§6.prop.12τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προσευρεῖν.
To three given straight lines to find a fourth proportional.
ἔστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ· δεῖ δὴ τῶν Α, Β, Γ τετάρτην ἀνάλογον προσευρεῖν.
Let the three given straight lines be A, B, C; it is required indeed to find a fourth proportional to A, B, C.
Ἐκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχουσαι τὴν ὑπὸ ΕΔΖ· καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ·
Let two straight lines DE, DZ be placed containing the angle EDZ; and let DH be made equal to A, and HE equal to B, and further DΘ equal to C.
καὶ ἐπιζευχθείσης τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ.
ἐπεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Γ· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ.
τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη ἀνάλογον προσεύρηται ἡ ΘΖ· ὅπερ ἔδει ποιῆσαι.
And, HΘ being joined, let EZ be drawn through E parallel to it. Since then, parallel to one of the sides of the triangle DEZ, namely EZ, HΘ has been drawn, therefore, as DH is to HE, so is DΘ to ΘZ. And DH is equal to A, and HE to B, and DΘ to C; therefore, as A is to B, so is C to ΘZ. Therefore, to three given straight lines A, B, C, a fourth proportional to them, ΘZ, has been found; which was to be done.
§6.prop.13δύο δοθεισῶν εὐθειῶν μέσην ἀνάλογον προσευρεῖν.
To two given straight lines to find a mean proportional.
ἔστωσαν αἱ δοθεῖσαι δύο εὐθεῖαι αἱ ΑΒ, ΒΓ· δεῖ δὴ τῶν ΑΒ, ΒΓ μέσην ἀνάλογον προσευρεῖν.
Let the two given straight lines be AB, BC; it is required indeed to find a mean proportional to AB, BC.
κείσθωσαν ἐπʼ εὐθείας, καὶ γεγράφθω ἐπὶ τῆς ΑΓ ἡμικύκλιον τὸ ΑΔΓ, καὶ ἤχθω ἀπὸ τοῦ Β σημείου τῇ ΑΓ εὐθείᾳ πρὸς ὀρθὰς ἡ ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΔ, ΔΓ.
ἐπεὶ ἐν ἡμικυκλίῳ γωνία ἐστὶν ἡ ὑπὸ ΑΔΓ, ὀρθή ἐστιν.
Let them be placed in a straight line, and let there be described on AC the semicircle ADC, and let BD be drawn from the point B at right angles to the straight line AC, and let AD, DC be joined. Since the angle ADC is an angle in a semicircle, it is right.
καὶ ἐπεὶ ἐν ὀρθογωνίῳ τριγώνῳ τῷ ΑΔΓ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἦκται ἡ ΔΒ, ἡ ΔΒ ἄρα τῶν τῆς βάσεως τμημάτων τῶν ΑΒ, ΒΓ μέση ἀνάλογόν ἐστιν.
And since, in the right-angled triangle ADC, from the right angle, the perpendicular DB has been drawn to the base, therefore DB is a mean proportional between the segments of the base, AB, BC.
δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΒΓ μέση ἀνάλογον προσεύρηται ἡ ΔΒ· ὅπερ ἔδει ποιῆσαι.
Therefore, to two given straight lines AB, BC, a mean proportional to them, DB, has been found; which was to be done.
§6.prop.14τῶν ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα.
In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional; and those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
ἔστω ἴσα τε καὶ ἰσογώνια παραλληλόγραμμα τὰ ΑΒ, ΒΓ ἴσας ἔχοντα τὰς πρὸς τῷ Β γωνίας, καὶ κείσθωσαν ἐπʼ εὐθείας αἱ ΔΒ, ΒΕ·
Let AB, BC be equal and equiangular parallelograms having the angles at B equal, and let DB, BE be placed in a straight line; therefore ZB, BH are also in a straight line.
ἐπʼ εὐθείας ἄρα εἰσὶ καὶ αἱ ΖΒ, ΒΗ. λέγω, ὅτι τῶν ΑΒ, ΒΓ ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ.
συμπεπληρώσθω γὰρ τὸ ΖΕ παραλληλόγραμμον.
I say that in AB, BC the sides about the equal angles are reciprocally proportional, that is, that as DB is to BE, so is HB to BZ. For let the parallelogram ZE be completed.
ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ, ἄλλο δέ τι τὸ ΖΕ, ἔστιν ἄρα ὡς τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ. ἀλλʼ ὡς μὲν τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως ἡ ΔΒ πρὸς τὴν ΒΕ, ὡς δὲ τὸ ΒΓ πρὸς τὸ ΖΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ·
Since then the parallelogram AB is equal to the parallelogram BC, and ZE is some other parallelogram, therefore, as AB is to ZE, so is BC to ZE. But, as AB is to ZE, so is DB to BE, and as BC is to ZE, so is HB to BZ; therefore, also, as DB is to BE, so is HB to BZ.
καὶ ὡς ἄρα ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ. τῶν ἄρα ΑΒ, ΒΓ παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας.
Therefore in the parallelograms AB, BC the sides about the equal angles are reciprocally proportional.
ἀλλὰ δὴ ἔστω ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ.
But indeed let as DB is to BE, so is HB to BZ; I say that the parallelogram AB is equal to the parallelogram BC.
ἐπεὶ γάρ ἐστιν ὡς ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως ἡ ΗΒ πρὸς τὴν ΒΖ, ἀλλʼ ὡς μὲν ἡ ΔΒ πρὸς τὴν ΒΕ, οὕτως τὸ ΑΒ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, ὡς δὲ ἡ ΗΒ πρὸς τὴν ΒΖ, οὕτως τὸ ΒΓ παραλληλόγραμμον πρὸς τὸ ΖΕ παραλληλόγραμμον, καὶ ὡς ἄρα τὸ ΑΒ πρὸς τὸ ΖΕ, οὕτως τὸ ΒΓ πρὸς τὸ ΖΕ· ἴσον ἄρα ἐστὶ τὸ ΑΒ παραλληλόγραμμον τῷ ΒΓ παραλληλογράμμῳ.
For since as DB is to BE, so is HB to BZ, but as DB is to BE, so is the parallelogram AB to the parallelogram ZE, and as HB is to BZ, so is the parallelogram BC to the parallelogram ZE, therefore, also, as AB is to ZE, so is BC to ZE; therefore the parallelogram AB is equal to the parallelogram BC.
τῶν ἄρα ἴσων τε καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα· ὅπερ ἔδει δεῖξαι.
Therefore, in equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional; and those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; which was to be demonstrated.