§1.prop.29ἡ εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας.
A straight line falling on parallel straight lines makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle on the same side, and the interior angles on the same side equal to two right angles.
εἰς γὰρ παραλλήλους εὐθείας τὰς ΑΒ, ΓΔ εὐθεῖα ἐμπιπτέτω ἡ ΕΖ· λέγω, ὅτι τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΑΗΘ, ΗΘΔ ἴσας ποιεῖ καὶ τὴν ἐκτὸς γωνίαν τὴν ὑπὸ ΕΗΒ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΗΘΔ ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη τὰς ὑπὸ ΒΗΘ, ΗΘΔ δυσὶν ὀρθαῖς ἴσας.
¦5 For let a straight line EF fall on parallel straight lines AB, CD; I say that it makes the alternate angles AHT, HTD equal, the exterior angle EHB equal to the interior and opposite angle HTD, and the interior angles on the same side, ¦10 BHT, HTD, equal to two right angles.
εἰ γὰρ ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ, μία αὐτῶν μείζων ἐστίν.
For if the angle AHT is unequal to the angle HTD, one of them is greater.
ἔστω μείζων ἡ ὑπὸ ΑΗΘ· κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ· αἱ ἄρα ὑπὸ ΑΗΘ, ΒΗΘ τῶν ὑπὸ ΒΗΘ, ΗΘΔ μείζονές εἰσιν.
Let the angle AHT be greater; let the angle BHT be added to both; therefore the ¦15 angles AHT, BHT are greater than the angles BHT, HTD.
ἀλλὰ αἱ ὑπὸ ΑΗΘ, ΒΗΘ δυσὶν ὀρθαῖς ἴσαι εἰσίν.
But the angles AHT, BHT are equal to two right angles.
αἱ ἄρα ὑπὸ ΒΗΘ, ΗΘΔ δύο ὀρθῶν ἐλάσσονές εἰσιν.
Therefore the angles BHT, HTD are less than two right angles.
αἱ δὲ ἀπʼ ἐλασσόνων ἢ δύο ὀρθῶν ἐκβαλλόμεναι εἰς ἄπειρον συμπίπτουσιν· αἱ ἄρα ΑΒ, ΓΔ ἐκβαλλόμεναι εἰς ἄπειρον συμπεσοῦνται· οὐ συμπίπτουσι δὲ διὰ τὸ παραλλήλους αὐτὰς ὑποκεῖσθαι· οὐκ ἄρα ἄνισός ἐστιν ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΗΘΔ· ἴση ἄρα.
And straight lines produced indefinitely from angles less than two right angles meet; therefore ¦20 AB, CD, being produced indefinitely, will meet; but they do not meet because they are assumed to be parallel; therefore the angle AHT is not unequal to the angle HTD; therefore it is equal.
ἀλλὰ ἡ ὑπὸ ΑΗΘ τῇ ὑπὸ ΕΗΒ ἐστιν ἴση· καὶ ἡ ὑπὸ ΕΗΒ ἄρα τῇ ὑπὸ ΗΘΔ ἐστιν ἴση.
But the angle AHT is equal to the angle EHB; therefore the angle EHB is also equal to the angle HTD.
κοινὴ προσκείσθω ἡ ὑπὸ ΒΗΘ· αἱ ἄρα ὑπὸ ΕΗΒ, ΒΗΘ ταῖς ὑπὸ ΒΗΘ, ΗΘΔ ἴσαι εἰσίν.
Let the angle BHT be added to both; ¦25 therefore the angles EHB, BHT are equal to the angles BHT, HTD.
ἀλλὰ αἱ ὑπὸ ΕΗΒ, ΒΗΘ δύο ὀρθαῖς ἴσαι εἰσίν· καὶ αἱ ὑπὸ ΒΗΘ, ΗΘΔ ἄρα δύο ὀρθαῖς ἴσαι εἰσίν.
But the angles EHB, BHT are equal to two right angles; therefore the angles BHT, HTD are also equal to two right angles.
ἡ ἄρα εἰς τὰς παραλλήλους εὐθείας εὐθεῖα ἐμπίπτουσα τάς τε ἐναλλὰξ γωνίας ἴσας ἀλλήλαις ποιεῖ καὶ τὴν ἐκτὸς τῇ ἐντὸς καὶ ἀπεναντίον ἴσην καὶ τὰς ἐντὸς καὶ ἐπὶ τὰ αὐτὰ μέρη δυσὶν ὀρθαῖς ἴσας· ὅπερ ἔδει δεῖξαι.
Therefore a straight line falling on parallel straight lines ¦30 makes the alternate angles equal to one another, the exterior angle equal to the interior and opposite angle on the same side, and the interior angles on the same side equal to two right angles; which was to be proved.
αἱ τῇ αὐτῇ εὐθείᾳ παράλληλοι καὶ ἀλλήλαις εἰσὶ παράλληλοι.
Straight lines parallel to the same straight line are also parallel to one another.
ἔστω ἑκατέρα τῶν ΑΒ, ΓΔ τῇ ΕΖ παράλληλος· λέγω, ὅτι καὶ ἡ ΑΒ τῇ ΓΔ ἐστι παράλληλος.
Let each of the straight lines AB, CD be parallel to EF; I say that AB is also parallel to CD.
ἐμπιπτέτω γὰρ εἰς αὐτὰς εὐθεῖα ἡ ΗΚ.
καὶ ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΑΒ, ΕΖ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἄρα ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ. πάλιν, ἐπεὶ εἰς παραλλήλους εὐθείας τὰς ΕΖ, ΓΔ εὐθεῖα ἐμπέπτωκεν ἡ ΗΚ, ἴση ἐστὶν ἡ ὑπὸ ΗΘΖ τῇ ὑπὸ ΗΚΔ. ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΗΚ τῇ ὑπὸ ΗΘΖ ἴση.
¦5 For let a straight line HK fall on them. And since the straight line HK has fallen on the parallel straight lines AB, EF, therefore the angle AHK is equal to the angle HTE. Again, since the straight line HK has fallen on the parallel straight lines EF, CD, the angle HTE is equal to the angle HKD. ¦10 And the angle AHK was also proved equal to the angle HTE.
καὶ ἡ ὑπὸ ΑΗΚ ἄρα τῇ ὑπὸ ΗΚΔ ἐστιν ἴση· καί εἰσιν ἐναλλάξ.
Therefore the angle AHK is also equal to the angle HKD; and they are alternate.
παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ.
ὅπερ ἔδει δεῖξαι.
Therefore AB is parallel to CD. which was to be proved.
§1.prop.31διὰ τοῦ δοθέντος σημείου τῇ δοθείσῃ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν.
Through a given point to draw a straight line parallel to a given straight line.
ἔστω τὸ μὲν δοθὲν σημεῖον τὸ Α, ἡ δὲ δοθεῖσα εὐθεῖα ἡ ΒΓ· δεῖ δὴ διὰ τοῦ Α σημείου τῇ ΒΓ εὐθείᾳ παράλληλον εὐθεῖαν γραμμὴν ἀγαγεῖν.
Let the given point be A, and the given straight line BC; it is required to draw through the point A a straight line parallel to the ¦5 straight line BC.
εἰλήφθω ἐπὶ τῆς ΒΓ τυχὸν σημεῖον τὸ Δ, καὶ ἐπεζεύχθω ἡ ΑΔ· καὶ συνεστάτω πρὸς τῇ ΔΑ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Α τῇ ὑπὸ ΑΔΓ γωνίᾳ ἴση ἡ ὑπὸ ΔΑΕ·
Let a chance point D be taken on BC, and let AD be joined; and let the angle DAE be constructed on the straight line DA and at the point ¦10 A on it, equal to the angle ADC; and let the straight line AF be produced in a straight line with EA.
καὶ ἐκβεβλήσθω ἐπʼ εὐθείας τῇ ΕΑ εὐθεῖα ἡ ΑΖ.
καὶ ἐπεὶ εἰς δύο εὐθείας τὰς ΒΓ, ΕΖ εὐθεῖα ἐμπίπτουσα ἡ ΑΔ τὰς ἐναλλὰξ γωνίας τὰς ὑπὸ ΕΑΔ, ΑΔΓ ἴσας ἀλλήλαις πεποίηκεν, παράλληλος ἄρα ἐστὶν ἡ ΕΑΖ τῇ ΒΓ.
διὰ τοῦ δοθέντος ἄρα σημείου τοῦ Α τῇ δοθείσῃ εὐθείᾳ τῇ ΒΓ παράλληλος εὐθεῖα γραμμὴ ἦκται ἡ ΕΑΖ· ὅπερ ἔδει ποιῆσαι.
And since the straight line AD falling on two straight lines BC, EF has made the alternate angles EAD, ADC equal to one another, therefore EAF is parallel to BC. ¦15 Therefore through the given point A the straight line EAF has been drawn parallel to the given straight line BC; which was to be done.