§1.prop.12ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον ἀπὸ τοῦ δοθέντος σημείου, ὃ μή ἐστιν ἐπʼ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.
To draw a straight line perpendicular to a given infinite straight line from a given point which is not on it.
ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἄπειρος ἡ ΑΒ τὸ δὲ δοθὲν σημεῖον, ὃ μή ἐστιν ἐπʼ αὐτῆς, τὸ Γ·
Let the given infinite straight line be AB, and the given point which is not on it C.
δεῖ δὴ ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπʼ αὐτῆς, κάθετον εὐθεῖαν γραμμὴν ἀγαγεῖν.
It is required to draw a straight line perpendicular to the given infinite straight line AB from the given point C which is not on it.
εἰλήφθω γὰρ ἐπὶ τὰ ἕτερα μέρη τῆς ΑΒ εὐθείας τυχὸν σημεῖον τὸ Δ, καὶ κέντρῳ μὲν τῷ Γ διαστήματι δὲ τῷ ΓΔ κύκλος γεγράφθω ὁ ΕΖΗ, καὶ τετμήσθω ἡ ΕΗ εὐθεῖα δίχα κατὰ τὸ Θ, καὶ ἐπεζεύχθωσαν αἱ ΓΗ, ΓΘ, ΓΕ εὐθεῖαι·
For let an arbitrary point D be taken on the other side of the straight line AB, and let a circle EFG be described with center C and distance CD, and let the straight line EG be bisected at H, and let the straight lines CG, CH, CE be joined.
λέγω, ὅτι ἐπὶ τὴν δοθεῖσαν εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπʼ αὐτῆς, κάθετος ἦκται ἡ ΓΘ.
ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΗΘ τῇ ΘΕ, κοινὴ δὲ ἡ ΘΓ, δύο δὴ αἱ ΗΘ, ΘΓ δύο ταῖς ΕΘ, ΘΓ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ·
I say that, to the given infinite straight line AB, from the given point C which is not on it, the perpendicular CH has been drawn.
καὶ βάσις ἡ ΓΗ βάσει τῇ ΓΕ ἐστιν ἴση· γωνία ἄρα ἡ ὑπὸ ΓΘΗ γωνίᾳ τῇ ὑπὸ ΕΘΓ ἐστιν ἴση.
For since GH is equal to HE, and HC is common, the two sides GH, HC are equal to the two sides EH, HC respectively; and the base CG is equal to the base CE; therefore the angle CHG is equal to the angle EHC.
καί εἰσιν ἐφεξῆς.
And they are adjacent.
ὅταν δὲ εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα τὰς ἐφεξῆς γωνίας ἴσας ἀλλήλαις ποιῇ, ὀρθὴ ἑκατέρα τῶν ἴσων γωνιῶν ἐστιν, καὶ ἡ ἐφεστηκυῖα εὐθεῖα κάθετος καλεῖται ἐφʼ ἣν ἐφέστηκεν.
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, and the straight line standing on it is called a perpendicular to that on which it stands.
ἐπὶ τὴν δοθεῖσαν ἄρα εὐθεῖαν ἄπειρον τὴν ΑΒ ἀπὸ τοῦ δοθέντος σημείου τοῦ Γ, ὃ μή ἐστιν ἐπʼ αὐτῆς, κάθετος ἦκται ἡ ΓΘ· ὅπερ ἔδει ποιῆσαι.
Therefore, to the given infinite straight line AB, from the given point C which is not on it, the perpendicular CH has been drawn. - Being what it was required to do.
§1.prop.13ἐὰν εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει.
If a straight line set up on a straight line makes angles, it will make either two right angles or angles equal to two right angles.
εὐθεῖα γάρ τις ἡ ΑΒ ἐπʼ εὐθεῖαν τὴν ΓΔ σταθεῖσα γωνίας ποιείτω τὰς ὑπὸ ΓΒΑ, ΑΒΔ·
For let some straight line AB set up on the straight line CD make the angles CBA, ABD.
λέγω, ὅτι αἱ ὑπὸ ΓΒΑ, ΑΒΔ γωνίαι ἤτοι δύο ὀρθαί εἰσιν ἢ δυσὶν ὀρθαῖς ἴσαι.
I say that the angles CBA, ABD are either two right angles or equal to two right angles.
εἰ μὲν οὖν ἴση ἐστὶν ἡ ὑπὸ ΓΒΑ τῇ ὑπὸ ΑΒΔ, δύο ὀρθαί εἰσιν.
Now if the angle CBA is equal to the angle ABD, they are two right angles.
εἰ δὲ οὔ, ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΓΔ πρὸς ὀρθὰς ἡ ΒΕ· αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν·
But if not, let BE be drawn from the point B at right angles to CD; therefore the angles CBE, EBD are two right angles.
καὶ ἐπεὶ ἡ ὑπὸ ΓΒΕ δυσὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΕΒΔ· αἱ ἄρα ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς ὑπὸ ΓΒΑ, ΑΒΕ, ΕΒΔ ἴσαι εἰσίν.
And since the angle CBE is equal to the two angles CBA, ABE, let the common angle EBD be added; therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD.
πάλιν, ἐπεὶ ἡ ὑπὸ ΔΒΑ δυσὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ ἴση ἐστίν, κοινὴ προσκείσθω ἡ ὑπὸ ΑΒΓ· αἱ ἄρα ὑπὸ ΔΒΑ, ΑΒΓ τρισὶ ταῖς ὑπὸ ΔΒΕ, ΕΒΑ, ΑΒΓ ἴσαι εἰσίν.
Again, since the angle DBA is equal to the two angles DBE, EBA, let the common angle ABC be added; therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC.
ἐδείχθησαν δὲ καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ τρισὶ ταῖς αὐταῖς ἴσαι· τὰ δὲ τῷ αὐτῷ ἴσα καὶ ἀλλήλοις ἐστὶν ἴσα· καὶ αἱ ὑπὸ ΓΒΕ, ΕΒΔ ἄρα ταῖς ὑπὸ ΔΒΑ, ΑΒΓ ἴσαι εἰσίν·
And the angles CBE, EBD were also proved equal to the same three; and things equal to the same thing are also equal to one another; therefore the angles CBE, EBD are also equal to the angles DBA, ABC.
ἀλλὰ αἱ ὑπὸ ΓΒΕ, ΕΒΔ δύο ὀρθαί εἰσιν· καὶ αἱ ὑπὸ ΔΒΑ, ΑΒΓ ἄρα δυσὶν ὀρθαῖς ἴσαι εἰσίν.
But the angles CBE, EBD are two right angles; therefore the angles DBA, ABC are also equal to two right angles.
ἐὰν ἄρα εὐθεῖα ἐπʼ εὐθεῖαν σταθεῖσα γωνίας ποιῇ, ἤτοι δύο ὀρθὰς ἢ δυσὶν ὀρθαῖς ἴσας ποιήσει· ὅπερ ἔδει δεῖξαι.
Therefore, if a straight line set up on a straight line makes angles, it will make either two right angles or angles equal to two right angles. - Being what it was required to prove.