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Euclid · Elements §1.prop.36-1.prop.38

Equivalence of Parallelograms and Triangles with Equal Bases

Passage 20 of 316 · Greek

Summary

Proposition 36 proves that parallelograms on equal bases and in the same parallels are equal, while Propositions 37 and 38 prove that triangles on the same or equal bases and in the same parallels are equal to one another.

§1.prop.36τὰ παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.
Parallelograms which are on equal bases and in the same parallels are equal to one another.
ἔστω παραλληλόγραμμα τὰ ΑΒΓΔ, ΕΖΗΘ ἐπὶ ἴσων βάσεων ὄντα τῶν ΒΓ, ΖΗ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΘ, ΒΗ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ. ἐπεζεύχθωσαν γὰρ αἱ ΒΕ, ΓΘ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΓ τῇ ΖΗ, ἀλλὰ ἡ ΖΗ τῇ ΕΘ ἐστιν ἴση, καὶ ἡ ΒΓ ἄρα τῇ ΕΘ ἐστιν ἴση.
Let ABCD, EFGH be parallelograms on equal bases BC, FG and in the same parallels AH, BG; I say that the parallelogram ABCD is equal to EFGH.
εἰσὶ δὲ καὶ παράλληλοι.
For let BE, CH be joined. And since BC is equal to FG, but FG is equal to EH, therefore BC is also equal to EH. And they are also parallel.
καὶ ἐπιζευγνύουσιν αὐτὰς αἱ ΕΒ, ΘΓ· αἱ δὲ τὰς ἴσας τε καὶ παραλλήλους ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι ἴσαι τε καὶ παράλληλοί εἰσι·
And EB, HC join them; but those joining equal and parallel on the same sides are themselves equal and parallel.
. παραλληλόγραμμον ἄρα ἐστὶ τὸ ΕΒΓΘ. καί ἐστιν ἴσον τῷ ΑΒΓΔ· βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει τὴν ΒΓ, καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστὶν αὐτῷ ταῖς ΒΓ, ΑΘ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΕΖΗΘ τῷ αὐτῷ τῷ ΕΒΓΘ ἐστιν ἴσον· ὥστε καὶ τὸ ΑΒΓΔ παραλληλόγραμμον τῷ ΕΖΗΘ ἐστιν ἴσον.
Therefore EBCH is a parallelogram. And it is equal to ABCD; for it has the same base BC with it, and is in the same parallels BC, AH with it. For the same reason, EFGH is also equal to the same EBCH; so that the parallelogram ABCD is also equal to EFGH.
τὰ ἄρα παραλληλόγραμμα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore, parallelograms which are on equal bases and in the same parallels are equal to one another; which was to be proved.
§1.prop.37τὰ τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.
Triangles which are on the same base and in the same parallels are equal to one another.
ἔστω τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΔ, ΒΓ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ.
Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC; I say that the triangle ABC is equal to the triangle DBC.
Ἐκβεβλήσθω ἡ ΑΔ ἐφʼ ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε, Ζ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΕ, διὰ δὲ τοῦ Γ τῇ ΒΔ παράλληλος ἤχθω ἡ ΓΖ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΕΒΓΑ, ΔΒΓΖ·
Let AD be produced on both sides to E, F; and through B let BE be drawn parallel to CA, and through C let CF be drawn parallel to BD.
καί εἰσιν ἴσα· ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΕΖ· καί ἐστι τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον· ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει· τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ἥμισυ τὸ ΔΒΓ τρίγωνον· ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει.
Therefore each of EBCA, DBCF is a parallelogram; and they are equal; for they are on the same base BC and in the same parallels BC, EF; and the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it; and the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it.
. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ.
Therefore the triangle ABC is equal to the triangle DBC.
τὰ ἄρα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore, triangles which are on the same base and in the same parallels are equal to one another; which was to be proved.
§1.prop.38τὰ τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν.
Triangles which are on equal bases and in the same parallels are equal to one another.
ἔστω τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΑΔ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF.
Ἐκβεβλήσθω γὰρ ἡ ΑΔ ἐφʼ ἑκάτερα τὰ μέρη ἐπὶ τὰ Η, θ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Ζ τῇ ΔΕ παράλληλος ἤχθω ἡ ΖΘ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΗΒΓΑ, ΔΕΖΘ·
For let AD be produced on both sides to G, H; and through B let BG be drawn parallel to CA, and through F let FH be drawn parallel to DE.
καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΗΘ· καί ἐστι τοῦ μὲν ΗΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον. ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει· τοῦ δὲ ΔΕΖΘ παραλληλογράμμου ἥμισυ τὸ ΖΕΔ τρίγωνον· ἡ γὰρ ΔΖ διάμετρος αὐτὸ δίχα τέμνει· .
Therefore each of GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH; for they are on equal bases BC, EF and in the same parallels BF, GH; and the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it; and the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it.
ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ.
Therefore the triangle ABC is equal to the triangle DEF.
τὰ ἄρα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore, triangles which are on equal bases and in the same parallels are equal to one another; which was to be proved.

Notes

  1. 1.prop.36αἱ δὲ τὰς ἴσας τε καὶ παραλλήλοις ἐπὶ τὰ αὐτὰ μέρη ἐπιζευγνύουσαι — Meaning 'and those [straight lines] joining equal and parallel [straight lines] on the same sides.' The feminine plural article and participle αἱ ... ἐπιζευγνύουσαι modify the omitted feminine noun εὐθεῖαι (straight lines), referring to the theorem proved in Proposition 33.
  2. 1.prop.37ἥμισυ — A neuter noun (or adjective) meaning 'half.' Here, a Common Notion (that the halves of equal things are equal to one another) is implicitly applied to deduce that the two triangles, which are halves of the equal parallelograms, are equal to one another.
  3. 1.prop.37Ἐκβεβλήσθω — Third-person singular present imperative passive of ἐκβάλλω (to produce, extend). A formulaic expression in geometrical constructions meaning 'let [a straight line] be produced.'

Cite this passage

Euclid, Elements §1.prop.36-1.prop.38. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.36-1.prop.38

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