§4.prop.15εἰς τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
To inscribe an equilateral and equiangular hexagon in the given circle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔΕΖ· δεῖ δὴ εἰς τὸν ΑΒΓΔΕΖ κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
Let the given circle be ABCDEF; it is required then to inscribe an equilateral and equiangular hexagon in the circle ABCDEF.
ἤχθω τοῦ ΑΒΓΔΕΖ κύκλου διάμετρος ἡ ΑΔ, καὶ εἰλήφθω τὸ κέντρον τοῦ κύκλου τὸ Η, καὶ κέντρῳ μὲν τῷ Δ διαστήματι δὲ τῷ ΔΗ κύκλος γεγράφθω ὁ ΕΗΓΘ, καὶ ἐπιζευχθεῖσαι αἱ ΕΗ, ΓΗ διήχθωσαν ἐπὶ τὰ Β, Ζ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ·
For let the diameter AD of the circle ABCDEF be drawn, let the center H of the circle be taken, and with center D and distance DH let the circle EGHC be described; let EH, CH, being joined, be carried through to the points B, F, and let AB, BC, CD, DE, EF, FA be joined.
λέγω, ὅτι τὸ ΑΒΓΔΕΖ ἑξάγωνον ἰσόπλευρόν τέ ἐστι καὶ ἰσογώνιον.
I say that the hexagon ABCDEF is equilateral and equiangular.
ἐπεὶ γὰρ τὸ Η σημεῖον κέντρον ἐστὶ τοῦ ΑΒΓΔΕΖ κύκλου, ἴση ἐστὶν ἡ ΗΕ τῇ ΗΔ. πάλιν, ἐπεὶ τὸ Δ σημεῖον κέντρον ἐστὶ τοῦ ΗΓΘ κύκλου, ἴση ἐστὶν ἡ ΔΕ τῇ ΔΗ. ἀλλʼ ἡ ΗΕ τῇ ΗΔ ἐδείχθη ἴση· καὶ ἡ ΗΕ ἄρα τῇ ΕΔ ἴση ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΕΗΔ τρίγωνον· καὶ αἱ τρεῖς ἄρα αὐτοῦ γωνίαι αἱ ὑπὸ ΕΗΔ, ΗΔΕ, ΔΕΗ ἴσαι ἀλλήλαις εἰσίν, ἐπειδήπερ τῶν ἰσοσκελῶν τριγώνων αἱ πρὸς τῇ βάσει γωνίαι ἴσαι ἀλλήλαις εἰσίν· καί εἰσιν αἱ τρεῖς τοῦ τριγώνου γωνίαι δυσὶν ὀρθαῖς ἴσαι· ἡ ἄρα ὑπὸ ΕΗΔ γωνία τρίτον ἐστὶ δύο ὀρθῶν.
For since the point H is the center of the circle ABCDEF, HE is equal to HD. Again, since the point D is the center of the circle EGHC, DE is equal to DH. But HE was proved equal to HD; therefore HE is also equal to ED; therefore the triangle EHD is equilateral; and therefore its three angles EHD, HDE, DEH are equal to one another, since in isosceles triangles the angles at the base are equal to one another; and the three angles of the triangle are equal to two right angles; therefore the angle EHD is one-third of two right angles.
ὁμοίως δὴ δειχθήσεται καὶ ἡ ὑπὸ ΔΗΓ τρίτον δύο ὀρθῶν.
Similarly it will be proved that the angle DHC is also one-third of two right angles.
καὶ ἐπεὶ ἡ ΓΗ εὐθεῖα ἐπὶ τὴν ΕΒ σταθεῖσα τὰς ἐφεξῆς γωνίας τὰς ὑπὸ ΕΗΓ, ΓΗΒ δυσὶν ὀρθαῖς ἴσας ποιεῖ, καὶ λοιπὴ ἄρα ἡ ὑπὸ ΓΗΒ τρίτον ἐστὶ δύο ὀρθῶν· αἱ ἄρα ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ γωνίαι ἴσαι ἀλλήλαις εἰσίν· ὥστε καὶ αἱ κατὰ κορυφὴν αὐταῖς αἱ ὑπὸ ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι εἰσίν.
And since the straight line CH standing on the straight line EB makes the adjacent angles EHC, CHB equal to two right angles, therefore the remaining angle CHB is also one-third of two right angles; therefore the angles EHD, DHC, CHB are equal to one another; so that their vertically opposite angles BHA, AHF, FHE are also equal.
αἱ ἓξ ἄρα γωνίαι αἱ ὑπὸ ΕΗΔ, ΔΗΓ, ΓΗΒ, ΒΗΑ, ΑΗΖ, ΖΗΕ ἴσαι ἀλλήλαις εἰσίν.
Therefore the six angles EHD, DHC, CHB, BHA, AHF, FHE are equal to one another.
αἱ δὲ ἴσαι γωνίαι ἐπὶ ἴσων περιφερειῶν βεβήκασιν· αἱ ἓξ ἄρα περιφέρειαι αἱ ΑΒ, ΒΓ, ΓΔ, ΔΕ, ΕΖ, ΖΑ ἴσαι ἀλλήλαις εἰσίν.
But equal angles stand on equal circumferences; therefore the six circumferences AB, BC, CD, DE, EF, FA are equal to one another.
ὑπὸ δὲ τὰς ἴσας περιφερείας αἱ ἴσαι εὐθεῖαι ὑποτείνουσιν· αἱ ἓξ ἄρα εὐθεῖαι ἴσαι ἀλλήλαις εἰσίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον.
And equal straight lines subtend equal circumferences; therefore the six straight lines are equal to one another; therefore the hexagon ABCDEF is equilateral.
λέγω δή, ὅτι καὶ ἰσογώνιον.
I say then that it is also equiangular.
ἐπεὶ γὰρ ἴση ἐστὶν ἡ ΖΑ περιφέρεια τῇ ΕΔ περιφερείᾳ, κοινὴ προσκείσθω ἡ ΑΒΓΔ περιφέρεια· ὅλη ἄρα ἡ ΖΑΒΓΔ ὅλῃ τῇ ΕΔΓΒΑ ἐστιν ἴση· καὶ βέβηκεν ἐπὶ μὲν τῆς ΖΑΒΓΔ περιφερείας ἡ ὑπὸ ΖΕΔ γωνία, ἐπὶ δὲ τῆς ΕΔΓΒΑ περιφερείας ἡ ὑπὸ ΑΖΕ γωνία·
For since the circumference AF is equal to the circumference ED, let the common circumference ABCD be added; therefore the whole circumference FABCD is equal to the whole circumference EDCBA; and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference EDCBA; therefore the angle AFE is equal to the angle FED.
ἴση ἄρα ἡ ὑπὸ ΑΖΕ γωνία τῇ ὑπὸ ΔΕΖ. ὁμοίως δὴ δειχθήσεται, ὅτι καὶ αἱ λοιπαὶ γωνίαι τοῦ ΑΒΓΔΕΖ ἑξαγώνου κατὰ μίαν ἴσαι εἰσὶν ἑκατέρᾳ τῶν ὑπὸ ΑΖΕ, ΖΕΔ γωνιῶν· ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔΕΖ ἑξάγωνον.
Similarly it will be proved that the remaining angles of the hexagon ABCDEF are also severally equal to each of the angles AFE, FED; therefore the hexagon ABCDEF is equiangular.
ἐδείχθη δὲ καὶ ἰσόπλευρον· καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔΕΖ κύκλον.
And it was also proved equilateral; and it has been inscribed in the circle ABCDEF.
εἰς ἄρα τὸν δοθέντα κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι.
Therefore in the given circle an equilateral and equiangular hexagon has been inscribed; which was required to do.
Πόρισμα
ἐκ δὴ τούτου φανερόν, ὅτι ἡ τοῦ ἑξαγώνου πλευρὰ ἴση ἐστὶ τῇ ἐκ τοῦ κέντρου τοῦ κύκλου.
Porism From this it is manifest that the side of the hexagon is equal to the radius of the circle.
ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον ἑξάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἀκολούθως τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις.
And similarly to the case of the pentagon, if through the divisions on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular hexagon, in accordance with what was said in the case of the pentagon.
καὶ ἔτι διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου εἰρημένοις εἰς τὸ δοθὲν ἑξάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν· ὅπερ ἔδει ποιῆσαι.
And further, by means similar to what was said in the case of the pentagon, we can both inscribe a circle in the given hexagon and circumscribe one about it; which was required to do.
§4.prop.16εἰς τὸν δοθέντα κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
To inscribe an equilateral and equiangular fifteen-sided figure in the given circle.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον ἐγγράψαι.
Let the given circle be ABCD; it is required then to inscribe an equilateral and equiangular fifteen-sided figure in the circle ABCD.
ἐγγεγράφθω εἰς τὸν ΑΒΓΔ κύκλον τριγώνου μὲν ἰσοπλεύρου τοῦ εἰς αὐτὸν ἐγγραφομένου πλευρὰ ἡ ΑΓ, πενταγώνου δὲ ἰσοπλεύρου ἡ ΑΒ· οἵων ἄρα ἐστὶν ὁ ΑΒΓΔ κύκλος ἴσων τμημάτων δεκαπέντε, τοιούτων ἡ μὲν ΑΒΓ περιφέρεια τρίτον οὖσα τοῦ κύκλου ἔσται πέντε, ἡ δὲ ΑΒ περιφέρεια πέμπτον οὖσα τοῦ κύκλου ἔσται τριῶν· λοιπὴ ἄρα ἡ ΒΓ τῶν ἴσων δύο.
Let there be inscribed in the circle ABCD the side AC of the equilateral triangle inscribed in it, and the side AB of the equilateral pentagon; therefore, of such equal segments as the circle ABCD is divided into fifteen, of such the circumference ABC, being a third of the circle, will contain five, and the circumference AB, being a fifth of the circle, will contain three; therefore the remaining circumference BC will contain two of those equal segments.
τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε· ἑκατέρα ἄρα τῶν ΒΕ, ΕΓ περιφερειῶν πεντεκαιδέκατόν ἐστι τοῦ ΑΒΓΔ κύκλου.
Let BC be bisected at E; therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD.
ἐὰν ἄρα ἐπιζεύξαντες τὰς ΒΕ, ΕΓ ἴσας αὐταῖς κατὰ τὸ συνεχὲς εὐθείας ἐναρμόσωμεν εἰς τὸν ΑΒΓΔ κύκλον, ἔσται εἰς αὐτὸν ἐγγεγραμμένον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον· ὅπερ ἔδει ποιῆσαι.
If therefore, joining BE, EC, we fit into the circle ABCD straight lines equal to them continuously, there will be inscribed in it an equilateral and equiangular fifteen-sided figure; which was required to do.
ὁμοίως δὲ τοῖς ἐπὶ τοῦ πενταγώνου ἐὰν διὰ τῶν κατὰ τὸν κύκλον διαιρέσεων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν, περιγραφήσεται περὶ τὸν κύκλον πεντεκαιδεκάγωνον ἰσόπλευρόν τε καὶ ἰσογώνιον.
And similarly to the case of the pentagon, if through the divisions on the circle we draw tangents to the circle, there will be circumscribed about the circle an equilateral and equiangular fifteen-sided figure.
ἔτι δὲ διὰ τῶν ὁμοίων τοῖς ἐπὶ τοῦ πενταγώνου δείξεων καὶ εἰς τὸ δοθὲν πεντεκαιδεκάγωνον κύκλον ἐγγράψομέν τε καὶ περιγράψομεν· ὅπερ ἔδει ποιῆσαι.
And further, by proofs similar to those in the case of the pentagon, we can both inscribe a circle in the given fifteen-sided figure and circumscribe one about it; which was required to do.