Page 498 - Euclid's Elements of Geometry

ST EW ibþ.

ELEMENTS BOOK 12

᾿Αλλὰ δὴ τῶν ΑΞ, ΕΟ κυλίνδρων ἀντιπεπονθέτωσαν αἱ (is) equal to height KL. Thus, as base ABCD is to base
βάσεις τοῖς ὕψεσιν, καὶ ἔστω ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν EFGH, so height MN (is) to height KL. Thus, the bases
ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος πρὸς τὸ ΚΛ ὕψος· λέγω, of cylinders AO and EP are reciprocally proportional to
ὅτι ἴσος ἐστὶν ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. their heights.
Τῶν γὰρ αὐτῶν κατασκευασθέντων ἐπεί ἐστιν ὡς ἡ And, so, let the bases of cylinders AO and EP be
ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ ΜΝ ὕψος reciprocally proportional to their heights, and (thus) let
πρὸς τὸ ΚΛ ὕψος, ἴσον δὲ τὸ ΚΛ ὕψος τῷ ΠΝ ὕψει, ἔσται base ABCD be to base EFGH, as height MN (is) to
ἄρα ὡς ἡ ΑΒΓΔ βάσις πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως τὸ height KL. I say that cylinder AO is equal to cylinder
ΜΝ ὕψος πρὸς τὸ ΠΝ ὕψος. ἀλλ᾿ ὡς μὲν ἡ ΑΒΓΔ βάσις EP.
πρὸς τὴν ΕΖΗΘ βάσιν, οὕτως ὁ ΑΞ κύλινδρος πρὸς τὸν For, with the same construction, since as base ABCD
ΕΣ κύλινδρον· ὑπὸ γὰρ τὸ αὐτὸ ὕψος εἰσίν· ὡς δὲ τὸ ΜΝ is to base EFGH, so height MN (is) to height KL, and
ὕψος πρὸς τὸ ΠΝ [ὕψος], οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν height KL (is) equal to height QN, thus, as base ABCD
ΕΣ κύλινδρον· ἔστιν ἄρα ὡς ὁ ΑΞ κύλινδρος πρὸς τὸν ΕΣ (is) to base EFGH, so height MN will be to height
κύλινδρον, οὕτως ὁ ΕΟ κύλινδρος πρὸς τὸν ΕΣ. ἴσος ἄρα QN. But, as base ABCD (is) to base EFGH, so cylin-
iþ as cylinder AO is to cylinder ES, so cylinder EP (is) to
ὁ ΑΞ κύλινδρος τῷ ΕΟ κυλίνδρῳ. ὡσαύτως δὲ καὶ ἐπὶ τῶν der AO (is) to cylinder ES. For they are the same height
κώνων· ὅπερ ἔδει δεῖξαι. [Prop. 12.11]. And as height MN (is) to [height] QN,
so cylinder EP (is) to cylinder ES [Prop. 12.13]. Thus,
(cylinder) ES [Prop. 5.11]. Thus, cylinder AO (is) equal
to cylinder EP [Prop. 5.9]. In the same manner, (the
proposition can) also (be demonstrated) for the cones.
(Which is) the very thing it was required to show.

.
Proposition 16
A
Δύο κύκλων περὶ τὸ αὐτὸ κέντρον ὄντων εἰς τὸν There being two circles about the same center, to
μείζονα κύκλον πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον inscribe an equilateral and even-sided polygon in the
ἐγγράψαι μὴ ψαῦον τοῦ ἐλάσσονος κύκλου. greater circle, not touching the lesser circle.
A
B H E D B E K G M D
H
L

Z G N

F
C
῎Εστωσαν οἱ δοθέντες δύο κύκλοι οἱ ΑΒΓΔ, ΕΖΗΘ Let ABCD and EFGH be the given two circles, about
περὶ τὸ αὐτὸ κέντρον τὸ Κ· δεῖ δὴ εἰς τὸν μείζονα κύκλον the same center, K. So, it is necessary to inscribe an
τὸν ΑΒΓΔ πολύγωνον ἰσόπλευρόν τε καὶ ἀρτιόπλευρον equilateral and even-sided polygon in the greater circle
ἐγγράψαι μὴ ψαῦον τοῦ ΕΖΗΘ κύκλου. ABCD, not touching circle EFGH.
῎Ηχθω γὰρ διὰ τοῦ Κ κέντρου εὐθεῖα ἡ ΒΚΔ, καὶ Let the straight-line BKD have been drawn through
ἀπὸ τοῦ Η σημείου τῇ ΒΔ εὐθείᾳ πρὸς ὀρθὰς ἤχθω ἡ the center K. And let GA have been drawn, at right-
ΗΑ καὶ διήχθω ἐπὶ τὸ Γ· ἡ ΑΓ ἄρα ἐφάπτεται τοῦ ΕΖΗΘ angles to the straight-line BD, through point G, and let it
κύκλου. τέμνοντες δὴ τὴν ΒΑΔ περιφέρειαν δίχα καὶ τὴν have been drawn through to C. Thus, AC touches circle
ἡμίσειαν αὐτῆς δίχα καὶ τοῦτο ἀεὶ ποιοῦντες καταλείψομεν EFGH [Prop. 3.16 corr.]. So, (by) cutting circumference
περιφέρειαν ἐλάσσονα τῆς ΑΔ. λελείφθω, καὶ ἔστω ἡ ΛΔ, BAD in half, and the half of it in half, and doing this con-
καὶ ἀπὸ τοῦ Λ ἐπὶ τὴν ΒΔ κάθετος ἤχθω ἡ ΛΜ καὶ διήχθω tinually, we will (eventually) leave a circumference less

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