Euclid · Elements §11.prop.29-11.prop.30

Equality of Parallelepipeds of Same Base and Height

Passage 257 of 316 · Greek

Summary

The author proves that two solid parallelepipeds on the same base and of the same height are equal to each other, dividing the proof into two cases: when their uprights are on the same straight lines (Proposition 29) and when they are not (Proposition 30).

§11.prop.29τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν.

Solid parallelepipeds which are on the same base and have the same height, and whose uprights are on the same straight lines, are equal to one another.

ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΖ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ ἐπὶ τῶν αὐτῶν εὐθειῶν ἔστωσαν τῶν ΖΝ, ΔΚ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ.

Let the solid parallelepipeds CM, CN be on the same base AB and have the same height, and let their uprights AH, AF, LM, LN, CD, CE, BT, BK be on the same straight lines FN, DK; I say, that the solid CM is equal to the solid CN.

ἐπεὶ γὰρ παραλληλόγραμμόν ἐστιν ἑκάτερον τῶν ΓΘ, ΓΚ, ἴση ἐστὶν ἡ ΓΒ ἑκατέρᾳ τῶν ΔΘ, ΕΚ· ὥστε καὶ ἡ ΔΘ τῇ ΕΚ ἐστιν ἴση.

For since each of CT, CK is a parallelogram, CB is equal to each of DT, EK; so that DT is also equal to EK.

κοινὴ ἀφῃρήσθω ἡ ΕΘ· λοιπὴ ἄρα ἡ ΔΕ λοιπῇ τῇ ΘΚ ἐστιν ἴση.

Let the common part ET be subtracted; therefore the remainder DE is equal to the remainder TK.

ὥστε καὶ τὸ μὲν ΔΓΕ τρίγωνον τῷ ΘΒΚ τριγώνῳ ἴσον ἐστίν, τὸ δὲ ΔΗ παραλληλόγραμμον τῷ ΘΝ παραλληλογράμμῳ.

So that the triangle DCE is also equal to the triangle TBK, and the parallelogram DG to the parallelogram TN.

διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΖΗ τρίγωνον τῷ ΜΛΝ τριγώνῳ ἴσον ἐστίν.

For the same reasons indeed, the triangle AFG is also equal to the triangle MLN.

ἔστι δὲ καὶ τὸ μὲν ΓΖ παραλληλόγραμμον τῷ ΒΜ παραλληλογράμμῳ ἴσον, τὸ δὲ ΓΗ τῷ ΒΝ· ἀπεναντίον γάρ· καὶ τὸ πρίσμα ἄρα τὸ περιεχόμενον ὑπὸ δύο μὲν τριγώνων τῶν ΑΖΗ, ΔΓΕ, τριῶν δὲ παραλληλογράμμων τῶν ΑΔ, ΔΗ, ΓΗ ἴσον ἐστὶ τῷ πρίσματι τῷ περιεχομένῳ ὑπὸ δύο μὲν τριγώνων τῶν ΜΛΝ, ΘΒΚ, τριῶν δὲ παραλληλογράμμων τῶν ΒΜ, ΘΝ, ΒΝ. κοινὸν προσκείσθω τὸ στερεόν, οὗ βάσις μὲν τὸ ΑΒ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΘΜ·

And the parallelogram CF is also equal to the parallelogram BM, and CG to BN; for they are opposite; therefore also the prism contained by two triangles AFG, DCE and three parallelograms AD, DG, CG is equal to the prism contained by two triangles MLN, TBK and three parallelograms BM, TN, BN.

ὅλον ἄρα τὸ ΓΜ στερεὸν παραλληλεπίπεδον ὅλῳ τῷ ΓΝ στερεῷ παραλληλεπιπέδῳ ἴσον ἐστίν.

Let there be added the common solid whose base is the parallelogram AB, and opposite to it GETM; therefore the whole solid parallelepiped CM is equal to the whole solid parallelepiped CN.

τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.

Therefore, solid parallelepipeds which are on the same base and have the same height, and whose uprights are on the same straight lines, are equal to one another; which was required to prove.

§11.prop.30τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν.

Solid parallelepipeds which are on the same base and have the same height, and whose uprights are not on the same straight lines, are equal to one another.

ἔστω ἐπὶ τῆς αὐτῆς βάσεως τῆς ΑΒ στερεὰ παραλληλεπίπεδα τὰ ΓΜ, ΓΝ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΗ, ΛΜ, ΛΝ, ΓΔ, ΓΕ, ΒΘ, ΒΚ μὴ ἔστωσαν ἐπὶ τῶν αὐτῶν εὐθειῶν· λέγω, ὅτι ἴσον ἐστὶ τὸ ΓΜ στερεὸν τῷ ΓΝ στερεῷ.

Let the solid parallelepipeds CM, CN be on the same base AB and have the same height, and let their uprights AF, AG, LM, LN, CD, CE, BT, BK not be on the same straight lines; I say that the solid CM is equal to the solid CN.

ἐκβεβλήσθωσαν γὰρ αἱ ΝΚ, ΔΘ καὶ συμπιπτέτωσαν ἀλλήλαις κατὰ τὸ Ρ, καὶ ἔτι ἐκβεβλήσθωσαν αἱ ΖΜ, ΗΕ ἐπὶ τὰ Ο, Π, καὶ ἐπεζεύχθωσαν αἱ ΑΞ, ΛΟ, ΓΠ, ΒΡ. ἴσον δή ἐστι τὸ ΓΜ στερεόν, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΖΔΘΜ, τῷ ΓΟ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ·

For let NK, DT be produced and meet one another at R, and let FM, GE be further produced to O, P, and let AX, LO, CP, BR be joined.

ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΖ, ΑΞ, ΛΜ, ΛΟ, ΓΔ, ΓΠ, ΒΘ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΖΟ, ΔΡ. ἀλλὰ τὸ ΓΟ στερεόν, οὗ βάσις μέν ἐστι τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΞΠΡΟ, ἴσον ἐστὶ τῷ ΓΝ στερεῷ, οὗ βάσις μὲν τὸ ΑΓΒΛ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΚΝ·

Indeed, the solid CM, whose base is the parallelogram ACBL, and opposite to it FDTM, is equal to the solid CO, whose base is the parallelogram ACBL, and opposite to it XPRO; for they are on the same base ACBL and have the same height, and their uprights AF, AX, LM, LO, CD, CP, BT, BR are on the same straight lines FO, DR.

ἐπί τε γὰρ πάλιν τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι αἱ ΑΗ, ΑΞ, ΓΕ, ΓΠ, ΛΝ, ΛΟ, ΒΚ, ΒΡ ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν τῶν ΗΠ, ΝΡ. ὥστε καὶ τὸ ΓΜ στερεὸν ἴσον ἐστὶ τῷ ΓΝ στερεῷ.

But the solid CO, whose base is the parallelogram ACBL, and opposite to it XPRO, is equal to the solid CN, whose base is the parallelogram ACBL, and opposite to it GEKN; for again they are on the same base ACBL and have the same height, and their uprights AG, AX, CE, CP, LN, LO, BK, BR are on the same straight lines GP, NR. So that the solid CM is also equal to the solid CN.

τὰ ἄρα ἐπὶ τῆς αὐτῆς βάσεως στερεὰ παραλληλεπίπεδα καὶ ὑπὸ τὸ αὐτὸ ὕψος, ὧν αἱ ἐφεστῶσαι οὐκ εἰσὶν ἐπὶ τῶν αὐτῶν εὐθειῶν, ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.

Therefore, solid parallelepipeds which are on the same base and have the same height, and whose uprights are not on the same straight lines, are equal to one another; which was required to prove.

Notes

prop.29ὧν αἱ ἐφεστῶσαι ἐπὶ τῶν αὐτῶν εἰσιν εὐθειῶν — The relative pronoun ὧν is a neuter plural genitive, taking the main subject τὰ ... στερεὰ παραλληλεπίπεδα as its antecedent, expressing a possessive relation ('of which solid parallelepipeds'). αἱ ἐφεστῶσαι is the feminine plural nominative of the perfect active participle of ἐφίστημι ('to stand up/straight'), functioning substantively here as the subject referring to the 'uprights' or 'standing edges'.
prop.29κοινὸν προσκείσθω τὸ στερεόν, οὗ βάσις μὲν τὸ ΑΒ παραλληλόγραμμον, ἀπεναντίον δὲ τὸ ΗΕΘΜ· — The relative pronoun οὗ (neuter singular genitive) is a possessive genitive with the preceding τὸ στερεόν as its antecedent. Within the relative clause, in conjunction with the contrastive particles μὲν... δέ..., the copula ἐστί is omitted, so that the structure is to be understood as: οὗ [ἐστι] βάσις μὲν τὸ AB ... [ἐστι] δὲ τὸ GETM (or ΗΕΘΜ in Greek letter systems).
prop.30ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΑΓΒΛ καὶ ὑπὸ τὸ αὐτὸ ὕψος — By the coordinating particles τε... καί..., two prepositional phrases governing different cases are juxtaposed. The former is ἐπί + genitive (τῆς αὐτῆς βάσεως) expressing location, and the latter is ὑπό + accusative (τὸ αὐτὸ ὕψος) expressing condition or property. Both function with equal grammatical standing as predicative complements of the shared verb εἰσι.

Cite this passage

Euclid, Elements §11.prop.29-11.prop.30. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:11.prop.29-11.prop.30

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