Page 473 - Euclid's Elements of Geometry

ST EW ibþ.

ELEMENTS BOOK 12

A
[ἔστωσαν] αἱ ΒΔ, ΖΘ· λέγω, ὅτι ἐστὶν ὡς ὁ ΑΒΓΔ κύκλος FH [be] their diameters. I say that as circle ABCD is to
R
X
πρὸς τὸν ΕΖΗΘ κύκλον, οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον circle EFGH, so the square on BD (is) to the square on
πρὸς τὸ ἀπὸ τῆς ΖΘ τετράγωνον. E FH. A E
B DZ O R K N

G H B P Q D F L G M H

S T S C T

Εἰ γὰρ μή ἐστιν ὡς ὁ ΑΒΓΔ κύκλος πρὸς τὸν ΕΖΗΘ, For if the circle ABCD is not to the (circle) EFGH,
οὕτως τὸ ἀπὸ τῆς ΒΔ τετράγωνον πρὸς τὸ ἀπὸ τῆς ΖΘ, as the square on BD (is) to the (square) on FH, then as
ἔσται ὡς τὸ ἀπὸ τῆς ΒΔ πρὸς τὸ ἀπὸ τῆς ΖΘ, οὕτως ὁ the (square) on BD (is) to the (square) on FH, so circle
ΑΒΓΔ κύκλος ἤτοι πρὸς ἔλασσόν τι τοῦ ΕΖΗΘ κύκλου ABCD will be to some area either less than, or greater
χωρίον ἢ πρὸς μεῖζον. ἔστω πρότερον πρὸς ἔλασσον τὸ than, circle EFGH. Let it, first of all, be (in that ratio) to
Σ. και ἐγγεγράφθω εἰς τὸν ΕΖΗΘ κύκλον τετράγωνον τὸ (some) lesser (area), S. And let the square EFGH have
ΕΖΗΘ. τὸ δὴ ἐγγεγραμμένον τετράγωνον μεῖζόν ἐστιν ἢ τὸ been inscribed in circle EFGH [Prop. 4.6]. So the in-
ἥμισυ τοῦ ΕΖΗΘ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν Ε, Ζ, Η, Θ scribed square is greater than half of circle EFGH, inas-
σημείων ἐφαπτομένας [εὐθείας] τοῦ κύκλου ἀγάγωμεν, τοῦ much as if we draw tangents to the circle through the
περιγραφομένου περὶ τὸν κύκλον τετραγώνου ἥμισύ ἐστι points E, F, G, and H, then square EFGH is half of the
τὸ ΕΖΗΘ τετράγωνον, τοῦ δὲ περιγραφέντος τετραγώνου square circumscribed about the circle [Prop. 1.47], and
ἐλάττων ἐστὶν ὁ κύκλος· ὥστε τὸ ΕΖΗΘ ἐγγεγραμμένον the circle is less than the circumscribed square. Hence,
τετράγωνον μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ ΕΖΗΘ κύκλου. the inscribed square EFGH is greater than half of cir-
τετμήσθωσαν δίχα αἱ ΕΖ, ΖΗ, ΗΘ, ΘΕ περιφέρειαι κατὰ cle EFGH. Let the circumferences EF, FG, GH, and
τὰ Κ, Λ, Μ, Ν σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΕΚ, ΚΖ, HE have been cut in half at points K, L, M, and N
ΖΛ, ΛΗ, ΗΜ, ΜΘ, ΘΝ, ΝΕ· καὶ ἕκαστον ἄρα τῶν ΕΚΖ, (respectively), and let EK, KF, FL, LG, GM, MH,
ΖΛΗ, ΗΜΘ, ΘΝΕ τριγώνων μεῖζόν ἐστιν ἢ τὸ ἥμισυ τοῦ HN, and NE have been joined. And, thus, each of
καθ᾿ ἑαυτὸ τμήματος τοῦ κύκλου, ἐπειδήπερ ἐὰν διὰ τῶν the triangles EKF, FLG, GMH, and HNE is greater
Κ, Λ, Μ, Ν σημείων ἐφαπτομένας τοῦ κύκλου ἀγάγωμεν than half of the segment of the circle about it, inasmuch
καὶ ἀναπληρώσωμεν τὰ ἐπὶ τῶν ΕΖ, ΖΗ, ΗΘ, ΘΕ εὐθειῶν as if we draw tangents to the circle through points K,
παραλληλόγραμμα, ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ L, M, and N, and complete the parallelograms on the
τριγώνων ἥμισυ ἔσται τοῦ καθ᾿ ἑαυτὸ παραλληλογράμμου, straight-lines EF, FG, GH, and HE, then each of the
ἀλλὰ τὸ καθ᾿ ἑαυτὸ τμῆμα ἔλαττόν ἐστι τοῦ παραλλη- triangles EKF, FLG, GMH, and HNE will be half
λογράμμου· ὥστε ἕκαστον τῶν ΕΚΖ, ΖΛΗ, ΗΜΘ, ΘΝΕ of the parallelogram about it, but the segment about it
τριγώνων μεῖζόν ἐστι τοῦ ἡμίσεως τοῦ καθ᾿ ἑαυτὸ τμήματος is less than the parallelogram. Hence, each of the tri-
τοῦ κύκλου. τέμνοντες δὴ τὰς ὑπολειπομένας περιφερείας angles EKF, FLG, GMH, and HNE is greater than
δίχα καὶ ἐπιζευγνύντες εὐθείας καὶ τοῦτο ἀεὶ ποιοῦντες κα- half of the segment of the circle about it. So, by cutting
ταλείψομέν τινα ἀποτμήματα τοῦ κύκλου, ἃ ἔσται ἐλάσσονα the circumferences remaining behind in half, and joining
τῆς ὑπεροχῆς, ᾗ ὑπερέχει ὁ ΕΖΗΘ κύκλος τοῦ Σ χωρίου. straight-lines, and doing this continually, we will (even-

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