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Euclid · Elements §1.prop.19-1.prop.21

Sides Subtending Angles and Triangle Inequality

Passage 12 of 316 · Greek

Summary

This chunk proves that in a triangle, the greater angle is subtended by the greater side (Proposition 19), that any two sides taken together are greater than the remaining side (Proposition 20), and that two lines constructed inside a triangle from the extremities of one side are less than the other two sides but contain a greater angle (Proposition 21).

§1.prop.19παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει.
In any triangle, the greater side subtends the greater angle.
ἔστω τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῆς ὑπὸ ΒΓΑ· λέγω, ὅτι καὶ πλευρὰ ἡ ΑΓ πλευρᾶς τῆς ΑΒ μείζων ἐστίν.
Let ABC be a triangle having the angle ABC greater than BCA; I say that the side AC is also greater than the side AB.
εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ἢ ἐλάσσων·
For if not, AC is either equal to AB or less.
ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ· ἴση γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ· οὐκ ἔστι δέ·
Now, AC is not equal to AB; for the angle ABC would also have been equal to ACB; but it is not; therefore AC is not equal to AB.
οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ. οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ· ἐλάσσων γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῆς ὑπὸ ΑΓΒ· οὐκ ἔστι δέ·
Nor indeed is AC less than AB; for the angle ABC would also have been less than ACB; but it is not; therefore AC is not less than AB.
οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση ἐστίν.
And it was proved that it is not equal either.
μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ. παντὸς ἄρα τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει·
Therefore AC is greater than AB.
ὅπερ ἔδει δεῖξαι.
Therefore, in any triangle, the greater side subtends the greater angle. - Being what it was required to prove.
§1.prop.20παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι.
In any triangle, two sides taken together in any manner are greater than the remaining side.
ἔστω γὰρ τρίγωνον τὸ ΑΒΓ· λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ΒΑ, ΑΓ τῆς ΒΓ, αἱ δὲ ΑΒ, ΒΓ τῆς ΑΓ, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. διήχθω γὰρ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον, καὶ κείσθω τῇ ΓΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ·
For let ABC be a triangle; I say that in the triangle ABC, two sides taken together in any manner are greater than the remaining side, namely, BA, AC than BC, AB, BC than AC, and BC, CA than AB. For let BA be produced to the point D, and let AD be made equal to CA, and let DC be joined.
μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ·
Since therefore DA is equal to AC, the angle ADC is also equal to ACD; therefore the angle BCD is greater than ADC.
καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν τῆς ὑπὸ ΒΔΓ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων.
And since DCB is a triangle having the angle BCD greater than BDC, and the greater side subtends the greater angle, therefore DB is greater than BC.
ἴση δὲ ἡ ΔΑ τῇ ΑΓ· μείζονες ἄρα αἱ ΒΑ, ΑΓ τῆς ΒΓ·
And DA is equal to AC; therefore BA, AC are greater than BC.
ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΒ, ΒΓ τῆς ΓΑ μείζονές εἰσιν, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι·
Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB.
ὅπερ ἔδει δεῖξαι.
Therefore, in any triangle, two sides taken together in any manner are greater than the remaining side. - Being what it was required to prove.
§1.prop.21ἐὰν τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μὲν ἔσονται, μείζονα δὲ γωνίαν περιέξουσιν.
If on one of the sides of a triangle, from its extremities, two straight lines are constructed meeting within the triangle, the constructed straight lines will be less than the remaining two sides of the triangle, but will contain a greater angle.
τριγώνου γὰρ τοῦ ΑΒΓ ἐπὶ μιᾶς τῶν πλευρῶν τῆς ΒΓ ἀπὸ τῶν περάτων τῶν Β, Γ δύο εὐθεῖαι ἐντὸς συνεστάτωσαν αἱ ΒΔ, ΔΓ· λέγω, ὅτι αἱ ΒΔ, ΔΓ τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν τῶν ΒΑ, ΑΓ ἐλάσσονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσι τὴν ὑπὸ ΒΔΓ τῆς ὑπὸ ΒΑΓ. διήχθω γὰρ ἡ ΒΔ ἐπὶ τὸ Ε. καὶ ἐπεὶ παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσιν, τοῦ ΑΒΕ ἄρα τριγώνου αἱ δύο πλευραὶ αἱ ΑΒ, ΑΕ τῆς ΒΕ μείζονές εἰσιν·
For on one side BC of the triangle ABC, from the extremities B, C, let two straight lines BD, DC be constructed meeting within the triangle; I say that BD, DC are less than the remaining two sides BA, AC of the triangle, but contain a greater angle, namely BDC than BAC. For let BD be produced to E. And since, in any triangle, two sides are greater than the remaining side, therefore in the triangle ABE, the two sides AB, AE are greater than BE.
κοινὴ προσκείσθω ἡ ΕΓ· αἱ ἄρα ΒΑ, ΑΓ τῶν ΒΕ, ΕΓ μείζονές εἰσιν.
Let EC be added to each; therefore BA, AC are greater than BE, EC.
πάλιν, ἐπεὶ τοῦ ΓΕΔ τριγώνου αἱ δύο πλευραὶ αἱ ΓΕ, ΕΔ τῆς ΓΔ μείζονές εἰσιν, κοινὴ προσκείσθω ἡ ΔΒ· αἱ ΓΕ, ΕΒ ἄρα τῶν ΓΔ, ΔΒ μείζονές εἰσιν.
Again, since in the triangle CED, the two sides CE, ED are greater than CD, let DB be added to each; therefore CE, EB are greater than CD, DB.
ἀλλὰ τῶν ΒΕ, ΕΓ μείζονες ἐδείχθησαν αἱ ΒΑ, ΑΓ· πολλῷ ἄρα αἱ ΒΑ, ΑΓ τῶν ΒΔ, ΔΓ μείζονές εἰσιν.
But BA, AC were proved greater than BE, EC; therefore BA, AC are much greater than BD, DC.
πάλιν, ἐπεὶ παντὸς τριγώνου ἡ ἐκτὸς γωνία τῆς ἐντὸς καὶ ἀπεναντίον μείζων ἐστίν, τοῦ ΓΔΕ ἄρα τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΓΕΔ. διὰ ταὐτὰ τοίνυν καὶ τοῦ ΑΒΕ τριγώνου ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΕΒ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἀλλὰ τῆς ὑπὸ ΓΕΒ μείζων ἐδείχθη ἡ ὑπὸ ΒΔΓ· πολλῷ ἄρα ἡ ὑπὸ ΒΔΓ μείζων ἐστὶ τῆς ὑπὸ ΒΑΓ. ἐὰν ἄρα τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν, αἱ συσταθεῖσαι τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες μέν εἰσιν, μείζονα δὲ γωνίαν περιέχουσιν· ὅπερ ἔδει δεῖξαι.
Again, since, in any triangle, the exterior angle is greater than the interior and opposite angle, therefore in the triangle CDE, the exterior angle BDC is greater than CED. For the same reason, therefore, in the triangle ABE, the exterior angle CEB is greater than BAC. But BDC was proved greater than CEB; therefore BDC is much greater than BAC. Therefore, if on one of the sides of a triangle, from its extremities, two straight lines are constructed meeting within the triangle, the constructed straight lines are less than the remaining two sides of the triangle, but contain a greater angle. - Being what it was required to prove.

Notes

  1. §1.prop.19ἴση γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ — The combination of the particle `ἄν` with the imperfect indicative `ἦν` forms a counterfactual apodosis ("for the angle ABC would have been equal to ACB [if AC were equal to AB]"). Since the conclusion is false, the assumption of equality is rejected by contradiction.
  2. §1.prop.20πάντῃ μεταλαμβανόμεναι — The present passive participle of `μεταλαμβάνω` (to take in turn, exchange) in the nominative feminine plural, agreeing with the subject `αἱ duo πλευραὶ`. It functions adverbially, meaning "taken together in any manner of combination" or "any two taken interchangeably."
  3. §1.prop.21ἐὰν τριγώνου ἐπὶ μιᾶς τῶν πλευρῶν ἀπὸ τῶν περάτων δύο εὐθεῖαι ἐντὸς συσταθῶσιν — A conditional protasis introduced by `ἐὰν` with the third-person plural aorist passive subjunctive `συσταθῶσιν` (from `συνίστημι`). Since the apodosis contains future-tense verbs (`ἔσονται` and `περιέξουσιν`), this forms a future-more-vivid or general conditional construction.
  4. §1.prop.21τῶν λοιπῶν τοῦ τριγώνου δύο πλευρῶν ἐλάττονες — The comparative adjective `ἐλάττονες` (and subsequently `μείζονα`) governs the genitive of comparison `τῶν λοιπῶν ... πλευρῶν` ("less than the remaining sides"). The phrase `τοῦ τριγώνου` is a possessive genitive qualifying `πλευρῶν` ("of the triangle").

Cite this passage

Euclid, Elements §1.prop.19-1.prop.21. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:1.prop.19-1.prop.21

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