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Euclid · Elements §4.prop.13-4.prop.14

Inscribing and Circumscribing a Circle in and About a Regular Pentagon

Passage 66 of 316 · Greek

Summary

This section demonstrates how to inscribe a circle within a given equilateral and equiangular pentagon (Proposition 13) and how to circumscribe a circle about it (Proposition 14).

§4.prop.13εἰς τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον ἐγγράψαι.
To inscribe a circle in the given pentagon, which is equilateral and equiangular.
ἔστω τὸ δοθὲν πεντάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον τὸ ΑΒΓΔΕ· δεῖ δὴ εἰς τὸ ΑΒΓΔΕ πεντάγωνον κύκλον ἐγγράψαι.
Let the given equilateral and equiangular pentagon be ABCDE; it is required then to inscribe a circle in the pentagon ABCDE.
τετμήσθω γὰρ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ εὐθειῶν· καὶ ἀπὸ τοῦ Ζ σημείου, καθʼ ὃ συμβάλλουσιν ἀλλήλαις αἱ ΓΖ, ΔΖ εὐθεῖαι, ἐπεζεύχθωσαν αἱ ΖΒ, ΖΑ, ΖΕ εὐθεῖαι.
For let each of the angles BCD, CDE be bisected by each of the straight lines CF, DF; and from the point F, at which the straight lines CF, DF meet one another, let the straight lines FB, FA, FE be joined.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΓΔ, κοινὴ δὲ ἡ ΓΖ, δύο δὴ αἱ ΒΓ, ΓΖ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν· καὶ γωνία ἡ ὑπὸ ΒΓΖ γωνίᾳ τῇ ὑπὸ ΔΓΖ ἴση· βάσις ἄρα ἡ ΒΖ βάσει τῇ ΔΖ ἐστιν ἴση, καὶ τὸ ΒΓΖ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἐστιν ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἡ ὑπὸ ΓΒΖ γωνία τῇ ὑπὸ ΓΔΖ. καὶ ἐπεὶ διπλῆ ἐστιν ἡ ὑπὸ ΓΔΕ τῆς ὑπὸ ΓΔΖ, ἴση δὲ ἡ μὲν ὑπὸ ΓΔΕ τῇ ὑπὸ ΑΒΓ, ἡ δὲ ὑπὸ ΓΔΖ τῇ ὑπὸ ΓΒΖ, καὶ ἡ ὑπὸ ΓΒΑ ἄρα τῆς ὑπὸ ΓΒΖ ἐστι διπλῆ·
And since BC is equal to CD, and CF is common, therefore the two straight lines BC, CF are equal to the two straight lines DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base DF, and the triangle BCF is equal to the triangle DCF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; therefore the angle CBF is equal to the angle CDF.
ἴση ἄρα ἡ ὑπὸ ΑΒΖ γωνία τῇ ὑπὸ ΖΒΓ· ἡ ἄρα ὑπὸ ΑΒΓ γωνία δίχα τέτμηται ὑπὸ τῆς ΒΖ εὐθείας.
And since the angle CDE is double of the angle CDF, and the angle CDE is equal to the angle ABC, and the angle CDF to the angle CBF, therefore the angle CBA is also double of the angle CBF; therefore the angle ABF is equal to the angle FBC; therefore the angle ABC is bisected by the straight line BF.
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκατέρα τῶν ὑπὸ ΒΑΕ, ΑΕΔ δίχα τέτμηται ὑπὸ ἑκατέρας τῶν ΖΑ, ΖΕ εὐθειῶν.
Similarly it will be proved that each of the angles BAE, AED is also bisected by each of the straight lines FA, FE.
ἤχθωσαν δὴ ἀπὸ τοῦ Ζ σημείου ἐπὶ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας κάθετοι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΘΓΖ γωνία τῇ ὑπὸ ΚΓΖ, ἐστὶ δὲ καὶ ὀρθὴ ἡ ὑπὸ ΖΘΓ τῇ ὑπὸ ΖΚΓ ἴση, δύο δὴ τρίγωνά ἐστι τὰ ΖΘΓ, ΖΚΓ τὰς δύο γωνίας δυσὶ γωνίαις ἴσας ἔχοντα καὶ μίαν πλευρὰν μιᾷ πλευρᾷ ἴσην κοινὴν αὐτῶν τὴν ΖΓ ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν·
Let then the perpendiculars FH, FG, FK, FL, FM be drawn from the point F to the straight lines AB, BC, CD, DE, EA.
καὶ τὰς λοιπὰς ἄρα πλευρὰς ταῖς λοιπαῖς πλευραῖς ἴσας ἕξει· ἴση ἄρα ἡ ΖΘ κάθετος τῇ ΖΚ καθέτῳ.
And since the angle GCF is equal to the angle KCF, and the right angle FGC is also equal to the right angle FKC, there are two triangles FGC, FKC having two angles equal to two angles and one side equal to one side, namely their common side FC subtending one of the equal angles; therefore they will also have the remaining sides equal to the remaining sides; therefore the perpendicular FG is equal to the perpendicular FK.
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΛ, ΖΜ, ΖΗ ἑκατέρᾳ τῶν ΖΘ, ΖΚ ἴση ἐστίν· αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΗ, ΖΘ, ΖΚ, ΖΛ, ΖΜ ἴσαι ἀλλήλαις εἰσίν.
Similarly it will be proved that each of FL, FM, FH is also equal to each of FG, FK; therefore the five straight lines FH, FG, FK, FL, FM are equal to one another.
ὁ ἄρα κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν Η, Θ, Κ, Λ, Μ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Η, Θ, Κ, Λ, Μ σημείοις γωνίας.
Therefore the circle described with center F and distance one of the points H, G, K, L, M will pass also through the remaining points, and will touch the straight lines AB, BC, CD, DE, EA, because the angles at the points H, G, K, L, M are right.
εἰ γὰρ οὐκ ἐφάψεται αὐτῶν, ἀλλὰ τεμεῖ αὐτάς, συμβήσεται τὴν τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπʼ ἄκρας ἀγομένην ἐντὸς πίπτειν τοῦ κύκλου· ὅπερ ἄτοπον ἐδείχθη.
For if it does not touch them, but cuts them, it will result that the straight line drawn at right angles to the diameter of the circle from its extremity falls within the circle; which was proved absurd.
οὐκ ἄρα ὁ κέντρῳ τῷ Ζ διαστήματι δὲ ἑνὶ τῶν η, Θ, Κ, Λ, Μ σημείων γραφόμενος κύκλος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΑ εὐθείας· ἐφάψεται ἄρα αὐτῶν.
Therefore the circle described with center F and distance one of the points H, G, K, L, M will not cut the straight lines AB, BC, CD, DE, EA; therefore it will touch them.
γεγράφθω ὡς ὁ ΗΘΚΛΜ. εἰς ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος ἐγγέγραπται·
Let it be described as GHKLM.
ὅπερ ἔδει ποιῆσαι.
Therefore in the given pentagon, which is equilateral and equiangular, a circle has been inscribed; which was required to do.
§4.prop.14περὶ τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλον περιγράψαι.
To circumscribe a circle about the given pentagon, which is equilateral and equiangular.
ἔστω τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, τὸ ΑΒΓΔΕ· δεῖ δὴ περὶ τὸ ΑΒΓΔΕ πεντάγωνον κύκλον περιγράψαι.
Let the given pentagon, which is equilateral and equiangular, be ABCDE; it is required then to circumscribe a circle about the pentagon ABCDE.
τετμήσθω δὴ ἑκατέρα τῶν ὑπὸ ΒΓΔ, ΓΔΕ γωνιῶν δίχα ὑπὸ ἑκατέρας τῶν ΓΖ, ΔΖ, καὶ ἀπὸ τοῦ Ζ σημείου, καθʼ ὃ συμβάλλουσιν αἱ εὐθεῖαι, ἐπὶ τὰ Β, Α, Ε σημεῖα ἐπεζεύχθωσαν εὐθεῖαι αἱ ΖΒ, ΖΑ, ΖΕ. ὁμοίως δὴ τῷ πρὸ τούτου δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΓΒΑ, ΒΑΕ, ΑΕΔ γωνιῶν δίχα τέτμηται ὑπὸ ἑκάστης τῶν ΖΒ, ΖΑ, ΖΕ εὐθειῶν.
For let each of the angles BCD, CDE be bisected by each of the straight lines CF, DF, and from the point F, at which the straight lines meet, let the straight lines FB, FA, FE be joined to the points B, A, E. Similarly to that before this, it will be proved that each of the angles CBA, BAE, AED is also bisected by each of the straight lines FB, FA, FE.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΒΓΔ γωνία τῇ ὑπὸ ΓΔΕ, καί ἐστι τῆς μὲν ὑπὸ ΒΓΔ ἡμίσεια ἡ ὑπὸ ΖΓΔ, τῆς δὲ ὑπὸ ΓΔΕ ἡμίσεια ἡ ὑπὸ ΓΔΖ, καὶ ἡ ὑπὸ ΖΓΔ ἄρα τῇ ὑπὸ ΖΔΓ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση.
And since the angle BCD is equal to the angle CDE, and the angle FCD is half of the angle BCD, and the angle FDC is half of the angle CDE, therefore the angle FCD is also equal to the angle FDC; so that the side FC is also equal to the side FD.
ὁμοίως δὴ δειχθήσεται, ὅτι καὶ ἑκάστη τῶν ΖΒ, ΖΑ, ΖΕ ἑκατέρᾳ τῶν ΖΓ, ΖΔ ἐστιν ἴση· αἱ πέντε ἄρα εὐθεῖαι αἱ ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ ἴσαι ἀλλήλαις εἰσίν.
Similarly it will be proved that each of FB, FA, FE is also equal to each of FC, FD; therefore the five straight lines FA, FB, FC, FD, FE are equal to one another.
ὁ ἄρα κέντρῳ τῷ Ζ καὶ διαστήματι ἑνὶ τῶν ΖΑ, ΖΒ, ΖΓ, ΖΔ, ΖΕ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος.
Therefore the circle described with center F and distance one of the straight lines FA, FB, FC, FD, FE will pass also through the remaining points and will be circumscribed.
περιγεγράφθω καὶ ἔστω ὁ ΑΒΓΔΕ. περὶ ἄρα τὸ δοθὲν πεντάγωνον, ὅ ἐστιν ἰσόπλευρόν τε καὶ ἰσογώνιον, κύκλος περιγέγραπται· ὅπερ ἔδει ποιῆσαι.
Let it be circumscribed, and let it be ABCDE. Therefore about the given pentagon, which is equilateral and equiangular, a circle has been circumscribed; which was required to do.

Notes

  1. 4.prop.13ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν — The antecedent of the relative pronoun ἃς is the preceding αἱ λοιπαὶ γωνίαι (the remaining angles). The preposition ὑπό with the accusative means 'to subtend' in a geometric context, indicating the relationship between equal sides and the angles they face.
  2. 4.prop.13ὑποτείνουσαν ὑπὸ μίαν τῶν ἴσων γωνιῶν — The feminine accusative singular participle ὑποτείνουσαν modifies τὴν ΖΓ (accusative) in the preceding clause. This attributive participle describes the side FC as subtending one of the equal angles.
  3. 4.prop.13διαστήματι δὲ ἑνὶ τῶν η, Θ, Κ, Λ, Μ σημείων — The lowercase η in the text is equivalent to the uppercase Η (the point H). The particles and word order place δὲ inside the dative phrase (διαστήματι ἑνὶ), which functions as a dative of means or measure, representing the equal distance to the points.
  4. 4.prop.14ὥστε καὶ πλευρὰ ἡ ΖΓ πλευρᾷ τῇ ΖΔ ἐστιν ἴση — This conclusion is based on Elements Book 1, Proposition 6 (if a triangle has two angles equal, the sides subtending them are also equal). The consecutive conjunction ὥστε introduces the logical consequence that side FC is equal to side FD because their opposite angles are equal.

Cite this passage

Euclid, Elements §4.prop.13-4.prop.14. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:4.prop.13-4.prop.14

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