By von Neumann J.

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It then follows from the computation that f (y) > 0, whence y ∈ Hl1 . If p ∈ Hl1 and q ∈ Hl2 , then f (p) > 0 and f (q) < 0. We can easily check that in this case λ0 p + (1 − λ0 )q ∈ l for λ0 = −f (q)/(f (p) − f (q)) . Note that 0 < λ0 < 1. Digression 4. In 1880, Pasch (1843–1930) proved the following theorem, which was later named after him. This theorem is a typical example of a property that, when drawn, leaves no doubt as to its correctness. It was used tacitly in Euclid’s Elements. Until the theorem was proved by Pasch, people were not aware of the fact that anything required proving.

Let us ﬁrst explain what a congruence is. 5. We call two ﬁgures F1 and F2 in the plane congruent, denoted by F1 ∼ = F2 , if there exists an isometric surjection H from F1 to F2 . In short, congruent ﬁgures have the same properties. Of course, by properties we mean those that can be stated using the notion of distance. The congruence criteria concern triangles. Let us ﬁrst recall the agreement to denote the side opposite a vertex by the corresponding lowercase letter; this letter often also denotes the length of the side.

Let l and m be lines in the coordinate plane, given by the parametric equations x = p + λa and x = q + μb, respectively. The lines l and m are perpendicular to each other if and only if a, b = 0. Proof. First note that if l // m, then a, b = 0. Indeed, if l // m, a and b have the same span. Since a = o = b, we then have b = νa for some ν = 0, and therefore a, b = ν a 2 , which is nonzero. To prove the theorem, we ﬁrst assume that l and m are perpendicular to each other. In that case the lines intersect, say at the point r; see Fig.