Humanitext Reader

Euclid · Elements §1.prop.42-1.prop.44

Construction of Equal Parallelograms and Equality of Complements

Passage 22 of 316 · Greek

Summary

Proposition 42 demonstrates the construction of a parallelogram equal to a given triangle with a given angle, Proposition 43 proves that the complements of parallelograms about the diameter are equal, and Proposition 44 shows how to apply a parallelogram equal to a given triangle to a given straight line in a given angle.

§1.prop.42τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.
To construct a parallelogram equal to a given triangle in a given rectilineal angle.
ἔστω τὸ μὲν δοθὲν τρίγωνον τὸ ΑΒΓ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ τῷ ΑΒΓ τριγώνῳ ἴσον παραλληλόγραμμον συστήσασθαι ἐν τῇ Δ γωνίᾳ εὐθυγράμμῳ.
Let the given triangle be ABC, and the given rectilineal angle D; it is required to construct a parallelogram equal to the triangle ABC in the rectilineal angle D.
τετμήσθω ἡ ΒΓ δίχα κατὰ τὸ Ε, καὶ ἐπεζεύχθω ἡ ΑΕ, καὶ συνεστάτω πρὸς τῇ ΕΓ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ σημείῳ τῷ Ε τῇ Δ γωνίᾳ ἴση ἡ ὑπὸ ΓΕΖ, καὶ διὰ μὲν τοῦ Α τῇ ΕΓ παράλληλος ἤχθω ἡ ΑΗ, διὰ δὲ τοῦ Γ τῇ ΕΖ παράλληλος ἤχθω ἡ ΓΗ·
Let BC be bisected at E, and let AE be joined; let the angle CEF be constructed equal to the angle D on the straight line EC and at the point E on it; and through A let AG be drawn parallel to EC, and through C let CG be drawn parallel to EF.
παραλληλόγραμμον ἄρα ἐστὶ τὸ ΖΕΓΗ. καὶ ἐπεὶ ἴση ἐστὶν ἡ ΒΕ τῇ ΕΓ, ἴσον ἐστὶ καὶ τὸ ΑΒΕ τρίγωνον τῷ ΑΕΓ τριγώνῳ· ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΕ, ΕΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΑΗ·
Therefore FECG is a parallelogram. And since BE is equal to EC, the triangle ABE is also equal to the triangle AEC; for they are on equal bases BE, EC and in the same parallels BC, AG.
διπλάσιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τοῦ ΑΕΓ τριγώνου.
Therefore the triangle ABC is double of the triangle AEC.
ἔστι δὲ καὶ τὸ ΖΕΓΗ παραλληλόγραμμον διπλάσιον τοῦ ΑΕΓ τριγώνου· βάσιν τε γὰρ αὐτῷ τὴν αὐτὴν ἔχει καὶ ἐν ταῖς αὐταῖς ἐστιν αὐτῷ παραλλήλοις·
But the parallelogram FECG is also double of the triangle AEC; for it has the same base with it and is in the same parallels with it.
ἴσον ἄρα ἐστὶ τὸ ΖΕΓΗ παραλληλόγραμμον τῷ ΑΒΓ τριγώνῳ.
Therefore the parallelogram FECG is equal to the triangle ABC.
καὶ ἔχει τὴν ὑπὸ ΓΕΖ γωνίαν ἴσην τῇ δοθείσῃ τῇ Δ. τῷ ἄρα δοθέντι τριγώνῳ τῷ ΑΒΓ ἴσον παραλληλόγραμμον συνέσταται τὸ ΖΕΓΗ ἐν γωνίᾳ τῇ ὑπὸ ΓΕΖ, ἥτις ἐστὶν ἴση τῇ Δ· ὅπερ ἔδει ποιῆσαι.
And it has the angle CEF equal to the given angle D. Therefore, the parallelogram FECG has been constructed equal to the given triangle ABC in the angle CEF, which is equal to D; which was to be done.
§1.prop.43παντὸς παραλληλογράμμου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν.
In any parallelogram the complements of the parallelograms about the diameter are equal to one another.
ἔστω παραλληλόγραμμον τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, περὶ δὲ τὴν ΑΓ παραλληλόγραμμα μὲν ἔστω τὰ ΖΘ, ΖΗ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΒΚ, ΚΔ· λέγω, ὅτι ἴσον ἐστὶ τὸ ΒΚ παραπλήρωμα τῷ ΚΔ παραπληρώματι.
Let ABCD be a parallelogram, and AC its diameter; about AC let FH, FG be parallelograms, and let the so-called complements be BK, KD; I say that the complement BK is equal to the complement KD.
ἐπεὶ γὰρ παραλληλόγραμμόν ἐστι τὸ ΑΒΓΔ, διάμετρος δὲ αὐτοῦ ἡ ΑΓ, ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΓΔ τριγώνῳ.
For, since ABCD is a parallelogram, and AC is its diameter, the triangle ABC is equal to the triangle ACD.
πάλιν, ἐπεὶ παραλληλόγραμμόν ἐστι τὸ ΕΘ, διάμετρος δὲ αὐτοῦ ἐστιν ἡ ΑΚ, ἴσον ἐστὶ τὸ ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ.
Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ΚΖΓ τρίγωνον τῷ ΚΗΓ ἐστιν ἴσον.
For the same reason the triangle KFC is also equal to KGC.
ἐπεὶ οὖν τὸ μὲν ΑΕΚ τρίγωνον τῷ ΑΘΚ τριγώνῳ ἐστὶν ἴσον, τὸ δὲ ΚΖΓ τῷ ΚΗΓ, τὸ ΑΕΚ τρίγωνον μετὰ τοῦ ΚΗΓ ἴσον ἐστὶ τῷ ΑΘΚ τριγώνῳ μετὰ τοῦ ΚΖΓ·
Therefore, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC.
ἔστι δὲ καὶ ὅλον τὸ ΑΒΓ τρίγωνον ὅλῳ τῷ ΑΔΓ ἴσον·
And the whole triangle ABC is also equal to the whole ADC.
λοιπὸν ἄρα τὸ ΒΚ παραπλήρωμα λοιπῷ τῷ ΚΔ παραπληρώματί ἐστιν ἴσον.
Therefore, the remaining complement BK is equal to the remaining complement KD.
παντὸς ἄρα παραλληλογράμμου χωρίου τῶν περὶ τὴν διάμετρον παραλληλογράμμων τὰ παραπληρώματα ἴσα ἀλλήλοις ἐστίν· ὅπερ ἔδει δεῖξαι.
Therefore, in any parallelogrammic area the complements of the parallelograms about the diameter are equal to one another; which was to be proved.
§1.prop.44παρὰ τὴν δοθεῖσαν εὐθεῖαν τῷ δοθέντι τριγώνῳ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν τῇ δοθείσῃ γωνίᾳ εὐθυγράμμῳ.
To a given straight line to apply a parallelogram equal to a given triangle in a given rectilineal angle.
ἔστω ἡ μὲν δοθεῖσα εὐθεῖα ἡ ΑΒ, τὸ δὲ δοθὲν τρίγωνον τὸ Γ, ἡ δὲ δοθεῖσα γωνία εὐθύγραμμος ἡ Δ· δεῖ δὴ παρὰ τὴν δοθεῖσαν εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβαλεῖν ἐν ἴσῃ τῇ Δ γωνίᾳ.
Let the given straight line be AB, the given triangle C, and the given rectilineal angle D; it is required to apply to the given straight line AB a parallelogram equal to the given triangle C in an angle equal to the angle D.
συνεστάτω τῷ Γ τριγώνῳ ἴσον παραλληλόγραμμον τὸ ΒΕΖΗ ἐν γωνίᾳ τῇ ὑπὸ ΕΒΗ, ἥ ἐστιν ἴση τῇ Δ· καὶ κείσθω ὥστε ἐπʼ εὐθείας εἶναι τὴν ΒΕ τῇ ΑΒ, καὶ διήχθω ἡ ΖΗ ἐπὶ τὸ Θ, καὶ διὰ τοῦ Α ὁποτέρᾳ τῶν ΒΗ, ΕΖ παράλληλος ἤχθω ἡ ΑΘ, καὶ ἐπεζεύχθω ἡ ΘΒ. καὶ ἐπεὶ εἰς παραλλήλους τὰς ΑΘ, ΕΖ εὐθεῖα ἐνέπεσεν ἡ ΘΖ, αἱ ἄρα ὑπὸ ΑΘΖ, ΘΖΕ γωνίαι δυσὶν ὀρθαῖς εἰσιν ἴσαι.
Let the parallelogram BEFG be constructed equal to the triangle C in the angle EBG which is equal to D; and let it be placed so that BE is in a straight line with AB; let FG be drawn through to H, and through A let AH be drawn parallel to either BG or EF, and let HB be joined. And since a straight line HF falls upon the parallel straight lines AH, EF, the angles AHF, HFE are equal to two right angles.
αἱ ἄρα ὑπὸ ΒΘΗ, ΗΖΕ δύο ὀρθῶν ἐλάσσονές εἰσιν· αἱ δὲ ἀπὸ ἐλασσόνων ἢ δύο ὀρθῶν εἰς ἄπειρον ἐκβαλλόμεναι συμπίπτουσιν· αἱ ΘΒ, ΖΕ ἄρα ἐκβαλλόμεναι συμπεσοῦνται.
Therefore the angles BHG, GFE are less than two right angles; and straight lines produced from angles less than two right angles meet if produced indefinitely; therefore HB, FE, if produced, will meet.
ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν κατὰ τὸ Κ, καὶ διὰ τοῦ Κ σημείου ὁποτέρᾳ τῶν ΕΑ, ΖΘ παράλληλος ἤχθω ἡ ΚΛ, καὶ ἐκβεβλήσθωσαν αἱ ΘΑ, ΗΒ ἐπὶ τὰ Λ, Μ σημεῖα.
Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, and let HA, GB be produced to the points L, M.
παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΛΚΖ, διάμετρος δὲ αὐτοῦ ἡ ΘΚ, περὶ δὲ τὴν ΘΚ παραλληλόγραμμα μὲν τὰ ΑΗ, ΜΕ, τὰ δὲ λεγόμενα παραπληρώματα τὰ ΛΒ, ΒΖ·
Therefore HLKF is a parallelogram, HK is its diameter, the parallelograms about HK are AG, ME, and the so-called complements are LB, BF; therefore LB is equal to BF.
ἴσον ἄρα ἐστὶ τὸ ΛΒ τῷ ΒΖ. ἀλλὰ τὸ ΒΖ τῷ Γ τριγώνῳ ἐστὶν ἴσον· καὶ τὸ ΛΒ ἄρα τῷ Γ ἐστιν ἴσον.
But BF is equal to the triangle C; therefore LB is also equal to C.
καὶ ἐπεὶ ἴση ἐστὶν ἡ ὑπὸ ΗΒΕ γωνία τῇ ὑπὸ ΑΒΜ, ἀλλὰ ἡ ὑπὸ ΗΒΕ τῇ Δ ἐστιν ἴση, καὶ ἡ ὑπὸ ΑΒΜ ἄρα τῇ Δ γωνίᾳ ἐστὶν ἴση.
And since the angle GBE is equal to the angle ABM, and the angle GBE is equal to D, the angle ABM is also equal to the angle D.
παρὰ τὴν δοθεῖσαν ἄρα εὐθεῖαν τὴν ΑΒ τῷ δοθέντι τριγώνῳ τῷ Γ ἴσον παραλληλόγραμμον παραβέβληται τὸ ΛΒ ἐν γωνίᾳ τῇ ὑπὸ ΑΒΜ, ἥ ἐστιν ἴση τῇ Δ· ὅπερ ἔδει ποιῆσαι.
Therefore, to the given straight line AB the parallelogram LB has been applied equal to the given triangle C in the angle ABM, which is equal to D; which was to be done.

Notes

  1. §1.prop.42συστήσασθαι — The infinitive συστήσασθαι is used in the title of the proposition to express the task to be performed in an imperative or objective sense.
  2. §1.prop.42τετμήσθω — The third-person singular perfect passive imperative τετμήσθω (let it be bisected) indicates that the state of bisection is to be established and maintained.
  3. §1.prop.43παντὸς παραλληλογράμμου — The genitive case of παντὸς παραλληλογράμμου functions as a partitive or locative genitive, indicating the whole domain in which the theorem holds.
  4. §1.prop.44δύο ὀρθῶν — Positioned right after the comparative adjective ἐλάσσονες, δύο ὀρθῶν is a genitive of comparison representing the standard 'than two right angles'.
  5. §1.prop.44ὁποτέρᾳ τῶν ΕΑ, ΖΘ — The dative pronoun ὁποτέρᾳ is governed by the adjective παράλληλος, meaning 'parallel to either of EA, ΖΘ'.

Cite this passage

Euclid, Elements §1.prop.42-1.prop.44. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.42-1.prop.44

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