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Euclid · Elements §4.prop.8-4.prop.9

Inscribing and Circumscribing a Circle in and About a Square

Passage 62 of 316 · Greek

Summary

Presents the constructions and proofs for inscribing a circle within a given square by bisecting the sides (Proposition 8) and circumscribing a circle about a given square using the intersection of the diagonals as the center (Proposition 9).

§4.prop.8εἰς τὸ δοθὲν τετράγωνον κύκλον ἐγγράψαι.
Into the given square to inscribe a circle.
ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ· δεῖ δὴ εἰς τὸ ΑΒΓΔ τετράγωνον κύκλον ἐγγράψαι.
Let the given square be ABCD; it is required then into the square ABCD to inscribe a circle.
τετμήσθω ἑκατέρα τῶν ΑΔ, ΑΒ δίχα κατὰ τὰ Ε, Ζ σημεῖα, καὶ διὰ μὲν τοῦ Ε ὁποτέρᾳ τῶν ΑΒ, ΓΔ παράλληλος ἤχθω ἡ ΕΘ, διὰ δὲ τοῦ Ζ ὁποτέρᾳ τῶν ΑΔ, ΒΓ παράλληλος ἤχθω ἡ ΖΚ· παραλληλόγραμμον ἄρα ἐστὶν ἕκαστον τῶν ΑΚ, ΚΒ, ΑΘ, ΘΔ, ΑΗ, ΗΓ, ΒΗ, ΗΔ, καὶ αἱ ἀπεναντίον αὐτῶν πλευραὶ δηλονότι ἴσαι.
Let each of AD, AB be bisected at the points E, Z, and through E let EΘ be drawn parallel to either of AB, CD, and through Z let ZK be drawn parallel to either of AD, BC; therefore each of AK, KB, AΘ, ΘΔ, AH, HΓ, BH, HΔ is a parallelogram, and their opposite sides are manifestly equal.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ΑΔ τῇ ΑΒ, καί ἐστι τῆς μὲν ΑΔ ἡμίσεια ἡ ΑΕ, τῆς δὲ ΑΒ ἡμίσεια ἡ ΑΖ, ἴση ἄρα καὶ ἡ ΑΕ τῇ ΑΖ· ὥστε καὶ αἱ ἀπεναντίον· ἴση ἄρα καὶ ἡ ΖΗ τῇ ΗΕ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΗΘ, ΗΚ ἑκατέρᾳ τῶν ΖΗ, ΗΕ ἐστιν ἴση· αἱ τέσσαρες ἄρα αἱ ΗΕ, ΗΖ, ΗΘ, ΗΚ ἴσαι ἀλλήλαις.
And since AD is equal to AB, and AE is half of AD, and AZ is half of AB, therefore AE is also equal to AZ; so that their opposite sides are also equal; therefore ZH is also equal to HE. Similarly then we will prove that each of HΘ, HK is also equal to each of ZH, HE; therefore the four straight lines HE, HZ, HΘ, HK are equal to one another.
ὁ ἄρα κέντρῳ μὲν τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων· καὶ ἐφάψεται τῶν ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθειῶν διὰ τὸ ὀρθὰς εἶναι τὰς πρὸς τοῖς Ε, Ζ, Θ, Κ γωνίας·
Therefore the circle described with center H and distance one of the points E, Z, Θ, K will also pass through the remaining points; and it will touch the straight lines AB, BC, CD, DA, because the angles at E, Z, Θ, K are right.
εἰ γὰρ τεμεῖ ὁ κύκλος τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ, ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπʼ ἄκρας ἀγομένη ἐντὸς πεσεῖται τοῦ κύκλου· ὅπερ ἄτοπον ἐδείχθη.
For if the circle cuts the straight lines AB, BC, CD, DA, the straight line drawn at right angles to the diameter of the circle from its extremity will fall within the circle; which was proved absurd.
οὐκ ἄρα ὁ κέντρῳ τῷ Η διαστήματι δὲ ἑνὶ τῶν Ε, Ζ, Θ, Κ κύκλος γραφόμενος τεμεῖ τὰς ΑΒ, ΒΓ, ΓΔ, ΔΑ εὐθείας.
Therefore the circle described with center H and distance one of the points E, Z, Θ, K will not cut the straight lines AB, BC, CD, DA.
ἐφάψεται ἄρα αὐτῶν καὶ ἔσται ἐγγεγραμμένος εἰς τὸ ΑΒΓΔ τετράγωνον.
Therefore it will touch them, and will be inscribed in the square ABCD.
εἰς ἄρα τὸ δοθὲν τετράγωνον κύκλος ἐγγέγραπται· ὅπερ ἔδει ποιῆσαι.
Therefore, into the given square a circle has been inscribed; which was required to do.
§4.prop.9περὶ τὸ δοθὲν τετράγωνον κύκλον περιγράψαι.
About the given square to circumscribe a circle.
ἔστω τὸ δοθὲν τετράγωνον τὸ ΑΒΓΔ· δεῖ δὴ περὶ τὸ ΑΒΓΔ τετράγωνον κύκλον περιγράψαι.
Let the given square be ABCD; it is required then about the square ABCD to circumscribe a circle.
ἐπιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν ἀλλήλας κατὰ τὸ Ε. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΔΑ τῇ ΑΒ, κοινὴ δὲ ἡ ΑΓ, δύο δὴ αἱ ΔΑ, ΑΓ δυσὶ ταῖς ΒΑ, ΑΓ ἴσαι εἰσίν· καὶ βάσις ἡ ΔΓ βάσει τῇ ΒΓ ἴση· γωνία ἄρα ἡ ὑπὸ ΔΑΓ γωνίᾳ τῇ ὑπὸ ΒΑΓ ἴση ἐστίν·
For let AC, BD be joined and let them cut one another at E. And since DA is equal to AB, and AC is common, therefore the two straight lines DA, AC are equal to the two straight lines BA, AC respectively; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC; therefore the angle DAB has been bisected by AC.
ἡ ἄρα ὑπὸ ΔΑΒ γωνία δίχα τέτμηται ὑπὸ τῆς ΑΓ. ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκάστη τῶν ὑπὸ ΑΒΓ, ΒΓΔ, ΓΔΑ δίχα τέτμηται ὑπὸ τῶν ΑΓ, ΔΒ εὐθειῶν.
Similarly then we will prove that each of the angles ABC, BCD, CDA has also been bisected by the straight lines AC, DB.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΔΑΒ γωνία τῇ ὑπὸ ΑΒΓ, καί ἐστι τῆς μὲν ὑπὸ ΔΑΒ ἡμίσεια ἡ ὑπὸ ΕΑΒ, τῆς δὲ ὑπὸ ΑΒΓ ἡμίσεια ἡ ὑπὸ ΕΒΑ, καὶ ἡ ὑπὸ ΕΑΒ ἄρα τῇ ὑπὸ ΕΒΑ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΕΑ τῇ ΕΒ ἐστιν ἴση.
And since the angle DAB is equal to the angle ABC, and the angle EAB is half of the angle DAB, and the angle EBA is half of the angle ABC, therefore the angle EAB is also equal to the angle EBA; so that the side EA is also equal to EB.
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἑκατέρα τῶν ΕΑ, ΕΒ ἑκατέρᾳ τῶν ΕΓ, ΕΔ ἴση ἐστίν.
Similarly then we will prove that each of EA, EB is also equal to each of EΓ, EΔ.
αἱ τέσσαρες ἄρα αἱ ΕΑ, ΕΒ, ΕΓ, ΕΔ ἴσαι ἀλλήλαις εἰσίν.
Therefore the four straight lines EA, EB, EΓ, EΔ are equal to one another.
ὁ ἄρα κέντρῳ τῷ Ε καὶ διαστήματι ἑνὶ τῶν Α, Β, Γ, Δ κύκλος γραφόμενος ἥξει καὶ διὰ τῶν λοιπῶν σημείων καὶ ἔσται περιγεγραμμένος περὶ τὸ ΑΒΓΔ τετράγωνον.
Therefore the circle described with center E and distance one of the points A, B, Γ, Δ will also pass through the remaining points, and will be circumscribed about the square ABCD.
περιγεγράφθω ὡς ὁ ΑΒΓΔ. περὶ τὸ δοθὲν ἄρα τετράγωνον κύκλος περιγέγραπται· ὅπερ ἔδει ποιῆσαι.
Let it be circumscribed as ABCD. Therefore, about the given square a circle has been circumscribed; which was required to do.

Notes

  1. §4.prop.8ὁποτέρᾳ — Dative singular feminine of the relative/interrogative pronoun meaning "either of the two." Here it is used in the sense of "parallel to either of AB, CD." Since the opposite sides of a square are already parallel, a line drawn parallel to one is geometrically and necessarily parallel to the other.
  2. §4.prop.8ὥστε καὶ αἱ ἀπεναντίον· — An elliptical construction with the verb (likely ἴσαι εἰσίν) omitted. Following the preceding statement that AE is equal to AZ, it means "so that the opposite [sides] [are also equal]," which directly leads to the next conclusion that ZH is equal to HE.
  3. §4.prop.8ἡ τῇ διαμέτρῳ τοῦ κύκλου πρὸς ὀρθὰς ἀπʼ ἄκρας ἀγομένη — A long attributive participial phrase acting as the subject. The article `ἡ` is completed by the passive participle `ἀγομένη` at the end, meaning "the [straight line] drawn at right angles to the diameter of the circle from its extremity." This refers to a proof by contradiction based on Book III, Proposition 16 (the line drawn perpendicular to the diameter at its end falls outside the circle).
  4. §4.prop.9ἐπιζευχθεῖσαι γὰρ αἱ ΑΓ, ΒΔ τεμνέτωσαν — The aorist passive participle `ἐπιζευχθεῖσαι` (having been joined) modifies the third-person plural imperative `τεμνέτωσαν` (let them cut), combining the sequential steps of drawing and intersecting the diagonals AC and BD into a single coherent action.

Cite this passage

Euclid, Elements §4.prop.8-4.prop.9. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:4.prop.8-4.prop.9

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