Humanitext Reader

Euclid · Elements §1.prop.9-1.prop.11

Bisection of Angles, Segments, and Perpendicular Construction

Passage 8 of 316 · Greek

Summary

In propositions 9 through 11, three geometric construction problems are presented and proved: bisecting a given angle, bisecting a given finite straight line, and drawing a perpendicular line from a given point on a given straight line.

§1.prop.9τὴν δοθεῖσαν γωνίαν εὐθύγραμμον δίχα τεμεῖν.
To bisect a given rectilineal angle.
ἔστω ἡ δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ. δεῖ δὴ αὐτὴν δίχα τεμεῖν.
Let the given rectilineal angle be the angle BAC. It is required to bisect it.
εἰλήφθω ἐπὶ τῆς ΑΒ τυχὸν σημεῖον τὸ Δ, καὶ ἀφῃρήσθω ἀπὸ τῆς ΑΓ τῇ ΑΔ ἴση ἡ ΑΕ, καὶ ἐπεζεύχθω ἡ ΔΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΔΕΖ, καὶ ἐπεζεύχθω ἡ ΑΖ·
Let an arbitrary point D be taken on AB, and let AE be cut off from AC equal to AD, and let DE be joined, and let an equilateral triangle DEF be constructed on DE, and let AF be joined.
λέγω, ὅτι ἡ ὑπὸ ΒΑΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας.
I say that the angle BAC has been bisected by the straight line AF.
ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΕ, κοινὴ δὲ ἡ ΑΖ, δύο δὴ αἱ ΔΑ, ΑΖ δυσὶ ταῖς ΕΑ, ΑΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ. καὶ βάσις ἡ ΔΖ βάσει τῇ ΕΖ ἴση ἐστίν· γωνία ἄρα ἡ ὑπὸ ΔΑΖ γωνίᾳ τῇ ὑπὸ ΕΑΖ ἴση ἐστίν.
For since AD is equal to AE, and AF is common, the two sides DA, AF are equal to the two sides EA, AF respectively; and the base DF is equal to the base EF; therefore the angle DAF is equal to the angle EAF.
ἡ ἄρα δοθεῖσα γωνία εὐθύγραμμος ἡ ὑπὸ ΒΑΓ δίχα τέτμηται ὑπὸ τῆς ΑΖ εὐθείας· ὅπερ ἔδει ποιῆσαι.
Therefore the given rectilineal angle BAC has been bisected by the straight line AF. - Being what it was required to do.
§1.prop.10τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην δίχα τεμεῖν.
To bisect a given finite straight line.
ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ·
Let the given finite straight line be AB.
δεῖ δὴ τὴν ΑΒ εὐθεῖαν πεπερασμένην δίχα τεμεῖν.
It is required to bisect the finite straight line AB.
συνεστάτω ἐπʼ αὐτῆς τρίγωνον ἰσόπλευρον τὸ ΑΒΓ, καὶ τετμήσθω ἡ ὑπὸ ΑΓΒ γωνία δίχα τῇ ΓΔ εὐθείᾳ·
Let there be constructed on it an equilateral triangle ABC, and let the angle ACB be bisected by the straight line CD.
λέγω, ὅτι ἡ ΑΒ εὐθεῖα δίχα τέτμηται κατὰ τὸ Δ σημεῖον.
I say that the straight line AB has been bisected at the point D.
ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΑΓ τῇ ΓΒ, κοινὴ δὲ ἡ ΓΔ, δύο δὴ αἱ ΑΓ, ΓΔ δύο ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΑΓΔ γωνίᾳ τῇ ὑπὸ ΒΓΔ ἴση ἐστίν· βάσις ἄρα ἡ ΑΔ βάσει τῇ ΒΔ ἴση ἐστίν.
For since AC is equal to CB, and CD is common, the two sides AC, CD are equal to the two sides BC, CD respectively; and the angle ACD is equal to the angle BCD; therefore the base AD is equal to the base BD.
ἡ ἄρα δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ δίχα τέτμηται κατὰ τὸ Δ· ὅπερ ἔδει ποιῆσαι.
Therefore the given finite straight line AB has been bisected at D. - Being what it was required to do.
§1.prop.11τῇ δοθείσῃ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν.
To draw a straight line at right angles to a given straight line from a given point on it.
ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον ἐπʼ αὐτῆς τὸ Γ·
Let the given straight line be AB, and the given point on it C.
δεῖ δὴ ἀπὸ τοῦ Γ σημείου τῇ ΑΒ εὐθείᾳ πρὸς ὀρθὰς γωνίας εὐθεῖαν γραμμὴν ἀγαγεῖν.
It is required to draw a straight line from the point C at right angles to the straight line AB.
εἰλήφθω ἐπὶ τῆς ΑΓ τυχὸν σημεῖον τὸ Δ, καὶ κείσθω τῇ ΓΔ ἴση ἡ ΓΕ, καὶ συνεστάτω ἐπὶ τῆς ΔΕ τρίγωνον ἰσόπλευρον τὸ ΖΔΕ, καὶ ἐπεζεύχθω ἡ ΖΓ·
Let an arbitrary point D be taken on AC, and let CE be made equal to CD, and let an equilateral triangle FDE be constructed on DE, and let FC be joined.
λέγω, ὅτι τῇ δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΖΓ. ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΔΓ τῇ ΓΕ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΔΓ, ΓΖ δυσὶ ταῖς ΕΓ, ΓΖ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ βάσις ἡ ΔΖ βάσει τῇ ΖΕ ἴση ἐστίν· γωνία ἄρα ἡ ὑπὸ ΔΓΖ γωνίᾳ τῇ ὑπὸ ΕΓΖ ἴση ἐστίν· καί εἰσιν ἐφεξῆς.
I say that, to the given straight line AB, from the given point C on it, the straight line FC has been drawn at right angles. For since DC is equal to CE, and CF is common, the two sides DC, CF are equal to the two sides EC, CF respectively; and the base DF is equal to the base FE; therefore the angle DCF is equal to the angle ECF; and they are adjacent.
ὅταν δὲ εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν· ὀρθὴ ἄρα ἐστὶν ἑκατέρα τῶν ὑπὸ ΔΓΖ, ΖΓΕ. τῇ ἄρα δοθείσῃ εὐθείᾳ τῇ ΑΒ ἀπὸ τοῦ πρὸς αὐτῇ δοθέντος σημείου τοῦ Γ πρὸς ὀρθὰς γωνίας εὐθεῖα γραμμὴ ἦκται ἡ ΓΖ· ὅπερ ἔδει ποιῆσαι.
But when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is a right angle; therefore each of the angles DCF, FCE is a right angle. Therefore, to the given straight line AB, from the given point C on it, the straight line FC has been drawn at right angles. - Being what it was required to do.

Notes

  1. §1.prop.9δεῖ δὴ αὐτὴν δίχα τεμεῖν — A standard expression of the "specification" (diorismos) indicating the goal of the construction, using the impersonal verb δεῖ with an accusative and infinitive construction (with αὐτήν as subject and τεμεῖν as infinitive).
  2. §1.prop.9εἰλήφθω — Third-person singular perfect imperative passive. A typical mathematical imperative used in Greek mathematics to command the establishment of an assumed geometric object ("let there be taken").
  3. §1.prop.11καί εἰσιν ἐφεξῆς — The subject is the two preceding angles. It states the fact that they are "adjacent" (ἐφεξῆς), providing the prerequisite condition to apply the definition of a right angle (Definition 10) that follows.

Cite this passage

Euclid, Elements §1.prop.9-1.prop.11. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.9-1.prop.11

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