Page 27 - Euclid's Elements of Geometry

ST EW aþ. kdþ ELEMENTS BOOK 1

Proposition 24
.
᾿Εὰν δύο τρίγωνα τὰς δύο πλευρὰς [ταῖς] δύο πλευραῖς is) the very thing it was required to do.
If two triangles have two sides equal to two sides, re-
ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας spectively, but (one) has the angle encompassed by the
μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ equal straight-lines greater than the (corresponding) an-
τὴν βάσιν τῆς βάσεως μείζονα ἕξει. gle (in the other), then (the former triangle) will also
have a base greater than the base (of the latter).
Α ∆ A D

Ε E
Β B

Ζ F
Η G
Γ C
῎Εστω δύο τρίγωνα τὰ ΑΒΓ, ΔΕΖ τὰς δύο πλευρὰς Let ABC and DEF be two triangles having the two
τὰς ΑΒ, ΑΓ ταῖς δύο πλευραῖς ταῖς ΔΕ, ΔΖ ἴσας ἔχοντα sides AB and AC equal to the two sides DE and DF,
ἑκατέραν ἑκατέρᾳ, τὴν μὲν ΑΒ τῇ ΔΕ τὴν δὲ ΑΓ τῇ ΔΖ, ἡ respectively. (That is), AB (equal) to DE, and AC to
δὲ πρὸς τῷ Α γωνία τῆς πρὸς τῷ Δ γωνίας μείζων ἔστω· DF. Let them also have the angle at A greater than the
λέγω, ὅτι καὶ βάσις ἡ ΒΓ βάσεως τῆς ΕΖ μείζων ἐστίν. angle at D. I say that the base BC is also greater than
᾿Επεὶ γὰρ μείζων ἡ ὑπὸ ΒΑΓ γωνία τῆς ὑπὸ ΕΔΖ the base EF.
γωνίας, συνεστάτω πρὸς τῇ ΔΕ εὐθείᾳ καὶ τῷ πρὸς αὐτῇ For since angle BAC is greater than angle EDF,
σημείῳ τῷ Δ τῇ ὑπὸ ΒΑΓ γωνίᾳ ἴση ἡ ὑπὸ ΕΔΗ, καὶ κείσθω let (angle) EDG, equal to angle BAC, have been
ὁποτέρᾳ τῶν ΑΓ, ΔΖ ἴση ἡ ΔΗ, καὶ ἐπεζεύχθωσαν αἱ ΕΗ, constructed at the point D on the straight-line DE
ΖΗ. [Prop. 1.23]. And let DG be made equal to either of
᾿Επεὶ οὖν ἴση ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΓ τῇ ΔΗ, AC or DF [Prop. 1.3], and let EG and FG have been
δύο δὴ αἱ ΒΑ, ΑΓ δυσὶ ταῖς ΕΔ, ΔΗ ἴσαι εἰσὶν ἑκατέρα joined.
ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΒΑΓ γωνίᾳ τῇ ὑπὸ ΕΔΗ ἴση· Therefore, since AB is equal to DE and AC to DG,
βάσις ἄρα ἡ ΒΓ βάσει τῇ ΕΗ ἐστιν ἴση. πάλιν, ἐπεὶ ἴση the two (straight-lines) BA, AC are equal to the two
ἐστὶν ἡ ΔΖ τῇ ΔΗ, ἴση ἐστὶ καὶ ἡ ὑπὸ ΔΗΖ γωνία τῇ ὑπὸ (straight-lines) ED, DG, respectively. Also the angle
ΔΖΗ· μείζων ἄρα ἡ ὑπὸ ΔΖΗ τῆς ὑπὸ ΕΗΖ· πολλῷ ἄρα BAC is equal to the angle EDG. Thus, the base BC
μείζων ἐστὶν ἡ ὑπὸ ΕΖΗ τῆς ὑπὸ ΕΗΖ. καὶ ἐπεὶ τρίγωνόν is equal to the base EG [Prop. 1.4]. Again, since DF
ἐστι τὸ ΕΖΗ μείζονα ἔχον τὴν ὑπὸ ΕΖΗ γωνίαν τῆς ὑπὸ is equal to DG, angle DGF is also equal to angle DFG
ΕΗΖ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, [Prop. 1.5]. Thus, DFG (is) greater than EGF. Thus,
μείζων ἄρα καὶ πλευρὰ ἡ ΕΗ τῆς ΕΖ. ἴση δὲ ἡ ΕΗ τῇ ΒΓ· EFG is much greater than EGF. And since triangle
μείζων ἄρα καὶ ἡ ΒΓ τῆς ΕΖ. EFG has angle EFG greater than EGF, and the greater
᾿Εὰν ἄρα δύο τρίγωνα τὰς δύο πλευρὰς δυσὶ πλευραῖς angle is subtended by the greater side [Prop. 1.19], side
ἴσας ἔχῃ ἑκατέραν ἑκατέρᾳ, τὴν δὲ γωνίαν τῆς γωνίας EG (is) thus also greater than EF. But EG (is) equal to
μείζονα ἔχῃ τὴν ὑπὸ τῶν ἴσων εὐθειῶν περιεχομένην, καὶ BC. Thus, BC (is) also greater than EF.
τὴν βάσιν τῆς βάσεως μείζονα ἕξει· ὅπερ ἔδει δεῖξαι. Thus, if two triangles have two sides equal to two
sides, respectively, but (one) has the angle encompassed
by the equal straight-lines greater than the (correspond-
ing) angle (in the other), then (the former triangle) will
also have a base greater than the base (of the latter).

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