§4.prop.6εἰς τὸν δοθέντα κύκλον τετράγωνον ἐγγράψαι.
Into the given circle to inscribe a square.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ εἰς τὸν ΑΒΓΔ κύκλον τετράγωνον ἐγγράψαι.
Let the given circle be ABCD; it is required then into the circle ABCD to inscribe a square.
ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ ἐπεζεύχθωσαν αἱ ΑΒ, ΒΓ, ΓΔ, ΔΑ.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΔ·
Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and let AB, BC, CD, DA be joined.
κέντρον γὰρ τὸ Ε· κοινὴ δὲ καὶ πρὸς ὀρθὰς ἡ ΕΑ, βάσις ἄρα ἡ ΑΒ βάσει τῇ ΑΔ ἴση ἐστίν.
And since BE is equal to ED, for E is the center, and EA is common and at right angles, therefore the base AB is equal to the base AD.
διὰ τὰ αὐτὰ δὴ καὶ ἑκατέρα τῶν ΒΓ, ΓΔ ἑκατέρᾳ τῶν ΑΒ, ΑΔ ἴση ἐστίν· ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον.
Therefore, for the same reasons, also each of BC, CD is equal to each of AB, AD; therefore the quadrilateral ABCD is equilateral.
λέγω δή, ὅτι καὶ ὀρθογώνιον.
I say then that it is also right-angled.
ἐπεὶ γὰρ ἡ ΒΔ εὐθεῖα διάμετρός ἐστι τοῦ ΑΒΓΔ κύκλου, ἡμικύκλιον ἄρα ἐστὶ τὸ ΒΑΔ· ὀρθὴ ἄρα ἡ ὑπὸ ΒΑΔ γωνία.
For since the straight line BD is a diameter of the circle ABCD, therefore BAD is a semicircle; therefore the angle BAD is right.
διὰ τὰ αὐτὰ δὴ καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ ὀρθή ἐστιν· ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΒΓΔ τετράπλευρον.
For the same reasons then, each of the angles ABC, BCD, CDA is also right; therefore the quadrilateral ABCD is right-angled.
ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν.
And it was also proved equilateral; therefore it is a square.
καὶ ἐγγέγραπται εἰς τὸν ΑΒΓΔ κύκλον.
And it has been inscribed in the circle ABCD.
εἰς ἄρα τὸν δοθέντα κύκλον τετράγωνον ἐγγέγραπται τὸ ΑΒΓΔ· ὅπερ ἔδει ποιῆσαι.
Therefore, into the given circle a square, ABCD, has been inscribed; which was required to do.
§4.prop.7περὶ τὸν δοθέντα κύκλον τετράγωνον περιγράψαι.
About the given circle to circumscribe a square.
ἔστω ὁ δοθεὶς κύκλος ὁ ΑΒΓΔ· δεῖ δὴ περὶ τὸν ΑΒΓΔ κύκλον τετράγωνον περιγράψαι.
Let the given circle be ABCD; it is required then about the circle ABCD to circumscribe a square.
ἤχθωσαν τοῦ ΑΒΓΔ κύκλου δύο διάμετροι πρὸς ὀρθὰς ἀλλήλαις αἱ ΑΓ, ΒΔ, καὶ διὰ τῶν Α, Β, Γ, Δ σημείων ἤχθωσαν ἐφαπτόμεναι τοῦ ΑΒΓΔ κύκλου αἱ ΖΗ, ΗΘ, ΘΚ, ΚΖ.
ἐπεὶ οὖν ἐφάπτεται ἡ ΖΗ τοῦ ΑΒΓΔ κύκλου, ἀπὸ δὲ τοῦ Ε κέντρου ἐπὶ τὴν κατὰ τὸ Α ἐπαφὴν ἐπέζευκται ἡ ΕΑ, αἱ ἄρα πρὸς τῷ Α γωνίαι ὀρθαί εἰσιν.
Let two diameters AC, BD of the circle ABCD be drawn at right angles to one another, and through the points A, B, C, D let straight lines ZH, HΘ, ΘK, KZ be drawn touching the circle ABCD. Since then ZH touches the circle ABCD, and EA has been joined from the center E to the point of contact at A, therefore the angles at A are right.
διὰ τὰ αὐτὰ δὴ καὶ αἱ πρὸς τοῖς Β, Γ, Δ σημείοις γωνίαι ὀρθαί εἰσιν.
For the same reasons then, the angles at the points B, C, D are also right.
καὶ ἐπεὶ ὀρθή ἐστιν ἡ ὑπὸ ΑΕΒ γωνία, ἐστὶ δὲ ὀρθὴ καὶ ἡ ὑπὸ ΕΒΗ, παράλληλος ἄρα ἐστὶν ἡ ΗΘ τῇ ΑΓ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΑΓ τῇ ΖΚ ἐστι παράλληλος. ὥστε καὶ ἡ ΗΘ τῇ ΖΚ ἐστι παράλληλος.
And since the angle AEB is right, and the angle EBH is also right, therefore HΘ is parallel to AC. For the same reasons then, AC is also parallel to ZK; so that HΘ is also parallel to ZK.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΖ, ΘΚ τῇ ΒΕΔ ἐστι παράλληλος.
Similarly then we will prove that each of HZ, ΘK is also parallel to BED.
παραλληλόγραμμα ἄρα ἐστὶ τὰ ΗΚ, ΗΓ, ΑΚ, ΖΒ, ΒΚ· ἴση ἄρα ἐστὶν ἡ μὲν ΗΖ τῇ ΘΚ, ἡ δὲ ΗΘ τῇ ΖΚ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΓ τῇ ΒΔ, ἀλλὰ καὶ ἡ μὲν ΑΓ ἑκατέρᾳ τῶν ΗΘ, ΖΚ, ἡ δὲ ΒΔ ἑκατέρᾳ τῶν ΗΖ, ΘΚ ἐστιν ἴση, ἰσόπλευρον ἄρα ἐστὶ τὸ ΖΗΘΚ τετράπλευρον.
Therefore HK, HC, AK, ZB, BK are parallegrams; therefore HZ is equal to ΘK, and HΘ to ZK. And since AC is equal to BD, while AC is equal to each of HΘ, ZK, and BD is equal to each of HZ, ΘK, therefore the quadrilateral ZHΘK is equilateral.
λέγω δή, ὅτι καὶ ὀρθογώνιον.
I say then that it is also right-angled.
ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΗΒΕΑ, καί ἐστιν ὀρθὴ ἡ ὑπὸ ΑΕΒ, ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΗΒ. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ πρὸς τοῖς Θ, Κ, Ζ γωνίαι ὀρθαί εἰσιν.
For since HBEA is a parallelogram, and the angle AEB is right, therefore the angle AHB is also right. Similarly then we will prove that the angles at Θ, K, Z are also right.
ὀρθογώνιον ἄρα ἐστὶ τὸ ΖΗΘΚ. ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν.
Therefore ZHΘK is right-angled. And it was also proved equilateral; therefore it is a square.
καὶ περιγέγραπται περὶ τὸν ΑΒΓΔ κύκλον.
And it has been circumscribed about the circle ABCD.
περὶ τὸν δοθέντα ἄρα κύκλον τετράγωνον περιγέγραπται· ὅπερ ἔδει ποιῆσαι.
Therefore, about the given circle a square has been circumscribed; which was required to do.