§2.prop.13ἐν τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφʼ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ.
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off inside by the perpendicular towards the acute angle.
ἔστω ὀξυγώνιον τρίγωνον τὸ ΑΒΓ ὀξεῖαν ἔχον τὴν πρὸς τῷ Β γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α σημείου ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ· λέγω, ὅτι τὸ ἀπὸ τῆς ΑΓ τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ.
Let ABG be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BG; I say that the square on AG is less than the squares on GB, BA by twice the rectangle contained by GB, BD.
ἐπεὶ γὰρ εὐθεῖα ἡ ΓΒ τέτμηται, ὡς ἔτυχεν, κατὰ τὸ Δ, τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τῷ ἀπὸ τῆς ΔΓ τετραγώνῳ.
For since the straight line GB has been cut, at random, at D, the squares on GB, BD are equal to twice the rectangle contained by GB, BD and the square on DG.
κοινὸν προσκείσθω τὸ ἀπὸ τῆς ΔΑ τετράγωνον· τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΔ, ΔΑ τετράγωνα ἴσα ἐστὶ τῷ τε δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ καὶ τοῖς ἀπὸ τῶν ΑΔ, ΔΓ τετραγώνοις.
Let the square on DA be added to both; therefore the squares on GB, BD, DA are equal to twice the rectangle contained by GB, BD and the squares on AD, DG.
ἀλλὰ τοῖς μὲν ἀπὸ τῶν ΒΔ, ΔΑ ἴσον τὸ ἀπὸ τῆς ΑΒ· ὀρθὴ γὰρ ἡ πρὸς τῷ Δ γωνίᾳ· τοῖς δὲ ἀπὸ τῶν ΑΔ, ΔΓ ἴσον τὸ ἀπὸ τῆς ΑΓ· τὰ ἄρα ἀπὸ τῶν ΓΒ, ΒΑ ἴσα ἐστὶ τῷ τε ἀπὸ τῆς ΑΓ καὶ τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ·
But the square on AB is equal to the squares on BD, DA; for the angle at D is right; and the square on AG is equal to the squares on AD, DG; therefore the squares on GB, BA are equal to the square on AG and twice the rectangle contained by GB, BD.
ὥστε μόνον τὸ ἀπὸ τῆς ΑΓ ἔλαττόν ἐστι τῶν ἀπὸ τῶν ΓΒ, ΒΑ τετραγώνων τῷ δὶς ὑπὸ τῶν ΓΒ, ΒΔ περιεχομένῳ ὀρθογωνίῳ.
So that the square on AG alone is less than the squares on GB, BA by twice the rectangle contained by GB, BD.
ἐν ἄρα τοῖς ὀξυγωνίοις τριγώνοις τὸ ἀπὸ τῆς τὴν ὀξεῖαν γωνίαν ὑποτεινούσης πλευρᾶς τετράγωνον ἔλαττόν ἐστι τῶν ἀπὸ τῶν τὴν ὀξεῖαν γωνίαν περιεχουσῶν πλευρῶν τετραγώνων τῷ περιεχομένῳ δὶς ὑπό τε μιᾶς τῶν περὶ τὴν ὀξεῖαν γωνίαν, ἐφʼ ἣν ἡ κάθετος πίπτει, καὶ τῆς ἀπολαμβανομένης ἐντὸς ὑπὸ τῆς καθέτου πρὸς τῇ ὀξείᾳ γωνίᾳ· ὅπερ ἔδει δεῖξαι.
Therefore in acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle, namely that on which the perpendicular falls, and the straight line cut off inside by the perpendicular towards the acute angle; which was to be proved.
§2.prop.14τῷ δοθέντι εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι.
To construct a square equal to a given rectilineal figure.
ἔστω τὸ δοθὲν εὐθύγραμμον τὸ Α· δεῖ δὴ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον συστήσασθαι.
Let A be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure A.
συνεστάτω γὰρ τῷ Α εὐθυγράμμῳ ἴσον παραλληλόγραμμον ὀρθογώνιον τὸ ΒΔ·
For let there be constructed the rectangular parallelogram BD equal to the rectilineal figure A.
εἰ μὲν οὖν ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ, γεγονὸς ἂν εἴη τὸ ἐπιταχθέν. συνέσταται γὰρ τῷ Α εὐθυγράμμῳ ἴσον τετράγωνον τὸ ΒΔ·
If then BE is equal to ED, that which was prescribed would be done; for the square BD has been constructed equal to the rectilineal figure A.
εἰ δὲ οὔ, μία τῶν ΒΕ, ΕΔ μείζων ἐστίν.
But if not, one of the straight lines BE, ED is greater.
ἔστω μείζων ἡ ΒΕ, καὶ ἐκβεβλήσθω ἐπὶ τὸ Ζ, καὶ κείσθω τῇ ΕΔ ἴση ἡ ΕΖ, καὶ τετμήσθω ἡ ΒΖ δίχα κατὰ τὸ Η, καὶ κέντρῳ τῷ Η, διαστήματι δὲ ἑνὶ τῶν ΗΒ, ΗΖ ἡμικύκλιον γεγράφθω τὸ ΒΘΖ, καὶ ἐκβεβλήσθω ἡ ΔΕ ἐπὶ τὸ Θ, καὶ ἐπεζεύχθω ἡ ΗΘ.
ἐπεὶ οὖν εὐθεῖα ἡ ΒΖ τέτμηται εἰς μὲν ἴσα κατὰ τὸ Η, εἰς δὲ ἄνισα κατὰ τὸ Ε, τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον μετὰ τοῦ ἀπὸ τῆς ΕΗ τετραγώνου ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΖ τετραγώνῳ.
Let BE be greater, and let it be produced to Z, and let EZ be made equal to ED, and let BZ be cut in half at H, and with center H and with one of the straight lines HB, HZ as distance let the semicircle BThZ be described, and let DE be produced to Th, and let HTh be joined. Since therefore the straight line BZ has been cut into equal parts at H and into unequal parts at E, the rectangle contained by BE, EZ with the square on EH is equal to the square on HZ.
ἴση δὲ ἡ ΗΖ τῇ ΗΘ· τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ τῆς ΗΕ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΗΘ. τῷ δὲ ἀπὸ τῆς ΗΘ ἴσα ἐστὶ τὰ ἀπὸ τῶν ΘΕ, ΕΗ τετράγωνα· τὸ ἄρα ὑπὸ τῶν ΒΕ, ΕΖ μετὰ τοῦ ἀπὸ ΗΕ ἴσα ἐστὶ τοῖς ἀπὸ τῶν ΘΕ, ΕΗ. κοινὸν ἀφῃρήσθω τὸ ἀπὸ τῆς ΗΕ τετράγωνον·
And HZ is equal to HTh; therefore the rectangle contained by BE, EZ with the square on HE is equal to the square on HTh. But the squares on ThE, EH are equal to the square on HTh; therefore the rectangle contained by BE, EZ with the square on HE is equal to the squares on ThE, EH.
λοιπὸν ἄρα τὸ ὑπὸ τῶν ΒΕ, ΕΖ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ τετραγώνῳ.
Let the square on HE be subtracted from both; therefore the remaining rectangle contained by BE, EZ is equal to the square on ETh.
ἀλλὰ τὸ ὑπὸ τῶν ΒΕ, ΕΖ τὸ ΒΔ ἐστιν· ἴση γὰρ ἡ ΕΖ τῇ ΕΔ· τὸ ἄρα ΒΔ παραλληλόγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΘΕ τετραγώνῳ.
But the rectangle contained by BE, EZ is BD; for EZ is equal to ED; therefore the parallelogram BD is equal to the square on ThE.
ἴσον δὲ τὸ ΒΔ τῷ Α εὐθυγράμμῳ. καὶ τὸ Α ἄρα εὐθύγραμμον ἴσον ἐστὶ τῷ ἀπὸ τῆς ΕΘ ἀναγραφησομένῳ τετραγώνῳ.
And BD is equal to the rectilineal figure A; therefore the rectilineal figure A is also equal to the square to be described on ETh.
τῷ ἄρα δοθέντι εὐθυγράμμῳ τῷ Α ἴσον τετράγωνον συνέσταται τὸ ἀπὸ τῆς ΕΘ ἀναγραφησόμενον· ὅπερ ἔδει ποιῆσαι.
Therefore, a square equal to the given rectilineal figure A has been constructed, namely that to be described on ETh; which was to be done.