§5.prop.1ἐὰν ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων.
If there be any number of magnitudes whatever which are, respectively, equal multiples of any magnitudes whatever, equal in multitude, then whatever multiple one of the magnitudes is of one, that multiple will all also be of all.
ἔστω ὁποσαοῦν μεγέθη τὰ ΑΒ, ΓΔ ὁποσωνοῦν μεγεθῶν τῶν Ε, Ζ ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον·
Let there be any number of magnitudes whatever, AB, ΓΔ, which are, respectively, equal multiples of any magnitudes whatever, E, Z, equal in multitude.
λέγω, ὅτι ὁσαπλάσιόν ἐστι τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ.
ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Ε καὶ τὸ ΓΔ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ μεγέθη ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τῷ ΓΔ ἴσα τῷ Ζ. διῃρήσθω τὸ μὲν ΑΒ εἰς τὰ τῷ Ε μεγέθη ἴσα τὰ ΑΗ, ΗΒ, τὸ δὲ ΓΔ εἰς τὰ τῷ Ζ ἴσα τὰ ΓΘ, ΘΔ·
I say that, whatever multiple AB is of E, that multiple will AB, ΓΔ also be of E, Z. For since AB is the same multiple of E that ΓΔ is of Z, therefore as many magnitudes as there are in AB equal to E, so many also are there in ΓΔ equal to Z. Let AB be divided into the magnitudes AH, HB equal to E, and ΓΔ into ΓΘ, ΘΔ equal to Z.
ἔσται δὴ ἴσον τὸ πλῆθος τῶν ΑΗ, ΗΒ τῷ πλήθει τῶν ΓΘ, ΘΔ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ μὲν ΑΗ τῷ Ε, τὸ δὲ ΓΘ τῷ Ζ, ἴσον ἄρα τὸ ΑΗ τῷ Ε, καὶ τὰ ΑΗ, ΓΘ τοῖς Ε, Ζ. διὰ τὰ αὐτὰ δὴ ἴσον ἐστὶ τὸ ΗΒ τῷ ε, καὶ τὰ ΗΒ, ΘΔ τοῖς Ε, Ζ·
The multitude of AH, HB will then be equal to the multitude of ΓΘ, ΘΔ. And since AH is equal to E, and ΓΘ to Z, therefore AH, ΓΘ are equal to E, Z. For the same reason HB is equal to E, and HB, ΘΔ to E, Z.
ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Ε, τοσαῦτα καὶ ἐν τοῖς ΑΒ, ΓΔ ἴσα τοῖς Ε, Ζ·
Therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, ΓΔ equal to E, Z.
ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΒ τοῦ Ε, τοσαυταπλάσια ἔσται καὶ τὰ ΑΒ, ΓΔ τῶν Ε, Ζ.
ἐὰν ἄρα ᾖ ὁποσαοῦν μεγέθη ὁποσωνοῦν μεγεθῶν ἴσων τὸ πλῆθος ἕκαστον ἑκάστου ἰσάκις πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ἓν τῶν μεγεθῶν ἑνός, τοσαυταπλάσια ἔσται καὶ τὰ πάντα τῶν πάντων·
Therefore, whatever multiple AB is of E, that multiple will AB, ΓΔ also be of E, Z. Therefore, if there be any number of magnitudes whatever which are, respectively, equal multiples of any magnitudes whatever, equal in multitude, then whatever multiple one of the magnitudes is of one, that multiple will all also be of all.
ὅπερ ἔδει δεῖξαι.
Which was to be proved.
§5.prop.2ἐὰν πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου.
If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, then the first and fifth combined will also be the same multiple of the second that the third and sixth are of the fourth.
πρῶτον γὰρ τὸ ΑΒ δευτέρου τοῦ Γ ἰσάκις ἔστω πολλαπλάσιον καὶ τρίτον τὸ ΔΕ τετάρτου τοῦ Ζ, ἔστω δὲ καὶ πέμπτον τὸ ΒΗ δευτέρου τοῦ Γ ἰσάκις πολλαπλάσιον καὶ ἕκτον τὸ ΕΘ τετάρτου τοῦ Ζ·
For let a first, AB, be the same multiple of a second, Γ, that a third, ΔΕ, is of a fourth, Z, and let a fifth, BH, also be the same multiple of the second, Γ, that a sixth, EΘ, is of the fourth, Z.
λέγω, ὅτι καὶ συντεθὲν πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ.
ἐπεὶ γὰρ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΒ τοῦ Γ καὶ τὸ ΔΕ τοῦ Ζ, ὅσα ἄρα ἐστὶν ἐν τῷ ΑΒ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΔΕ ἴσα τῷ Ζ. διὰ τὰ αὐτὰ δὴ καὶ ὅσα ἐστὶν ἐν τῷ ΒΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν τῷ ΕΘ ἴσα τῷ Ζ·
I say that the first and fifth combined, AH, will also be the same multiple of the second, Γ, that the third and sixth, ΔΘ, are of the fourth, Z. For since AB is the same multiple of Γ that ΔΕ is of Z, therefore, as many magnitudes as there are in AB equal to Γ, so many also are there in ΔΕ equal to Z. For the same reason, as many as there are in BH equal to Γ, so many also are there in EΘ equal to Z.
ὅσα ἄρα ἐστὶν ἐν ὅλῳ τῷ ΑΗ ἴσα τῷ Γ, τοσαῦτα καὶ ἐν ὅλῳ τῷ ΔΘ ἴσα τῷ Ζ·
Therefore, as many as there are in the whole AH equal to Γ, so many also are there in the whole ΔΘ equal to Z.
ὁσαπλάσιον ἄρα ἐστὶ τὸ ΑΗ τοῦ Γ, τοσαυταπλάσιον ἔσται καὶ τὸ ΔΘ τοῦ Ζ. καὶ συντεθὲν ἄρα πρῶτον καὶ πέμπτον τὸ ΑΗ δευτέρου τοῦ Γ ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τὸ ΔΘ τετάρτου τοῦ Ζ.
ἐὰν ἄρα πρῶτον δευτέρου ἰσάκις ᾖ πολλαπλάσιον καὶ τρίτον τετάρτου, ᾖ δὲ καὶ πέμπτον δευτέρου ἰσάκις πολλαπλάσιον καὶ ἕκτον τετάρτου, καὶ συντεθὲν πρῶτον καὶ πέμπτον δευτέρου ἰσάκις ἔσται πολλαπλάσιον καὶ τρίτον καὶ ἕκτον τετάρτου·
Therefore, whatever multiple AH is of Γ, that multiple will ΔΘ also be of Z. Therefore, the first and fifth combined, AH, will be the same multiple of the second, Γ, that the third and sixth, ΔΘ, are of the fourth, Z. Therefore, if a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, then the first and fifth combined will also be the same multiple of the second that the third and sixth are of the fourth.
ὅπερ ἔδει δεῖξαι.
Which was to be proved.