \(y\mathrel {\not \! I}n\). By the Euclidean parallel property, there exists a unique line \(n_y\) such that \(y\,\mathrm {I}\,n_y\) and \(n_y\parallel n\).

We claim that \(n_y\) is distinct and not parallel to \(\ell \). First, \(n_y\neq \ell \). Indeed, if \(n_y=\ell \), then \(n_y\) and \(n\) would meet at \(a\), because \(a\,\mathrm {I}\,\ell \) and \(a\,\mathrm {I}\,n\). This contradicts \(n_y\parallel n\). Therefore \(n_y\neq \ell \).

Next, suppose for contradiction that \(n_y\parallel \ell \). Since \(n_y\parallel n\), Problem 1 implies that \(\ell =n\) or \(\ell \parallel n\). But \(\ell \neq n\) (since \(b\mathrel {\not \! I} \ell \)), and \(\ell \) meets \(n\) at \(a\). Thus \(\ell \) is not parallel to \(n\). This is a contradiction. Hence \(n_y\) is not parallel to \(\ell \).

Since \(n_y\) is distinct and not parallel to \(\ell \), the lines \(n_y\) and \(\ell \) meet in a unique point. Let \(x\) be this point, so \(R(n_y)\cap R(\ell )=\{x\}\).

We claim that \(x\neq a\). Suppose, for contradiction, that \(x=a\). Then \(a\,\mathrm {I}\,n_y\). But \(a\,\mathrm {I}\,n\), and \(n_y\parallel n\). This contradicts the fact that parallel lines do not meet. Therefore \(x\neq a\).

Since \(x\in R(\ell )\) and \(x\neq a\), the line \(n_x\) is defined as the unique line through \(x\) parallel to \(n\). But \(n_y\) also passes through \(x\), because \(x\,\mathrm {I}\,n_y\), and \(n_y\parallel n\). By uniqueness in the Euclidean parallel property, we get \(n_y=n_x\).

Now \(y\,\mathrm {I}\,n_y\), \(y\in R(m)\), and \(n_y=n_x\). Hence \(y\in R(n_x)\cap R(m)\). By definition, \(f(x)\) is the unique point in \(R(n_x)\cap R(m)\). Therefore \(f(x)=y\). Thus \(f\) is surjective.

Since \(f\) is both injective and surjective, \(f\) is a bijection.