§18.1## 18.
## 18.
Uulgo prop.
Commonly Prop.
LXXXVII.
Ἐὰν δύο εὐθεῖαι δοθὲν χωρίον περιέχωσιν ἐν δεδομένῃ γωνίᾳ, τὸ δὲ ἀπὸ τῆς μείζονος τοῦ ἀπὸ τῆς ἐλάσσονος δοθέντι μεῖζον ᾖ, καὶ ἑκατέρα αὐτῶν ἔσται δοθεῖσα.
LXXXVII. If two straight lines contain a given area in a given angle, and the square on the greater is greater than the square on the less by a given area, each of them will be given.
δύο γὰρ εὐθεῖαι αἱ ΑΒ, ΒΓ δοθὲν περιεχέτωσαν χωρίον τὸ ΑΓ ἐν δεδομένῃ γωνίᾳ τῇ ὑπὸ τῶν ΑΒΓ, τὸ δὲ ἀπὸ τῆς ΑΒ δοθέντι μεῖζον ἔστω τοῦ ἀπὸ τῆς ΒΓ·
For let two straight lines AB, BC contain a given area AC in a given angle ABC, and let the square on AB be greater than the square on BC by a given area.
λέγω, ὅτι δοθεῖσά ἐστιν ἑκατέρα τῶν ΑΒ, ΒΓ.
ἐπεὶ γὰρ τὸ ἀπὸ τῆς ΑΒ τοῦ ἀπὸ τῆς ΒΓ δοθέντι μεῖζόν ἐστιν, ἀφῃρήσθω τὸ δοθὲν τὸ ὑπὸ τῶν ΑΒ, ΒΔ· λοιπὸν ἄρα τὸ ὑπὸ τῶν BA, ΑΔ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΒΓ. καὶ ἐπεὶ δοθέν ἐστι τὸ ὑπὸ τῶν ΑΒ, ΒΓ, ἔστι δὲ καὶ τὸ ὑπὸ τῶν ΑΒ, ΒΔ δοθέν, λόγος ἄρα τοῦ ὑπὸ τῶν ΑΒ, ΒΔ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ δοθείς.
I say that each of AB, BC is given. For since the square on AB is greater than the square on BC by a given area, let there be subtracted the given area contained by AB, BD; therefore the remaining area contained by BA, AD is equal to the square on BC. And since the area contained by AB, BC is given, and the area contained by AB, BD is also given, therefore the ratio of the area contained by AB, BD to the area contained by AB, BC is given.
καί ἐστιν, ὡς τὸ ὑπὸ τῶν ΑΒ, ΒΔ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΓ, οὕτως ἡ ΔΒ πρὸς ΒΓ· λόγος ἄρα καὶ τῆς ΔΒ πρὸς ΒΓ δοθείς· λόγος ἄρα καὶ τοῦ ἀπὸ τῆς ΔΒ πρὸς τὸ ἀπὸ τῆς ΒΓ δοθείς.
And as the area contained by AB, BD is to the area contained by AB, BC, so is DB to BC; therefore the ratio of DB to BC is also given; therefore the ratio of the square on DB to the square on BC is also given.
τῷ δὲ ἀπὸ τῆς ΓΒ ἴσον τὸ ὑπὸ τῶν ΒΑ, ΑΔ· λόγος ἄρα καὶ τοῦ ὑπὸ τῶν ΒΑ, ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΒ δοθείς· καὶ τοῦ τετράκις ἄρα ὑπὸ τῶν ΒΑ, ΑΔ μετὰ τοῦ ἀπὸ τῆς ΔΒ πρὸς τὸ ἀπὸ τῆς ΒΔ λόγος δοθείς.
And the area contained by BA, AD is equal to the square on GB; therefore the ratio of the area contained by BA, AD to the square on DB is also given; therefore the ratio of four times the area contained by BA, AD together with the square on DB to the square on BD is given.
ἀλλὰ τὸ τετράκις ὑπὸ τῶν ΒΑ, ΑΔ μετὰ τοῦ ἀπὸ τῆς ΒΔ τὸ ἀπὸ συναμφοτέρου τῆς ΒΑ, ΑΔ ἐστιν· λόγος ἄρα καὶ τοῦ ἀπὸ συναμφοτέρου τῆς ΒΑ, ΑΔ πρὸς τὸ ἀπὸ τῆς ΔΒ δοθείς· λόγος ἄρα καὶ συναμφοτέρου τῆς ΒΑ, ΑΔ πρὸς ΔΒ δοθείς.
But four times the area contained by BA, AD together with the square on BD is the square on BA and AD together; therefore the ratio of the square on BA and AD together to the square on DB is given; therefore the ratio of BA and AD together to DB is given.
καὶ συνθέντι συναμφοτέρου τῆς ΒΑ, ΑΔ μετὰ τῆς ΔΒ, τουτέστι δύο τῶν ΑΒ πρὸς ΒΔ λόγος ἐστὶ δοθείς· καὶ τῆς ΑΒ ἄρα πρὸς ΒΔ λόγος ἐστὶ δοθείς.
And by addition, the ratio of BA and AD together with DB, that is, twice AB, to BD is given; therefore the ratio of AB to BD is given.
τῆς δὲ ΔΒ πρὸς τὴν ΒΓ λόγος ἐστὶ δοθείς· καὶ τῆς ΑΒ ἄρα πρὸς ΒΓ λόγος δοθείς.
And the ratio of DB to BC is given; therefore the ratio of AB to BC is given.
καὶ ἐπεὶ λόγος τῆς ΑΒ πρὸς ΒΔ δοθείς, καί ἐστιν, ὡς ἡ ΑΒ πρὸς ΒΔ, οὕτως τὸ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΔ, λόγος ἄρα καὶ τοῦ ἀπὸ τῆς ΑΒ πρὸς τὸ ὑπὸ τῶν ΑΒ, ΒΔ δοθείς.
And since the ratio of AB to BD is given, and as AB is to BD, so is the square on AB to the area contained by AB, BD, therefore the ratio of the square on AB to the area contained by AB, BD is given.
δοθὲν δὲ τὸ ὑπὸ τῶν ΑΒ, ΒΔ· οὕτως γὰρ δοθὲν ἀφῄρηται· δοθὲν ἄρα καὶ τὸ ἀπὸ τῆς ΑΒ· δοθεῖσα ἄρα ἡ ΑΒ. καί ἐστι λόγος τῆς ΑΒ πρὸς ΒΓ δοθείς· δοθεῖσα ἄρα καὶ ἡ ΒΓ.
And the area contained by AB, BD is given; for it was subtracted as a given area; therefore also the square on AB is given; therefore AB is given. And the ratio of AB to BC is given; therefore BC is also given.
§18.2## Λῆμμα τοῦ ἐπάνω.
## Lemma for the above.
Πῶς δοθέν ἐστι τὸ ὑπὸ τῶν ΑΒΓ ὀρθογώνιον ἀμβλείας ὑποκειμένης τῆς ὑπὸ ΑΒΓ γωνίας;
How is the rectangle contained by AB, BC given when the angle ABC is assumed to be obtuse?
ἤχθω ἀπὸ τοῦ Β σημείου κάθετος ἡ ΒΔ, καὶ ἐκβεβλήσθω ἡ ΓΔ ἐπὶ τὸ Θ, καὶ συμπεπληρώσθω τὸ ΒΔΘΑ ὀρθογώνιον·
Let a perpendicular BD be drawn from point B, and let CD be produced to Θ, and let the rectangle BDΘA be completed; therefore it is equal to AC.
ἴσον ἄρα ἐστὶ τῷ ΑΓ. καὶ ἐκβεβλήσθω ἡ ΔΒ ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΒΓ ἴση ἡ ΒΖ, καὶ συμπεπληρώσθω τὸ ΑΖ ὀρθογώνιον.
And let DB be produced to Z, and let BZ be placed equal to BC, and let the rectangle AZ be completed.
ἐπεὶ οὖν δοθεῖσά ἐστιν ἡ ὑπὸ ΑΒΓ ὑπόκειται γάρ· δοθεῖσα δὲ καὶ ἡ ὑπὸ ΑΒΔ· ὀρθὴ γάρ·
Since, then, the angle ABC is given (for it is assumed), and the angle ABD is also given (for it is right), therefore the remaining angle DBC is given.
λοιπὴ ἄρα ἡ ὑπὸ ΔΒΓ δοθεῖσά ἐστιν. καὶ ὀρθὴ ἡ Δ· λοιπὴ ἄρα, ἡ Γ δοθεῖσά ἐστιν· δοθὲν ἄρα τὸ ΒΓΔ τρίγωνον τῷ εἴδει· λόγος ἄρα τῆς ΔΒ πρὸς ΒΓ δοθείς.
And the angle D is right; therefore the remaining angle C is given; therefore the triangle BCD is given in species; therefore the ratio of DB to BC is given.
ἴση δὲ ἡ ΒΓ τῇ ΒΖ· λόγος ἄρα καὶ τῆς ΔΒ πρὸς ΒΖ δοθείς· ὥστε καὶ τοῦ ΒΘ πρὸς ΖΑ λόγος δοθείς.
And BC is equal to BZ; therefore the ratio of DB to BZ is also given; so that the ratio of BΘ to ZA is also given.
ἴσον δὲ τὸ ΒΘ τῷ ΑΓ· λόγος ἄρα τοῦ ΑΓ πρὸς ΑΖ δοθείς.
And BΘ is equal to AC; therefore the ratio of AC to AZ is given.
καὶ δοθὲν τὸ ΑΓ· δοθὲν ἄρα καὶ τὸ ΑΖ, τουτέστι τὸ ὑπὸ ΑΒΖ, τουτέστι τὸ ὑπὸ AΒΓ.
And AC is given; therefore also AZ is given, that is, the rectangle contained by AB, BZ, that is, the rectangle contained by AB, BC.