Humanitext Reader

Euclid · Elements §13.prop.7

Equiangularity of Equilateral Pentagon with Three Equal Angles

Passage 299 of 316 · Greek

Summary

This proposition proves that if three angles (either consecutive or not) of an equilateral pentagon are equal, the pentagon is equiangular, using the congruence of triangles formed by the intersection of diagonals.

§13.prop.7ἐὰν πενταγώνου ἰσοπλεύρου αἱ τρεῖς γωνίαι ἤτοι αἱ κατὰ τὸ ἑξῆς ἢ αἱ μὴ κατὰ τὸ ἑξῆς ἴσαι ὦσιν, ἰσογώνιον ἔσται τὸ πεντάγωνον.
If three angles of an equilateral pentagon, either consecutive or not consecutive, be equal, the pentagon will be equiangular.
πενταγώνου γὰρ ἰσοπλεύρου τοῦ ΑΒΓΔΕ αἱ τρεῖς γωνίαι πρότερον αἱ κατὰ τὸ ἑξῆς αἱ πρὸς τοῖς Α, Β, Γ ἴσαι ἀλλήλαις ἔστωσαν· λέγω, ὅτι ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον.
For let the three consecutive angles of the equilateral pentagon ABCDE, namely those at A, B, C, be equal to one another; I say that the pentagon ABCDE is equiangular.
ἐπεζεύχθωσαν γὰρ αἱ ΑΓ, ΒΕ, ΖΔ. καὶ ἐπεὶ δύο αἱ ΓΒ, ΒΑ δυσὶ ταῖς ΒΑ, ΑΕ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΓΒΑ γωνίᾳ τῇ ὑπὸ ΒΑΕ ἐστιν ἴση, βάσις ἄρα ἡ ΑΓ βάσει τῇ ΒΕ ἐστιν ἴση, καὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΕ τριγώνῳ ἴσον, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν, ἡ μὲν ὑπὸ ΒΓΑ τῇ ὑπὸ ΒΕΑ, ἡ δὲ ὑπὸ ΑΒΕ τῇ ὑπὸ ΓΑΒ· ὥστε καὶ πλευρὰ ἡ ΑΖ πλευρᾷ τῇ ΒΖ ἐστιν ἴση.
For let AC, BE, ZD be joined. And since the two straight lines CB, BA are equal to the two BA, AE, each to each, and the angle CBA is equal to the angle BAE, therefore the base AC is equal to the base BE, the triangle ABC is equal to the triangle ABE, and the remaining angles will be equal to the remaining angles, namely those subtended by the equal sides, that is, BCA to BEA, and ABE to CAB; hence the side AZ is also equal to the side BZ.
ἐδείχθη δὲ καὶ ὅλη ἡ ΑΓ ὅλῃ τῇ ΒΕ ἴση· καὶ λοιπὴ ἄρα ἡ ΖΓ λοιπῇ τῇ ΖΕ ἐστιν ἴση.
But the whole AC was also shown to be equal to the whole BE; therefore the remainder ZC is also equal to the remainder ZE.
ἔστι δὲ καὶ ἡ ΓΔ τῇ ΔΕ ἴση.
But CD is also equal to DE.
δύο δὴ αἱ ΖΓ, ΓΔ δυσὶ ταῖς ΖΕ, ΕΔ ἴσαι εἰσίν· καὶ βάσις αὐτῶν κοινὴ ἡ ΖΔ· γωνία ἄρα ἡ ὑπὸ ΖΓΔ γωνίᾳ τῇ ὑπὸ ΖΕΔ ἐστιν ἴση.
Therefore the two ZC, CD are equal to the two ZE, ED; and their base ZD is common; therefore the angle ZCD is equal to the angle ZED.
ἐδείχθη δὲ καὶ ἡ ὑπὸ ΒΓΑ τῇ ὑπὸ ΑΕΒ ἴση· καὶ ὅλη ἄρα ἡ ὑπὸ ΒΓΔ ὅλῃ τῇ ὑπὸ ΑΕΔ ἴση.
And the angle BCA was also shown to be equal to the angle AEB; therefore the whole angle BCD is also equal to the whole angle AED.
ἀλλʼ ἡ ὑπὸ ΒΓΔ ἴση ὑπόκειται ταῖς πρὸς τοῖς Α, Β γωνίαις· καὶ ἡ ὑπὸ ΑΕΔ ἄρα ταῖς πρὸς τοῖς Α, Β γωνίαις ἴση ἐστίν.
But the angle BCD is by hypothesis equal to the angles at A, B; therefore the angle AED is also equal to the angles at A, B.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἡ ὑπὸ ΓΔΕ γωνία ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Β, Γ γωνίαις· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον.
Similarly we shall show that the angle CDE is also equal to the angles at A, B, C; therefore the pentagon ABCDE is equiangular.
ἀλλὰ δὴ μὴ ἔστωσαν ἴσαι αἱ κατὰ τὸ ἑξῆς γωνίαι, ἀλλʼ ἔστωσαν ἴσαι αἱ πρὸς τοῖς Α, Γ, Δ σημείοις· λέγω, ὅτι καὶ οὕτως ἰσογώνιόν ἐστι τὸ ΑΒΓΔΕ πεντάγωνον.
But now let the consecutive angles not be equal, but let those at the points A, C, D be equal; I say that in this case too the pentagon ABCDE is equiangular.
ἐπεζεύχθω γὰρ ἡ ΒΔ. καὶ ἐπεὶ δύο αἱ ΒΑ, ΑΕ δυσὶ ταῖς ΒΓ, ΓΔ ἴσαι εἰσὶ καὶ γωνίας ἴσας περιέχουσιν, βάσις ἄρα ἡ ΒΕ βάσει τῇ ΒΔ ἴση ἐστίν, καὶ τὸ ΑΒΕ τρίγωνον τῷ ΒΓΔ τριγώνῳ ἴσον ἐστίν, καὶ αἱ λοιπαὶ γωνίαι ταῖς λοιπαῖς γωνίαις ἴσαι ἔσονται, ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν· ἴση ἄρα ἐστὶν ἡ ὑπὸ ΑΕΒ γωνία τῇ ὑπὸ ΓΔΒ. ἔστι δὲ καὶ ἡ ὑπὸ ΒΕΔ γωνία τῇ ὑπὸ ΒΔΕ ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΕ πλευρᾷ τῇ ΒΔ ἐστιν ἴση.
For let BD be joined. And since the two BA, AE are equal to the two BC, CD, and contain equal angles, therefore the base BE is equal to the base BD, the triangle ABE is equal to the triangle BCD, and the remaining angles will be equal to the remaining angles, namely those subtended by the equal sides; therefore the angle AEB is equal to the angle CDB. And the angle BED is also equal to the angle BDE, since the side BE is also equal to the side BD.
καὶ ὅλη ἄρα ἡ ὑπὸ ΑΕΔ γωνία ὅλῃ τῇ ὑπὸ ΓΔΕ ἐστιν ἴση.
Therefore the whole angle AED is equal to the whole angle CDE.
ἀλλὰ ἡ ὑπὸ ΓΔΕ ταῖς πρὸς τοῖς Α, Γ γωνίαις ὑπόκειται ἴση· καὶ ἡ ὑπὸ ΑΕΔ ἄρα γωνία ταῖς πρὸς τοῖς Α, Γ ἴση ἐστίν.
But the angle CDE is by hypothesis equal to the angles at A, C; therefore the angle AED is also equal to the angles at A, C.
διὰ τὰ αὐτὰ δὴ καὶ ἡ ὑπὸ ΑΒΓ ἴση ἐστὶ ταῖς πρὸς τοῖς Α, Γ, Δ γωνίαις.
For the same reason, the angle ABC is also equal to the angles at A, C, D.
ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕ πεντάγωνον· ὅπερ ἔδει δεῖξαι.
Therefore the pentagon ABCDE is equiangular; which it was required to prove.

Notes

  1. §13.prop.7ΖΔ — Point Z is not explicitly defined in the text, but is introduced as the intersection of AC and BE. This intersection Z is then used to join the straight line ZD.
  2. ¦5¦αἱ πρὸς τοῖς Α, Β, Γ — "The [angles] at A, B, C." Here, γωνίαι (angles, plural) is omitted, referring back to αἱ τρεῖς γωνίαι (the three angles) in the preceding sentence. The preposition πρός with the dative case is used to mean "the angles at ...".
  3. ¦10¦ὑφʼ ἃς αἱ ἴσαι πλευραὶ ὑποτείνουσιν — The antecedent of the relative pronoun ἃς is "the remaining angles" (αἱ λοιπαὶ γωνίαι). It means "subtended by the equal sides", indicating that in congruent triangles, the angles opposite to the corresponding equal sides are equal to each other.

Cite this passage

Euclid, Elements §13.prop.7. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:13.prop.7

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