§13.prop.5ἔστω εὐθεῖα ἡ ΑΒ, καὶ τετμήσθω ἄκρον καὶ μέσον λόγον κατὰ τὸ Γ, καὶ ἔστω μεῖζον τμῆμα τὸ ΑΓ· λέγω, ὅτι τὰ ἀπὸ τῶν ΑΒ, ΒΓ τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΓΑ.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ καταγεγράφθω τὸ σχῆμα.
Let AB be a straight line, and let it be cut in extreme and mean ratio at C, and let AC be the greater segment; I say that the squares on AB, BC are triple of the square on CA. For let the square ADEB be described on AB, and let the figure be drawn.
ἐπεὶ οὖν ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΓ, τὸ ἄρα ὑπὸ τῶν ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΓ. καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒΓ τὸ ΑΚ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΘΗ· ἴσον ἄρα ἐστὶ τὸ ΑΚ τῷ ΘΗ. καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΖ τῷ ΖΕ, κοινὸν προσκείσθω τὸ ΓΚ· ὅλον ἄρα τὸ ΑΚ ὅλῳ τῷ ΓΕ ἐστιν ἴσον· τὰ ἄρα ΑΚ, ΓΕ τοῦ ΑΚ ἐστι διπλάσια.
Since then AB has been cut in extreme and mean ratio at C, and AC is the greater segment, therefore the rectangle contained by AB, BC is equal to the square on AC. And AK is the rectangle contained by AB, BC, and ThH is the square on AC; therefore AK is equal to ThH. And since AZ is equal to ZE, let the common GK be added; therefore the whole AK is equal to the whole GE; therefore AK, GE are double of AK.
ἀλλὰ τὰ ΑΚ, ΓΕ ὁ ΛΜΝ γνώμων ἐστὶ καὶ τὸ ΓΚ τετράγωνον· ὁ ἄρα ΛΜΝ γνώμων καὶ τὸ ΓΚ τετράγωνον διπλάσιά ἐστι τοῦ ΑΚ. ἀλλὰ μὴν καὶ τὸ ΑΚ τῷ ΘΗ ἐδείχθη ἴσον· ὁ ἄρα ΛΜΝ γνώμων καὶ τὰ ΓΚ, ΘΗ τετράγωνα τριπλάσιά ἐστι τοῦ ΘΗ τετραγώνου.
But AK, GE are the gnomon LMN and the square GK; therefore the gnomon LMN and the square GK are double of AK. But indeed AK was also shown to be equal to ThH; therefore the gnomon LMN and the squares GK, ThH are triple of the square ThH.
καί ἐστιν ὁ ΛΜΝ γνώμων καὶ τὰ ΓΚ, ΘΗ τετράγωνα ὅλον τὸ ΑΕ καὶ τὸ ΓΚ, ἅπερ ἐστὶ τὰ ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα, τὸ δὲ ΗΘ τὸ ἀπὸ τῆς ΑΓ τετράγωνον.
And the gnomon LMN and the squares GK, ThH are the whole AE and GK, which are the squares on AB, BC, and HTh is the square on AC.
τὰ ἄρα ἀπὸ τῶν ΑΒ, ΒΓ τετράγωνα τριπλάσιά ἐστι τοῦ ἀπὸ τῆς ΑΓ τετραγώνου· ὅπερ ἔδει δεῖξαι.
Therefore the squares on AB, BC are triple of the square on AC; which it was required to prove.
ἐὰν εὐθεῖα γραμμὴ ἄκρον καὶ μέσον λόγον τμηθῇ, καὶ προστεθῇ αὐτῇ ἴση τῷ μείζονι τμήματι, ἡ ὅλη εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα.
If a straight line be cut in extreme and mean ratio, and there be added to it a straight line equal to the greater segment, the whole straight line has been cut in extreme and mean ratio, and the greater segment is the original straight line.
εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τετμήσθω κατὰ τὸ Γ σημεῖον, καὶ ἔστω μεῖζον τμῆμα ἡ ΑΓ, καὶ τῇ ΑΓ ἴση ἡ ΑΔ. λέγω, ὅτι ἡ ΔΒ εὐθεῖα ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ἐξ ἀρχῆς εὐθεῖα ἡ ΑΒ.
Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΕ, καὶ καταγεγράφθω τὸ σχῆμα.
For let some straight line AB be cut in extreme and mean ratio at the point C, and let AC be the greater segment, and let AD be equal to AC; I say that the straight line DB has been cut in extreme and mean ratio at A, and the greater segment is the original straight line AB. For let the square AE be described on AB, and let the figure be drawn.
ἐπεὶ ἡ ΑΒ ἄκρον καὶ μέσον λόγον τέτμηται κατὰ τὸ Γ, τὸ ἄρα ὑπὸ ΑΒΓ ἴσον ἐστὶ τῷ ἀπὸ ΑΓ. καί ἐστι τὸ μὲν ὑπὸ ΑΒΓ τὸ ΓΕ, τὸ δὲ ἀπὸ τῆς ΑΓ τὸ ΓΘ· ἴσον ἄρα τὸ ΓΕ τῷ ΘΓ. ἀλλὰ τῷ μὲν ΓΕ ἴσον ἐστὶ τὸ ΘΕ, τῷ δὲ ΘΓ ἴσον τὸ ΔΘ· καὶ τὸ ΔΘ ἄρα ἴσον ἐστὶ τῷ ΘΕ.
Since AB has been cut in extreme and mean ratio at C, therefore the rectangle contained by AB, BC is equal to the square on AC. And GE is the rectangle contained by AB, BC, and GTh is the square on AC; therefore GE is equal to ThG. But ThE is equal to GE, and DTh is equal to ThG; therefore DTh is also equal to ThE.
ὅλον ἄρα τὸ ΔΚ ὅλῳ τῷ ΑΕ ἐστιν ἴσον.
Therefore the whole DK is equal to the whole AE.
καί ἐστι τὸ μὲν ΔΚ τὸ ὑπὸ τῶν ΒΔ, ΔΑ· ἴση γὰρ ἡ ΑΔ τῇ ΔΛ· τὸ δὲ ΑΕ τὸ ἀπὸ τῆς ΑΒ· τὸ ἄρα ὑπὸ τῶν ΒΔΑ ἴσον ἐστὶ τῷ ἀπὸ τῆς ΑΒ. ἔστιν ἄρα ὡς ἡ ΔΒ πρὸς τὴν ΒΑ, οὕτως ἡ ΒΑ πρὸς τὴν ΑΔ. μείζων δὲ ἡ ΔΒ τῆς ΒΑ· μείζων ἄρα καὶ ἡ ΒΑ τῆς ΑΔ.
ἡ ἄρα ΔΒ ἄκρον καὶ μέσον λόγον τέμηται κατὰ τὸ Α, καὶ τὸ μεῖζον τμῆμά ἐστιν ἡ ΑΒ· ὅπερ ἔδει δεῖξαι.
And DK is the rectangle contained by BD, DA, for AD is equal to DL; and AE is the square on AB; therefore the rectangle contained by BD, DA is equal to the square on AB. Therefore, as DB is to BA, so is BA to AD. And DB is greater than BA; therefore BA is also greater than AD. Therefore DB has been cut in extreme and mean ratio at A, and the greater segment is AB; which it was required to prove.