Euclid 1.proposition.1
Christopher Blackwell /
- Created on 2018-03-23 13:21:22
- Translated by Heath
- Aligned by Christopher Blackwell
Euclid Book 1 Proposition 1
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urn:cts:greekLit:tlg1799.tlg001.heiberg:1.prop.1
urn:cts:greekLit:tlg1799.tlg001.heath:1.prop.1
PROPOSITION 1
ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συστήσασθαι .
ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ . δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον συστήσασθαι .
κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΓΔ , καὶ πάλιν κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΑ κύκλος γεγράφθω ὁ ΑΓΕ , καὶ ἀπὸ τοῦ Γ σημείου , καθ᾽ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι , ἐπὶ τὰ Α , Β σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΓΑ , ΓΒ .
καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου , ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ : πάλιν , ἐπεὶ τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου , ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ . ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση : ἑκατέρα ἄρα τῶν ΓΑ , ΓΒ τῇ ΑΒ ἐστὶν ἴση .
τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα : καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστὶν ἴση : αἱ τρεῖς ἄρα αἱ ΓΑ , ΑΒ , ΒΓ ἴσαι ἀλλήλαις εἰσίν .
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον , καὶ συνέσταται ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τῆς ΑΒ .
Ἐπὶ τῆς δοθείσης ἄρα εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συνέσταται : ὅπερ ἔδει ποιῆσαι .
ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συστήσασθαι .
ἔστω ἡ δοθεῖσα εὐθεῖα πεπερασμένη ἡ ΑΒ . δεῖ δὴ ἐπὶ τῆς ΑΒ εὐθείας τρίγωνον ἰσόπλευρον συστήσασθαι .
κέντρῳ μὲν τῷ Α διαστήματι δὲ τῷ ΑΒ κύκλος γεγράφθω ὁ ΒΓΔ , καὶ πάλιν κέντρῳ μὲν τῷ Β διαστήματι δὲ τῷ ΒΑ κύκλος γεγράφθω ὁ ΑΓΕ , καὶ ἀπὸ τοῦ Γ σημείου , καθ᾽ ὃ τέμνουσιν ἀλλήλους οἱ κύκλοι , ἐπὶ τὰ Α , Β σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΓΑ , ΓΒ .
καὶ ἐπεὶ τὸ Α σημεῖον κέντρον ἐστὶ τοῦ ΓΔΒ κύκλου , ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ : πάλιν , ἐπεὶ τὸ Β σημεῖον κέντρον ἐστὶ τοῦ ΓΑΕ κύκλου , ἴση ἐστὶν ἡ ΒΓ τῇ ΒΑ . ἐδείχθη δὲ καὶ ἡ ΓΑ τῇ ΑΒ ἴση : ἑκατέρα ἄρα τῶν ΓΑ , ΓΒ τῇ ΑΒ ἐστὶν ἴση .
τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα : καὶ ἡ ΓΑ ἄρα τῇ ΓΒ ἐστὶν ἴση : αἱ τρεῖς ἄρα αἱ ΓΑ , ΑΒ , ΒΓ ἴσαι ἀλλήλαις εἰσίν .
ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον , καὶ συνέσταται ἐπὶ τῆς δοθείσης εὐθείας πεπερασμένης τῆς ΑΒ .
Ἐπὶ τῆς δοθείσης ἄρα εὐθείας πεπερασμένης τρίγωνον ἰσόπλευρον συνέσταται : ὅπερ ἔδει ποιῆσαι .
PROPOSITION
1
On a given finite straight line to construct an equilateral triangle .
Let AB be the given finite straight line . Thus it is required to construct an equilateral triangle on the straight line AB .
With centre A and distance AB let the circle BCD be described ; [ Post . 3 ] again , with centre B and distance BA let the circle ACE be described ; [ Post . 3 ] and from the point C , in which the circles cut one another , to the points A , B let the straight lines CA , CB be joined . [ Post . 1 ]
Now , since the point A is the centre of the circle CDB , AC is equal to AB . [ Def . 15 ] Again , since the point B is the centre of the circle CAE , BC is equal to BA . [ Def . 15 ] But CA was also proved equal to AB ; therefore each of the straight lines CA , CB is equal to AB .
And things which are equal to the same thing are also equal to one another ; [ C . N . 1 ] therefore CA is also equal to CB . Therefore the three straight lines CA , AB , BC are equal to one another .
Therefore the triangle ABC is equilateral ; and it has been constructed on the given finite straight line AB .
Being what it was required to do .
On a given finite straight line to construct an equilateral triangle .
Let AB be the given finite straight line . Thus it is required to construct an equilateral triangle on the straight line AB .
With centre A and distance AB let the circle BCD be described ; [ Post . 3 ] again , with centre B and distance BA let the circle ACE be described ; [ Post . 3 ] and from the point C , in which the circles cut one another , to the points A , B let the straight lines CA , CB be joined . [ Post . 1 ]
Now , since the point A is the centre of the circle CDB , AC is equal to AB . [ Def . 15 ] Again , since the point B is the centre of the circle CAE , BC is equal to BA . [ Def . 15 ] But CA was also proved equal to AB ; therefore each of the straight lines CA , CB is equal to AB .
And things which are equal to the same thing are also equal to one another ; [ C . N . 1 ] therefore CA is also equal to CB . Therefore the three straight lines CA , AB , BC are equal to one another .
Therefore the triangle ABC is equilateral ; and it has been constructed on the given finite straight line AB .
Being what it was required to do .