15. In equal triangles which have one angle equal to one angle the sides about the equal angles are reciprocally proportional; and those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal. Let ABC, ADE be equal triangles having one angle equal to one angle, namely the angle BAC to the angle DAE; I say that in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional, that is to say, that, as CA is to AD, so is EA to AB. For let them be placed so that CA is in a straight line with AD; therefore EA is also in a straight line with AB. [] Let BD be joined. Since then the triangle ABC is equal to the triangle ADE, and BAD is another area, therefore, as the triangle CAB is to the triangle BAD, so is the triangle EAD to the triangle BAD. [] But, as CAB is to BAD, so is CA to AD, [] and, as EAD is to BAD, so is EA to AB. [id.] Therefore also, as CA is to AD, so is EA to AB. [] Therefore in the triangles ABC, ADE the sides about the equal angles are reciprocally proportional. Next, let the sides of the triangles ABC, ADE be reciprocally proportional, that is to say, let EA be to AB as CA to AD; I say that the triangle ABC is equal to the triangle ADE. For, if BD be again joined, since, as CA is to AD, so is EA to AB, while, as CA is to AD, so is the triangle ABC to the triangle BAD, and, as EA is to AB, so is the triangle EAD to the triangle BAD, [] therefore, as the triangle ABC is to the triangle BAD, so is the triangle EAD to the triangle BAD. [] Therefore each of the triangles ABC, EAD has the same ratio to BAD. Therefore the triangle ABC is equal to the triangle EAD. [] Therefore etc. Q. E. D.
15. τῶν ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΑΔΕ μίαν μιᾷ ἴσην ἔχοντα γωνίαν τὴν ὑπὸ ΒΑΓ τῇ ὑπὸ ΔΑΕ· λέγω, ὅτι τῶν ΑΒΓ, ΑΔΕ τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, τουτέστιν, ὅτι ἐστὶν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. κείσθω γὰρ ὥστε ἐπʼ εὐθείας εἶναι τὴν ΓΑ τῇ ΑΔ· ἐπʼ εὐθείας ἄρα ἐστὶ καὶ ἡ ΕΑ τῇ ΑΒ. καὶ ἐπεζεύχθω ἡ ΒΔ. ἐπεὶ οὖν ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ, ἄλλο δέ τι τὸ ΒΑΔ, ἔστιν ἄρα ὡς τὸ ΓΑΒ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἀλλʼ ὡς μὲν τὸ ΓΑΒ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΓΑ πρὸς τὴν ΑΔ, ὡς δὲ τὸ ΕΑΔ πρὸς τὸ ΒΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. καὶ ὡς ἄρα ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ. τῶν ΑΒΓ, ΑΔΕ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἀλλὰ δὴ ἀντιπεπονθέτωσαν αἱ πλευραὶ τῶν ΑΒΓ, ΑΔΕ τριγώνων, καὶ ἔστω ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΔΕ τριγώνῳ. Ἐπιζευχθείσης γὰρ πάλιν τῆς ΒΔ, ἐπεί ἐστιν ὡς ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως ἡ ΕΑ πρὸς τὴν ΑΒ, ἀλλʼ ὡς μὲν ἡ ΓΑ πρὸς τὴν ΑΔ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς δὲ ἡ ΕΑ πρὸς τὴν ΑΒ, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, ὡς ἄρα τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον, οὕτως τὸ ΕΑΔ τρίγωνον πρὸς τὸ ΒΑΔ τρίγωνον. ἑκάτερον ἄρα τῶν ΑΒΓ, ΕΑΔ πρὸς τὸ ΒΑΔ τὸν αὐτὸν ἔχει λόγον. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ [τρίγωνον] τῷ ΕΑΔ τριγώνῳ. τῶν ἄρα ἴσων καὶ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας· καὶ ὧν μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἐκεῖνα ἴσα ἐστίν· ὅπερ ἔδει δεῖξαι.
16. If four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means; and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines will be proportional. Let the four straight lines AB, CD, E, F be proportional, so that, as AB is to CD, so is E to F; I say that the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Let AG, CH be drawn from the points A, C at right angles to the straight lines AB, CD, and let AG be made equal to F, and CH equal to E. Let the parallelograms BG, DH be completed. Then since, as AB is to CD, so is E to F, while E is equal to CH, and F to AG, therefore, as AB is to CD, so is CH to AG. Therefore in the parallelograms BG, DH the sides about the equal angles are reciprocally proportional. But those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal; [] therefore the parallelogram BG is equal to the parallelogram DH. And BG is the rectangle AB, F, for AG is equal to F; and DH is the rectangle CD, E, for E is equal to CH; therefore the rectangle contained by AB, F is equal to the rectangle contained by CD, E. Next, let the rectangle contained by AB, F be equal to the rectangle contained by CD, E; I say that the four straight lines will be proportional, so that, as AB is to CD, so is E to F. For, with the same construction, since the rectangle AB, F is equal to the rectangle CD, E, and the rectangle AB, F is BG, for AG is equal to F, and the rectangle CD, E is DH, for CH is equal to E, therefore BG is equal to DH. And they are equiangular But in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional. [] Therefore, as AB is to CD, so is CH to AG. But CH is equal to E, and AG to F; therefore, as AB is to CD, so is E to F. Therefore etc. Q. E. D.
16. ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται. ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, Ε, Ζ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ· λέγω, ὅτι τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. ἤχθωσαν [γὰρ] ἀπὸ τῶν Α, Γ σημείων ταῖς ΑΒ, ΓΔ εὐθείαις πρὸς ὀρθὰς αἱ ΑΗ, ΓΘ, καὶ κείσθω τῇ μὲν Ζ ἴση ἡ ΑΓ, τῇ δὲ Ε ἴση ἡ ΓΘ. καὶ συμπεπληρώσθω τὰ ΒΗ, ΔΘ παραλληλόγραμμα. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ, ἴση δὲ ἡ μὲν Ε τῇ ΓΘ, ἡ δὲ Ζ τῇ ΑΗ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. τῶν ΒΗ, ΔΘ ἄρα παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα· ἴσον ἄρα ἐστὶ τὸ ΒΗ παραλληλόγραμμον τῷ ΔΘ παραλληλογράμμῳ. καί ἐστι τὸ μὲν ΒΗ τὸ ὑπὸ τῶν ΑΒ, Ζ· ἴση γὰρ ἡ ΑΗ τῇ Ζ· τὸ δὲ ΔΘ τὸ ὑπὸ τῶν ΓΔ, Ε· ἴση γὰρ ἡ Ε τῇ ΓΘ· τὸ ἄρα ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ. ἀλλὰ δὴ τὸ ὑπὸ τῶν ΑΒ, Ζ περιεχόμενον ὀρθογώνιον ἴσον ἔστω τῷ ὑπὸ τῶν ΓΔ, Ε περιεχομένῳ ὀρθογωνίῳ· λέγω, ὅτι αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν ΑΒ, Ζ ἴσον ἐστὶ τῷ ὑπὸ τῶν ΓΔ, Ε, καί ἐστι τὸ μὲν ὑπὸ τῶν ΑΒ, Ζ τὸ ΒΗ· ἴση γάρ ἐστιν ἡ ΑΗ τῇ Ζ· τὸ δὲ ὑπὸ τῶν ΓΔ, Ε τὸ ΔΘ· ἴση γὰρ ἡ ΓΘ τῇ Ε· τὸ ἄρα ΒΗ ἴσον ἐστὶ τῷ ΔΘ. καί ἐστιν ἰσογώνια. τῶν δὲ ἴσων καὶ ἰσογωνίων παραλληλογράμμων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΓΘ πρὸς τὴν ΑΗ. ἴση δὲ ἡ μὲν ΓΘ τῇ Ε, ἡ δὲ ΑΗ τῇ Ζ· ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ Ε πρὸς τὴν Ζ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ, αἱ τέσσαρες εὐθεῖαι ἀνάλογον ἔσονται· ὅπερ ἔδει δεῖξαι.
17. If three straight lines be proportional, the rectangle contained by the extremes is equal to the square on the mean; and, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines will be proportional. Let the three straight lines A, B, C be proportional, so that, as A is to B, so is B to C; I say that the rectangle contained by A, C is equal to the square on B. Let D be made equal to B. Then, since, as A is to B, so is B to C, and B is equal to D, therefore, as A is to B, so is D to C. But, if four straight lines be proportional, the rectangle contained by the extremes is equal to the rectangle contained by the means. [] Therefore the rectangle A, C is equal to the rectangle B, D. But the rectangle B, D is the square on B, for B is equal to D; therefore the rectangle contained by A, C is equal to the square on B. Next, let the rectangle A, C be equal to the square on B; I say that, as A is to B, so is B to C. For, with the same construction, since the rectangle A, C is equal to the square on B, while the square on B is the rectangle B, D, for B is equal to D, therefore the rectangle A, C is equal to the rectangle B, D. But, if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportional. [] Therefore, as A is to B, so is D to C. But B is equal to D; therefore, as A is to B, so is B to C. Therefore etc. Q. E. D.
17. ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται. ἔστωσαν τρεῖς εὐθεῖαι ἀνάλογον αἱ Α, Β, Γ, ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ· λέγω, ὅτι τὸ ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς β τετραγώνῳ. κείσθω τῇ Β ἴση ἡ Δ. καὶ ἐπεί ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ, ἴση δὲ ἡ Β τῇ Δ, ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, ἡ Δ πρὸς τὴν Γ. ἐὰν δὲ τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον [ὀρθογώνιον] ἴσον ἐστὶ τῷ ὑπὸ τῶν μέσων περιεχομένῳ ὀρθογωνίῳ. τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἀλλὰ τὸ ὑπὸ τῶν Β, Δ τὸ ἀπὸ τῆς Β ἐστιν· ἴση γὰρ ἡ Β τῇ Δ· τὸ ἄρα ὑπὸ τῶν Α, Γ περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς Β τετραγώνῳ. ἀλλὰ δὴ τὸ ὑπὸ τῶν Α, Γ ἴσον ἔστω τῷ ἀπὸ τῆς Β· λέγω, ὅτι ἐστὶν ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. τῶν γὰρ αὐτῶν κατασκευασθέντων, ἐπεὶ τὸ ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ἀπὸ τῆς Β, ἀλλὰ τὸ ἀπὸ τῆς Β τὸ ὑπὸ τῶν Β, Δ ἐστιν· ἴση γὰρ ἡ Β τῇ Δ· τὸ ἄρα ὑπὸ τῶν Α, Γ ἴσον ἐστὶ τῷ ὑπὸ τῶν Β, Δ. ἐὰν δὲ τὸ ὑπὸ τῶν ἄκρων ἴσον ᾖ τῷ ὑπὸ τῶν μέσων, αἱ τέσσαρες εὐθεῖαι ἀνάλογόν εἰσιν. ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Δ πρὸς τὴν Γ. ἴση δὲ ἡ Β τῇ Δ· ὡς ἄρα ἡ Α πρὸς τὴν Β, οὕτως ἡ Β πρὸς τὴν Γ. ἐὰν ἄρα τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ἐστὶ τῷ ἀπὸ τῆς μέσης τετραγώνῳ· κἂν τὸ ὑπὸ τῶν ἄκρων περιεχόμενον ὀρθογώνιον ἴσον ᾖ τῷ ἀπὸ τῆς μέσης τετραγώνῳ, αἱ τρεῖς εὐθεῖαι ἀνάλογον ἔσονται· ὅπερ ἔδει δεῖξαι.
18. On a given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure. Let AB be the given straight line and CE the given rectilineal figure; thus it is required to describe on the straight line AB a rectilineal figure similar and similarly situated to the rectilineal figure CE. Let DF be joined, and on the straight line AB, and at the points A, B on it, let the angle GAB be constructed equal to the angle at C, and the angle ABG equal to the angle CDF. [] Therefore the remaining angle CFD is equal to the angle AGB; [] therefore the triangle FCD is equiangular with the triangle GAB. Therefore, proportionally, as FD is to GB, so is FC to GA, and CD to AB. Again, on the straight line BG, and at the points B, G on it, let the angle BGH be constructed equal to the angle DFE, and the angle GBH equal to the angle FDE. [] Therefore the remaining angle at E is equal to the remaining angle at H; [] therefore the triangle FDE is equiangular with the triangle GBH; therefore, proportionally, as FD is to GB, so is FE to GH, and ED to HB. [] But it was also proved that, as FD is to GB, so is FC to GA, and CD to AB; therefore also, as FC is to AG, so is CD to AB, and FE to GH, and further ED to HB. And, since the angle CFD is equal to the angle AGB, and the angle DFE to the angle BGH, therefore the whole angle CFE is equal to the whole angle AGH. For the same reason the angle CDE is also equal to the angle ABH. And the angle at C is also equal to the angle at A, and the angle at E to the angle at H. Therefore AH is equiangular with CE; and they have the sides about their equal angles proportional; therefore the rectilineal figure AH is similar to the rectilineal figure CE. [] Therefore on the given straight line AB the rectilineal figure AH has been described similar and similarly situated to the given rectilineal figure CE. Q. E. F.
18. ἀπὸ τῆς δοθείσης εὐθείας τῷ δοθέντι εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι. ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν εὐθύγραμμον τὸ ΓΕ· δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τῷ ΓΕ εὐθυγράμμῳ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγράψαι. ἐπεζεύχθω ἡ ΔΖ, καὶ συνεστάτω πρὸς τῇ ΑΒ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Α, Β τῇ μὲν πρὸς τῷ Γ γωνίᾳ ἴση ἡ ὑπὸ ΗΑΒ, τῇ δὲ ὑπὸ ΓΔΖ ἴση ἡ ὑπὸ ΑΒΗ. λοιπὴ ἄρα ἡ ὑπὸ ΓΖΔ τῇ ὑπὸ ΑΗΒ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΖΓΔ τρίγωνον τῷ ΗΑΒ τριγώνῳ. ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ, καὶ ἡ ΓΔ πρὸς τὴν ΑΒ. πάλιν συνεστάτω πρὸς τῇ ΒΗ εὐθείᾳ καὶ τοῖς πρὸς αὐτῇ σημείοις τοῖς Β, Η τῇ μὲν ὑπὸ ΔΖΕ γωνίᾳ ἴση ἡ ὑπὸ ΒΗΘ, τῇ δὲ ὑπὸ ΖΔΕ ἴση ἡ ὑπὸ ΗΒΘ. λοιπὴ ἄρα ἡ πρὸς τῷ Ε λοιπῇ τῇ πρὸς τῷ Θ ἐστιν ἴση· ἰσογώνιον ἄρα ἐστὶ τὸ ΖΔΕ τρίγωνον τῷ ΗΘΒ τριγώνῳ· ἀνάλογον ἄρα ἐστὶν ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἡ ΕΔ πρὸς τὴν ΘΒ. ἐδείχθη δὲ καὶ ὡς ἡ ΖΔ πρὸς τὴν ΗΒ, οὕτως ἡ ΖΓ πρὸς τὴν ΗΑ καὶ ἡ ΓΔ πρὸς τὴν ΑΒ· καὶ ὡς ἄρα ἡ ΖΓ πρὸς τὴν ΑΗ, οὕτως ἥ τε ΓΔ πρὸς τὴν ΑΒ καὶ ἡ ΖΕ πρὸς τὴν ΗΘ καὶ ἔτι ἡ ΕΔ πρὸς τὴν ΘΒ. καὶ ἐπεὶ ἴση ἐστὶν ἡ μὲν ὑπὸ ΓΖΔ γωνία τῇ ὑπὸ ΑΗΒ, ἡ δὲ ὑπὸ ΔΖΕ τῇ ὑπὸ ΒΗΘ, ὅλη ἄρα ἡ ὑπὸ ΓΖΕ ὅλῃ τῇ ὑπὸ ΑΗΘ ἐστιν ἴση. διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΘ ἐστιν ἴση. ἔστι δὲ καὶ ἡ μὲν πρὸς τῷ Γ τῇ πρὸς τῷ Α ἴση, ἡ δὲ πρὸς τῷ Ε τῇ πρὸς τῷ Θ. ἰσογώνιον ἄρα ἐστὶ τὸ ΑΘ τῷ ΓΕ· καὶ τὰς περὶ τὰς ἴσας γωνίας αὐτῶν πλευρὰς ἀνάλογον ἔχει· ὅμοιον ἄρα ἐστὶ τὸ ΑΘ εὐθύγραμμον τῷ ΓΕ εὐθυγράμμῳ. ἀπὸ τῆς δοθείσης ἄρα εὐθείας τῆς ΑΒ τῷ δοθέντι εὐθυγράμμῳ τῷ ΓΕ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον ἀναγέγραπται τὸ ΑΘ· ὅπερ ἔδει ποιῆσαι.
19. Similar triangles are to one another in the duplicate ratio of the corresponding sides. Let ABC, DEF be similar triangles having the angle at B equal to the angle at E, and such that, as AB is to BC, so is DE to EF, so that BC corresponds to EF; [] I say that the triangle ABC has to the triangle DEF a ratio duplicate of that which BC has to EF. For let a third proportional BG be taken to BC, EF, so that, as BC is to EF, so is EF to BG; [] and let AG be joined. Since then, as AB is to BC, so is DE to EF, therefore, alternately, as AB is to DE, so is BC to EF. [] But, as BC is to EF, so is EF to BG; therefore also, as AB is to DE, so is EF to BG. [] Therefore in the triangles ABG, DEF the sides about the equal angles are reciprocally proportional. But those triangles which have one angle equal to one angle, and in which the sides about the equal angles are reciprocally proportional, are equal; [] therefore the triangle ABG is equal to the triangle DEF. Now since, as BC is to EF, so is EF to BG, and, if three straight lines be proportional, the first has to the third a ratio duplicate of that which it has to the second, [] therefore BC has to BG a ratio duplicate of that which CB has to EF. But, as CB is to BG, so is the triangle ABC to the triangle ABG; [] therefore the triangle ABC also has to the triangle ABG a ratio duplicate of that which BC has to EF. But the triangle ABG is equal to the triangle DEF; therefore the triangle ABC also has to the triangle DEF a ratio duplicate of that which BC has to EF. Therefore etc. Porism. From this it is manifest that, if three straight lines be proportional, then, as the first is to the third, so is the figure described on the first to that which is similar and similarly described on the second. Q. E. D.
19. τὰ ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. ἔστω ὅμοια τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἴσην ἔχοντα τὴν πρὸς τῷ Β γωνίαν τῇ πρὸς τῷ Ε, ὡς δὲ τὴν ΑΒ πρὸς τὴν ΒΓ, οὕτως τὴν ΔΕ πρὸς τὴν ΕΖ, ὥστε ὁμόλογον εἶναι τὴν ΒΓ τῇ ΕΖ· λέγω, ὅτι τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. εἰλήφθω γὰρ τῶν ΒΓ, ΕΖ τρίτη ἀνάλογον ἡ ΒΗ, ὥστε εἶναι ὡς τὴν ΒΓ πρὸς τὴν ΕΖ, οὕτως τὴν ΕΖ πρὸς τὴν ΒΗ· καὶ ἐπεζεύχθω ἡ ΑΗ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΒΓ, οὕτως ἡ ΔΕ πρὸς τὴν ΕΖ, ἐναλλὰξ ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν ΔΕ, οὕτως ἡ ΒΓ πρὸς τὴν ΕΖ. ἀλλʼ ὡς ἡ ΒΓ πρὸς ΕΖ, οὕτως ἐστὶν ἡ ΕΖ πρὸς ΒΗ. καὶ ὡς ἄρα ἡ ΑΒ πρὸς ΔΕ, οὕτως ἡ ΕΖ πρὸς ΒΗ· τῶν ΑΒΗ, ΔΕΖ ἄρα τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας. ὧν δὲ μίαν μιᾷ ἴσην ἐχόντων γωνίαν τριγώνων ἀντιπεπόνθασιν αἱ πλευραὶ αἱ περὶ τὰς ἴσας γωνίας, ἴσα ἐστὶν ἐκεῖνα. ἴσον ἄρα ἐστὶ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ. καὶ ἐπεί ἐστιν ὡς ἡ ΒΓ πρὸς τὴν ΕΖ, οὕτως ἡ ΕΖ πρὸς τὴν ΒΗ, ἐὰν δὲ τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἡ πρώτη πρὸς τὴν τρίτην διπλασίονα λόγον ἔχει ἤπερ πρὸς τὴν δευτέραν, ἡ ΒΓ ἄρα πρὸς τὴν ΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΓΒ πρὸς τὴν ΕΖ. ὡς δὲ ἡ ΓΒ πρὸς τὴν ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον· καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΑΒΗ διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. ἴσον δὲ τὸ ΑΒΗ τρίγωνον τῷ ΔΕΖ τριγώνῳ· καὶ τὸ ΑΒΓ ἄρα τρίγωνον πρὸς τὸ ΔΕΖ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΒΓ πρὸς τὴν ΕΖ. τὰ ἄρα ὅμοια τρίγωνα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν· [ὅπερ ἔδει δεῖξαι]. Πόρισμα ἐκ δὴ τούτου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔστιν ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον [ἐπείπερ ἐδείχθη, ὡς ἡ ΓΒ πρὸς ΒΗ, οὕτως τὸ ΑΒΓ τρίγωνον πρὸς τὸ ΑΒΗ τρίγωνον, τουτέστι τὸ ΔΕΖ ]· ὅπερ ἔδει δεῖξαι.
20. Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to the corresponding side. Let ABCDE, FGHKL be similar polygons, and let AB correspond to FG; I say that the polygons ABCDE, FGHKL are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which AB has to FG. Let BE, EC, GL, LH be joined. Now, since the polygon ABCDE is similar to the polygon FGHKL, the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. [] Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; [] so that it is also similar; [] therefore the angle ABE is equal to the angle FGL. But the whole angle ABC is also equal to the whole angle FGH because of the similarity of the polygons; therefore the remaining angle EBC is equal to the angle LGH. And, since, because of the similarity of the triangles ABE, FGL, as EB is to BA, so is LG to GF, and moreover also, because of the similarity of the polygons, as AB is to BC, so is FG to GH, therefore, ex aequali, as EB is to BC, so is LG to GH; [] that is, the sides about the equal angles EBC, LGH are proportional; therefore the triangle EBC is equiangular with the triangle LGH, [] so that the triangle EBC is also similar to the triangle LGH. [] For the same reason the triangle ECD is also similar to the triangle LHK. Therefore the similar polygons ABCDE, FGHKL have been divided into similar triangles, and into triangles equal in multitude. I say that they are also in the same ratio as the wholes, that is, in such manner that the triangles are proportional, and ABE, EBC, ECD are antecedents, while FGL, LGH, LHK are their consequents, and that the polygon ABCDE has to the polygon FGHKL a ratio duplicate of that which the corresponding side has to the corresponding side, that is AB to FG. For let AC, FH be joined. Then since, because of the similarity of the polygons, the angle ABC is equal to the angle FGH, and, as AB is to BC, so is FG to GH, the triangle ABC is equiangular with the triangle FGH; [] therefore the angle BAC is equal to the angle GFH, and the angle BCA to the angle GHF. And, since the angle BAM is equal to the angle GFN, and the angle ABM is also equal to the angle FGN, therefore the remaining angle AMB is also equal to the remaining angle FNG; [] therefore the triangle ABM is equiangular with the triangle FGN. Similarly we can prove that the triangle BMC is also equiangular with the triangle GNH. Therefore, proportionally, as AM is to MB, so is FN to NG, and, as BM is to MC, so is GN to NH; so that, in addition, ex aequali, as AM is to MC, so is FN to NH. But, as AM is to MC, so is the triangle ABM to MBC, and AME to EMC; for they are to one another as their bases. [] Therefore also, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; [] therefore, as the triangle AMB is to BMC, so is ABE to CBE. But, as AMB is to BMC, so is AM to MC; therefore also, as AM is to MC, so is the triangle ABE to the triangle EBC. For the same reason also, as FN is to NH, so is the triangle FGL to the triangle GLH. And, as AM is to MC, so is FN to NH; therefore also, as the triangle ABE is to the triangle BEC, so is the triangle FGL to the triangle GLH; and, alternately, as the triangle ABE is to the triangle FGL, so is the triangle BEC to the triangle GLH. Similarly we can prove, if BD, GK be joined, that, as the triangle BEC is to the triangle LGH, so also is the triangle ECD to the triangle LHK. And since, as the triangle ABE is to the triangle FGL, so is EBC to LGH, and further ECD to LHK, therefore also, as one of the antecedents is to one of the consequents so are all the antecedents to all the consequents; [therefore, as the triangle ABE is to the triangle FGL, so is the polygon ABCDE to the polygon FGHKL. But the triangle ABE has to the triangle FGL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG; for similar triangles are in the duplicate ratio of the corresponding sides. [] Therefore the polygon ABCDE also has to the polygon FGHKL a ratio duplicate of that which the corresponding side AB has to the corresponding side FG. Therefore etc. Porism. Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides. Q. E. D.
20. τὰ ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν. ἔστω ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ, ὁμόλογος δὲ ἔστω ἡ ΑΒ τῇ ΖΗ· λέγω, ὅτι τὰ ΑΒΓΔΕ, ΖΗΘΚΛ πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν αἱ ΒΕ, ΕΓ, ΗΛ, ΛΘ. καὶ ἐπεὶ ὅμοιόν ἐστι τὸ ΑΒΓΔΕ πολύγωνον τῷ ΖΗΘΚΛ πολυγώνῳ, ἴση ἐστὶν ἡ ὑπὸ ΒΑΕ γωνία τῇ ὑπὸ ΗΖΛ. καί ἐστιν ὡς ἡ ΒΑ πρὸς ΑΕ, οὕτως ἡ ΗΖ πρὸς ΖΛ. ἐπεὶ οὖν δύο τρίγωνά ἐστι τὰ ΑΒΕ, ΖΗΛ μίαν γωνίαν μιᾷ γωνίᾳ ἴσην ἔχοντα, περὶ δὲ τὰς ἴσας γωνίας τὰς πλευρὰς ἀνάλογον, ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΕ τρίγωνον τῷ ΖΗΛ τριγώνῳ· ὥστε καὶ ὅμοιον· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΒΕ γωνία τῇ ὑπὸ ΖΗΛ. ἔστι δὲ καὶ ὅλη ἡ ὑπὸ ΑΒΓ ὅλῃ τῇ ὑπὸ ΖΗΘ ἴση διὰ τὴν ὁμοιότητα τῶν πολυγώνων· λοιπὴ ἄρα ἡ ὑπὸ ΕΒΓ γωνία τῇ ὑπὸ ΛΗΘ ἐστιν ἴση. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν ΑΒΕ, ΖΗΛ τριγώνων ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΑ, οὕτως ἡ ΛΗ πρὸς ΗΖ, ἀλλὰ μὴν καὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἐστὶν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, διʼ ἴσου ἄρα ἐστὶν ὡς ἡ ΕΒ πρὸς ΒΓ, οὕτως ἡ ΛΗ πρὸς ΗΘ, καὶ περὶ τὰς ἴσας γωνίας τὰς ὑπὸ ΕΒΓ, ΛΗΘ αἱ πλευραὶ ἀνάλογόν εἰσιν· ἰσογώνιον ἄρα ἐστὶ τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ· ὥστε καὶ ὅμοιόν ἐστι τὸ ΕΒΓ τρίγωνον τῷ ΛΗΘ τριγώνῳ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΓΔ τρίγωνον ὅμοιόν ἐστι τῷ ΛΘΚ τριγώνῳ. τὰ ἄρα ὅμοια πολύγωνα τὰ ΑΒΓΔΕ, ΖΗΘΚΛ εἴς τε ὅμοια τρίγωνα διῄρηται καὶ εἰς ἴσα τὸ πλῆθος. λέγω, ὅτι καὶ ὁμόλογα τοῖς ὅλοις, τουτέστιν ὥστε ἀνάλογον εἶναι τὰ τρίγωνα, καὶ ἡγούμενα μὲν εἶναι τὰ ΑΒΕ, ΕΒΓ, ΕΓΔ, ἑπόμενα δὲ αὐτῶν τὰ ΖΗΛ, ΛΗΘ, ΛΘΚ, καὶ ὅτι τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ. ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΖΘ. καὶ ἐπεὶ διὰ τὴν ὁμοιότητα τῶν πολυγώνων ἴση ἐστὶν ἡ ὑπὸ ΑΒΓ γωνία τῇ ὑπὸ ΖΗΘ, καί ἐστιν ὡς ἡ ΑΒ πρὸς ΒΓ, οὕτως ἡ ΖΗ πρὸς ΗΘ, ἰσογώνιόν ἐστι τὸ ΑΒΓ τρίγωνον τῷ ΖΗΘ τριγώνῳ· ἴση ἄρα ἐστὶν ἡ μὲν ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΗΖΘ, ἡ δὲ ὑπὸ ΒΓΑ τῇ ὑπὸ ΗΘΖ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΑΜ γωνία τῇ ὑπὸ ΗΖΝ, ἔστι δὲ καὶ ἡ ὑπὸ ΑΒΜ τῇ ὑπὸ ΖΗΝ ἴση, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΑΜΒ λοιπῇ τῇ ὑπὸ ΖΝΗ ἴση ἐστίν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΜ τρίγωνον τῷ ΖΗΝ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ ΒΜΓ τρίγωνον ἰσογώνιόν ἐστι τῷ ΗΝΘ τριγώνῳ. ἀνάλογον ἄρα ἐστίν, ὡς μὲν ἡ ΑΜ πρὸς ΜΒ, οὕτως ἡ ΖΝ πρὸς ΝΗ, ὡς δὲ ἡ ΒΜ πρὸς ΜΓ, οὕτως ἡ ΗΝ πρὸς ΝΘ· ὥστε καὶ διʼ ἴσου, ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ. ἀλλʼ ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΜ [τρίγωνον] πρὸς τὸ ΜΒΓ, καὶ τὸ ΑΜΕ πρὸς τὸ ΕΜΓ· πρὸς ἄλληλα γάρ εἰσιν ὡς αἱ βάσεις. καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ὡς ἄρα τὸ ΑΜΒ τρίγωνον πρὸς τὸ ΒΜΓ, οὕτως τὸ ΑΒΕ πρὸς τὸ ΓΒΕ. ἀλλʼ ὡς τὸ ΑΜΒ πρὸς τὸ ΒΜΓ, οὕτως ἡ ΑΜ πρὸς ΜΓ· καὶ ὡς ἄρα ἡ ΑΜ πρὸς ΜΓ, οὕτως τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΕΒΓ τρίγωνον. διὰ τὰ αὐτὰ δὴ καὶ ὡς ἡ ΖΝ πρὸς ΝΘ, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. καί ἐστιν ὡς ἡ ΑΜ πρὸς ΜΓ, οὕτως ἡ ΖΝ πρὸς ΝΘ· καὶ ὡς ἄρα τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΒΕΓ τρίγωνον, οὕτως τὸ ΖΗΛ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον, καὶ ἐναλλὰξ, ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΗΛΘ τρίγωνον. ὁμοίως δὴ δείξομεν ἐπιζευχθεισῶν τῶν ΒΔ, ΗΚ, ὅτι καὶ ὡς τὸ ΒΕΓ τρίγωνον πρὸς τὸ ΛΗΘ τρίγωνον, οὕτως τὸ ΕΓΔ τρίγωνον πρὸς τὸ ΛΘΚ τρίγωνον. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΕΒΓ πρὸς τὸ ΛΗΘ, καὶ ἔτι τὸ ΕΓΔ πρὸς τὸ ΛΘΚ, καὶ ὡς ἄρα ἓν τῶν ἡγουμένων πρὸς ἓν τῶν ἑπομένων, οὕτως ἅπαντα τὰ ἡγούμενα πρὸς ἅπαντα τὰ ἑπόμενα· ἔστιν ἄρα ὡς τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον, οὕτως τὸ ΑΒΓΔΕ πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον. ἀλλὰ τὸ ΑΒΕ τρίγωνον πρὸς τὸ ΖΗΛ τρίγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν· τὰ γὰρ ὅμοια τρίγωνα ἐν διπλασίονι λόγῳ ἐστὶ τῶν ὁμολόγων πλευρῶν. καὶ τὸ ΑΒΓΔΕ ἄρα πολύγωνον πρὸς τὸ ΖΗΘΚΛ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ ὁμόλογος πλευρὰ πρὸς τὴν ΖΗ ὁμόλογον πλευράν. τὰ ἄρα ὅμοια πολύγωνα εἴς τε ὅμοια τρίγωνα διαιρεῖται καὶ εἰς ἴσα τὸ πλῆθος καὶ ὁμόλογα τοῖς ὅλοις, καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον διπλασίονα λόγον ἔχει ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν· [ὅπερ ἔδει δεῖξαι]. Πόρισμα ὡσαύτως δὲ καὶ ἐπὶ τῶν [ὁμοίων] τετραπλεύρων δειχθήσεται, ὅτι ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ἐδείχθη δὲ καὶ ἐπὶ τῶν τριγώνων· ὥστε καὶ καθόλου τὰ ὅμοια εὐθύγραμμα σχήματα πρὸς ἄλληλα ἐν διπλασίονι λόγῳ εἰσὶ τῶν ὁμολόγων πλευρῶν. ὅπερ ἔδει δεῖξαι. [πόρισμα β# ] [καὶ ἐὰν τῶν ΑΒ, ΖΗ τρίτην ἀνάλογον λάβωμεν τὴν Ξ, ἡ ΒΑ πρὸς τὴν Ξ διπλασίονα λόγον ἔχει ἤπερ ἡ ΑΒ πρὸς τὴν ΖΗ. ἔχει δὲ καὶ τὸ πολύγωνον πρὸς τὸ πολύγωνον ἢ τὸ τετράπλευρον πρὸς τὸ τετράπλευρον διπλασίονα λόγον ἤπερ ἡ ὁμόλογος πλευρὰ πρὸς τὴν ὁμόλογον πλευράν, τουτέστιν ἡ ΑΒ πρὸς τὴν ΖΗ· ἐδείχθη δὲ τοῦτο καὶ ἐπὶ τῶν τριγώνων· ὥστε καὶ καθόλου φανερόν, ὅτι, ἐὰν τρεῖς εὐθεῖαι ἀνάλογον ὦσιν, ἔσται ὡς ἡ πρώτη πρὸς τὴν τρίτην, οὕτως τὸ ἀπὸ τῆς πρώτης εἶδος πρὸς τὸ ἀπὸ τῆς δευτέρας τὸ ὅμοιον καὶ ὁμοίως ἀναγραφόμενον.]
21. Figures which are similar to the same rectilineal figure are also similar to one another. For let each of the rectilineal figures A, B be similar to C; I say that A is also similar to B. For, since A is similar to C, it is equiangular with it and has the sides about the equal angles proportional. [] Again, since B is similar to C, it is equiangular with it and has the sides about the equal angles proportional. Therefore each of the figures A, B is equiangular with C and with C has the sides about the equal angles proportional; therefore A is similar to B. Q. E. D.
21. τὰ τῷ αὐτῷ εὐθυγράμμῳ ὅμοια καὶ ἀλλήλοις ἐστὶν ὅμοια. ἔστω γὰρ ἑκάτερον τῶν Α, Β εὐθυγράμμων τῷ Γ ὅμοιον· λέγω, ὅτι καὶ τὸ Α τῷ Β ἐστιν ὅμοιον. ἐπεὶ γὰρ ὅμοιόν ἐστι τὸ Α τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. πάλιν, ἐπεὶ ὅμοιόν ἐστι τὸ Β τῷ Γ, ἰσογώνιόν τέ ἐστιν αὐτῷ καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ἑκάτερον ἄρα τῶν Α, Β τῷ Γ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει [ὥστε καὶ τὸ Α τῷ Β ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει]. ὅμοιον ἄρα ἐστὶ τὸ Α τῷ Β· ὅπερ ἔδει δεῖξαι.
22. If four straight lines be proportional, the rectilineal figures similar and similarly described upon them will also be proportional; and, if the rectilineal figures similar and similarly described upon them be proportional, the straight lines will themselves also be proportional. Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH. For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [] Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P. [] But, as AB is to O, so is KAB to LCD, [] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [] Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH. For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH. [] Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR. But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [] Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR. [] But it is also similar and similarly situated to it; therefore GH is equal to QR. And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH. Therefore etc. Q. E. D.
22. ἐὰν τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπʼ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται· κἂν τὰ ἀπʼ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται. ἔστωσαν τέσσαρες εὐθεῖαι ἀνάλογον αἱ ΑΒ, ΓΔ, ΕΖ, ΗΘ, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, καὶ ἀναγεγράφθωσαν ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΗΘ ὅμοιά τε καὶ ὁμοίως κείμενα εὐθύγραμμα τὰ ΜΖ, ΝΘ· λέγω, ὅτι ἐστὶν ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. εἰλήφθω γὰρ τῶν μὲν ΑΒ, ΓΔ τρίτη ἀνάλογον ἡ Ξ, τῶν δὲ ΕΖ, ΗΘ τρίτη ἀνάλογον ἡ Ο. καὶ ἐπεί ἐστιν ὡς μὲν ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ὡς δὲ ἡ ΓΔ πρὸς τὴν Ξ, οὕτως ἡ ΗΘ πρὸς τὴν Ο, διʼ ἴσου ἄρα ἐστὶν ὡς ἡ ΑΒ πρὸς τὴν Ξ, οὕτως ἡ ΕΖ πρὸς τὴν Ο. ἀλλʼ ὡς μὲν ἡ ΑΒ πρὸς τὴν Ξ, οὕτως [καὶ] τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, ὡς δὲ ἡ ΕΖ πρὸς τὴν Ο, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ· καὶ ὡς ἄρα τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. ἀλλὰ δὴ ἔστω ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ· λέγω, ὅτι ἐστὶ καὶ ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. εἰ γὰρ μή ἐστιν, ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ, ἔστω ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγεγράφθω ἀπὸ τῆς ΠΡ ὁποτέρῳ τῶν ΜΖ, ΝΘ ὅμοιόν τε καὶ ὁμοίως κείμενον εὐθύγραμμον τὸ ΣΡ. ἐπεὶ οὖν ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, καὶ ἀναγέγραπται ἀπὸ μὲν τῶν ΑΒ, ΓΔ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΚΑΒ, ΛΓΔ, ἀπὸ δὲ τῶν ΕΖ, ΠΡ ὅμοιά τε καὶ ὁμοίως κείμενα τὰ ΜΖ, ΣΡ, ἔστιν ἄρα ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΣΡ. ὑπόκειται δὲ καὶ ὡς τὸ ΚΑΒ πρὸς τὸ ΛΓΔ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ· καὶ ὡς ἄρα τὸ ΜΖ πρὸς τὸ ΣΡ, οὕτως τὸ ΜΖ πρὸς τὸ ΝΘ. τὸ ΜΖ ἄρα πρὸς ἑκάτερον τῶν ΝΘ, ΣΡ τὸν αὐτὸν ἔχει λόγον· ἴσον ἄρα ἐστὶ τὸ ΝΘ τῷ ΣΡ. ἔστι δὲ αὐτῷ καὶ ὅμοιον καὶ ὁμοίως κείμενον· ἴση ἄρα ἡ ΗΘ τῇ ΠΡ. καὶ ἐπεί ἐστιν ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΠΡ, ἴση δὲ ἡ ΠΡ τῇ ΗΘ, ἔστιν ἄρα ὡς ἡ ΑΒ πρὸς τὴν ΓΔ, οὕτως ἡ ΕΖ πρὸς τὴν ΗΘ. ἐὰν ἄρα τέσσαρες εὐθεῖαι ἀνάλογον ὦσιν, καὶ τὰ ἀπʼ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ἔσται· κἂν τὰ ἀπʼ αὐτῶν εὐθύγραμμα ὅμοιά τε καὶ ὁμοίως ἀναγεγραμμένα ἀνάλογον ᾖ, καὶ αὐταὶ αἱ εὐθεῖαι ἀνάλογον ἔσονται· ὅπερ ἔδει δεῖξαι. [λῆμμα] [ὅτι δέ, ἐὰν εὐθύγραμμα ἴσα ᾖ καὶ ὅμοια, αἱ ὁμόλογοι αὐτῶν πλευραὶ ἴσαι ἀλλήλαις εἰσίν, δείξομεν οὕτως.] [ἔστω ἴσα καὶ ὅμοια εὐθύγραμμα τὰ ΝΘ, ΣΡ, καὶ ἔστω ὡς ἡ ΘΗ πρὸς τὴν ΗΝ, οὕτως ἡ ΡΠ πρὸς τὴν ΠΣ· λέγω, ὅτι ἴση ἐστὶν ἡ ΡΠ τῇ ΘΗ.] [εἰ γὰρ ἄνισοί εἰσιν, μία αὐτῶν μείζων ἐστίν. ἔστω μείζων ἡ ΡΠ τῆς ΘΗ. καὶ ἐπεί ἐστιν ὡς ἡ ΡΠ πρὸς ΠΣ, οὕτως ἡ ΘΗ πρὸς τὴν ΗΝ, καὶ ἐναλλάξ, ὡς ἡ ΡΠ πρὸς τὴν ΘΗ, οὕτως ἡ ΠΣ πρὸς τὴν ΗΝ, μείζων δὲ ἡ ΠΡ τῆς ΘΗ, μείζων ἄρα καὶ ἡ ΠΣ τῆς ΗΝ· ὥστε καὶ τὸ ΡΣ μεῖζόν ἐστι τοῦ ΘΝ. ἀλλὰ καὶ ἴσον· ὅπερ ἀδύνατον. οὐκ ἄρα ἄνισός ἐστιν ἡ ΠΡ τῇ ΗΘ· ἴση ἄρα· ὅπερ ἔδει δεῖξαι.]