Page 450 - Euclid's Elements of Geometry

ST EW iaþ.

ELEMENTS BOOK 11

παράλληλος ἄρα ἐστὶν ἡ ΒΓ τῇ ΑΔ. ἐδείχθη δὲ καὶ ἡ ΑΒ AC, their common sections are parallel [Prop. 11.16].
τῇ ΔΓ παράλληλος· παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΓ. Thus, BC is parallel to AD. And AB was also shown (to
ὁμοίως δὴ δείξομεν, ὅτι καὶ ἕκαστον τῶν ΔΖ, ΖΗ, ΗΒ, ΒΖ, be) parallel to DC. Thus, AC is a parallelogram. So, sim-
ΑΕ παραλληλόγραμμόν ἐστιν. ilarly, we can also show that DF, FG, GB, BF, and AE
᾿Επεζεύχθωσαν αἱ ΑΘ, ΔΖ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ are each parallelograms.
μὲν ΑΒ τῇ ΔΓ, ἡ δὲ ΒΘ τῇ ΓΖ, δύο δὴ αἱ ΑΒ, ΒΘ ἁπτόμεναι Let AH and DF have been joined. And since AB is
ἀλλήλων παρὰ δύο εὐθείας τὰς ΔΓ, ΓΖ ἁπτομένας ἀλλήλων parallel to DC, and BH to CF, so the two (straight-lines)
εἰσὶν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ· ἴσας ἄρα γωνίας περιέξουσιν· joining one another, AB and BH, are parallel to the two
ἴση ἄρα ἡ ὑπὸ ΑΒΘ γωνία τῇ ὑπὸ ΔΓΖ. καὶ ἐπεὶ δύο αἱ ΑΒ, straight-lines joining one another, DC and CF (respec-
ΒΘ δυσὶ ταῖς ΔΓ, ΓΖ ἴσαι εἰσίν, καὶ γωνία ἡ ὑπὸ ΑΒΘ γωνίᾳ tively), not (being) in the same plane. Thus, they will
τῇ ὑπὸ ΔΓΖ ἐστιν ἴση, βάσις ἄρα ἡ ΑΘ βάσει τῇ ΔΖ ἐστιν contain equal angles [Prop. 11.10]. Thus, angle ABH
ἴση, καὶ τὸ ΑΒΘ τρίγωνον τῷ ΔΓΖ τριγώνῳ ἴσον ἐστίν. καί (is) equal to (angle) DCF. And since the two (straight-
ἐστι τοῦ μὲν ΑΒΘ διπλάσιον τὸ ΒΗ παραλληλόγραμμον, lines) AB and BH are equal to the two (straight-lines)
τοῦ δὲ ΔΓΖ διπλάσιον τὸ ΓΕ παραλληλόγραμμον· ἴσον DC and CF (respectively) [Prop. 1.34], and angle ABH
ἄρα τὸ ΒΗ παραλληλόγραμμον τῷ ΓΕ παραλληλογράμμῳ· is equal to angle DCF, the base AH is thus equal to the
ὁμοίως δὴ δείξομεν, ὅτι καὶ τὸ μὲν ΑΓ τῷ ΗΖ ἐστιν ἴσον, base DF, and triangle ABH is equal to triangle DCF
τὸ δὲ ΑΕ τῷ ΒΖ. [Prop. 1.4]. And parallelogram BG is double (triangle)
keþ planes then its opposite planes are both equal and paral-
᾿Εὰν ἄρα στερεὸν ὑπὸ παραλλήλων ἐπιπέδων περιέχηται, ABH, and parallelogram CE double (triangle) DCF
τὰ ἀπεναντίον αὐτοῦ ἐπίπεδα ἴσα τε καὶ παραλληλόγραμμά [Prop. 1.34]. Thus, parallelogram BG (is) equal to paral-
ἐστιν· ὅπερ ἔδει δεῖξαι. lelogram CE. So, similarly, we can show that AC is also
equal to GF, and AE to BF.
Thus, if a solid (figure) is contained by (six) parallel

lelogrammic. (Which is) the very thing it was required to
show.

.
Proposition 25
᾿Εὰν στερεὸν παραλληλεπίπεδον ἐπιπέδῳ τμηθῇ πα- If a parallelipiped solid is cut by a plane which is par-
R U D W T
Y
ραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις, ἔσται ὡς ἡ βάσις allel to the opposite planes (of the parallelipiped) then as
πρὸς τὴν βάσιν, οὕτως τὸ στερεὸν πρὸς τὸ στερεόν. the base (is) to the base, so the solid will be to the solid.
X B H O X Q B R G U D I Y T

Z G S
F

A E P V F C W S

L
K
A
M
H
E
N
Στερεὸν γὰρ παραλληλεπίπεδον τὸ ΑΒΓΔ ἐπιπέδῳ τῷ For let the parallelipiped solid ABCD have been cut
ΖΗ τετμήσθω παραλλήλῳ ὄντι τοῖς ἀπεναντίον ἐπιπέδοις by the plane FG which is parallel to the opposite planes
τοῖς ΡΑ, ΔΘ· λέγω, ὅτι ἐστὶν ὡς ἡ ΑΕΖΦ βάσις πρὸς τὴν RA and DH. I say that as the base AEFV (is) to the base
ΕΘΓΖ βάσιν, οὕτως τὸ ΑΒΖΥ στερεὸν πρὸς τὸ ΕΗΓΔ EHCF, so the solid ABFU (is) to the solid EGCD.
στερεόν. For let AH have been produced in each direction. And
᾿Εκβεβλήσθω γὰρ ἡ ΑΘ ἐφ᾿ ἑκάτερα τὰ μέρη, καὶ let any number whatsoever (of lengths), AK and KL,
κείσθωσαν τῇ μὲν ΑΕ ἴσαι ὁσαιδηποτοῦν αἱ ΑΚ, ΚΛ, τῇ δὲ be made equal to AE, and any number whatsoever (of
ΕΘ ἴσαι ὁσαιδηποτοῦν αἱ ΘΜ, ΜΝ, καὶ συμπεπληρώσθω lengths), HM and MN, equal to EH. And let the paral-
τὰ ΛΟ, ΚΦ, ΘΧ, ΜΣ παραλληλόγραμμα καὶ τὰ ΛΠ, ΚΡ, lelograms LP, KV , HW, and MS have been completed,

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