Euclid · Elements §2.prop.4

Square on a Divided Line and the Square of a Sum

Passage 27 of 316 · Greek

Summary

Demonstrates that when a straight line is cut at random, the square on the whole is equal to the sum of the squares on the segments and twice the rectangle contained by the segments (the geometric proof of the square of a sum).

§2.prop.4ἐὰν εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ.

If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments.

εὐθεῖα γὰρ γραμμὴ ἡ ΑΒ τετμήσθω, ὡς ἔτυχεν, κατὰ τὸ Γ. λέγω, ὅτι τὸ ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ.

For let the straight line AB be cut at random at C; I say that the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

Ἀναγεγράφθω γὰρ ἀπὸ τῆς ΑΒ τετράγωνον τὸ ΑΔΕΒ, καὶ ἐπεζεύχθω ἡ ΒΔ, καὶ διὰ μὲν τοῦ Γ ὁποτέρᾳ τῶν ΑΔ, ΕΒ παράλληλος ἤχθω ἡ ΓΖ, διὰ δὲ τοῦ Η ὁποτέρᾳ τῶν ΑΒ, ΔΕ παράλληλος ἤχθω ἡ ΘΚ. καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΓΖ τῇ ΑΔ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΓΗΒ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΔΒ. ἀλλʼ ἡ ὑπὸ ΑΔΒ τῇ ὑπὸ ΑΒΔ ἐστιν ἴση, ἐπεὶ καὶ πλευρὰ ἡ ΒΑ τῇ ΑΔ ἐστιν ἴση· καὶ ἡ ὑπὸ ΓΗΒ ἄρα γωνία τῇ ὑπὸ ΗΒΓ ἐστιν ἴση· ὥστε καὶ πλευρὰ ἡ ΒΓ πλευρᾷ τῇ ΓΗ ἐστιν ἴση·

For let the square ADEB be described on AB, and let BD be joined, and through C let CZ be drawn parallel to either AD or EB, and through H let QK be drawn parallel to either AB or DE. And since CZ is parallel to AD, and BD has fallen upon them, the exterior angle GHB is equal to the interior and opposite angle ADB. But the angle ADB is equal to the angle ABD, since the side BA is also equal to AD; therefore the angle GHB is also equal to the angle HBG; so that the side BC is also equal to the side GH.

ἀλλʼ ἡ μὲν ΓΒ τῇ ΗΚ ἐστιν ἴση, ἡ δὲ ΓΗ τῇ ΚΒ· καὶ ἡ ΗΚ ἄρα τῇ ΚΒ ἐστιν ἴση·

But, on the one hand, GB is equal to HK, and on the other, GH to KB; therefore HK is also equal to KB.

ἰσόπλευρον ἄρα ἐστὶ τὸ ΓΗΚΒ. λέγω δή, ὅτι καὶ ὀρθογώνιον.

Therefore GHKB is equilateral. I say indeed that it is also right-angled.

ἐπεὶ γὰρ παράλληλός ἐστιν ἡ ΓΗ τῇ ΒΚ, αἱ ἄρα ὑπὸ ΚΒΓ, ΗΓΒ γωνίαι δύο ὀρθαῖς εἰσιν ἴσαι.

For since GH is parallel to BK, the angles KBG, HGB are equal to two right angles.

ὀρθὴ δὲ ἡ ὑπὸ ΚΒΓ· ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΒΓΗ· ὥστε καὶ αἱ ἀπεναντίον αἱ ὑπὸ ΓΗΚ, ΗΚΒ ὀρθαί εἰσιν.

And the angle KBG is right; therefore the angle BGH is also right; so that the opposite angles GHK, HKB are also right.

ὀρθογώνιον ἄρα ἐστὶ τὸ ΓΗΚΒ· ἐδείχθη δὲ καὶ ἰσόπλευρον· τετράγωνον ἄρα ἐστίν· καί ἐστιν ἀπὸ τῆς ΓΒ. διὰ τὰ αὐτὰ δὴ καὶ τὸ ΘΖ τετράγωνόν ἐστιν·

Therefore GHKB is right-angled; and it was also proved equilateral; therefore it is a square; and it is on GB.

καί ἐστιν ἀπὸ τῆς ΘΗ, τουτέστιν τῆς ΑΓ· τὰ ἄρα ΘΖ, ΚΓ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ εἰσιν.

For the same reasons indeed, QZ is also a square; and it is on QH, that is, on AC; therefore QZ, KC are squares on AC, CB.

καὶ ἐπεὶ ἴσον ἐστὶ τὸ ΑΗ τῷ ΗΕ, καί ἐστι τὸ ΑΗ τὸ ὑπὸ τῶν ΑΓ, ΓΒ· ἴση γὰρ ἡ ΗΓ τῇ ΓΒ· καὶ τὸ ΗΕ ἄρα ἴσον ἐστὶ τῷ ὑπὸ ΑΓ, ΓΒ· τὰ ἄρα ΑΗ, ΗΕ ἴσα ἐστὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ. ἔστι δὲ καὶ τὰ ΘΖ, ΓΚ τετράγωνα ἀπὸ τῶν ΑΓ, ΓΒ·

And since AH is equal to HE, and AH is the rectangle contained by AC, CB — for HC is equal to CB — therefore HE is also equal to the rectangle contained by AC, CB; therefore AH, HE are equal to twice the rectangle contained by AC, CB.

τὰ ἄρα τέσσαρα τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ἴσα ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ.

And QZ, GK are also squares on AC, CB; therefore the four QZ, GK, AH, HE are equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

ἀλλὰ τὰ ΘΖ, ΓΚ, ΑΗ, ΗΕ ὅλον ἐστὶ τὸ ΑΔΕΒ, ὅ ἐστιν ἀπὸ τῆς ΑΒ τετράγωνον· τὸ ἄρα ἀπὸ τῆς ΑΒ τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν ΑΓ, ΓΒ τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ.

But QZ, GK, AH, HE are the whole ADEB, which is the square on AB; therefore the square on AB is equal to the squares on AC, CB and twice the rectangle contained by AC, CB.

ἐὰν ἄρα εὐθεῖα γραμμὴ τμηθῇ, ὡς ἔτυχεν, τὸ ἀπὸ τῆς ὅλης τετράγωνον ἴσον ἐστὶ τοῖς τε ἀπὸ τῶν τμημάτων τετραγώνοις καὶ τῷ δὶς ὑπὸ τῶν τμημάτων περιεχομένῳ ὀρθογωνίῳ· ὅπερ ἔδει δεῖξαι. .

If therefore a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments; which was to be proved.

Notes

15ἡ ὑπὸ ΓΗΒ — The prepositional phrase with the feminine article "the [angle] under G, H, B" represents the angle GHB. Although the noun `γωνία` (angle) is explicitly written in this instance, it is more commonly omitted in geometric texts, with only the article `ἡ` indicating the angle.
30τουτέστιν τῆς ΑΓ — The genitive `τῆς ΑΓ` following `τουτέστιν` (that is) refers to "the square on AC", standing parallel to `τῆς ΘΗ` in the preceding clause. It is governed by the omitted preposition `ἀπὸ`. QH is equal to AC because they are opposite sides of the parallelogram AH.
7τῷ δὶς ὑπὸ τῶν ΑΓ, ΓΒ περιεχομένῳ ὀρθογωνίῳ — This phrase means "twice the rectangle contained by AC, CB". The adverb `δὶς` (twice) is placed between the article `τῷ` and the participle `περιεχομένῳ`, functioning attributively to modify the entire concept of the rectangle.

Cite this passage

Euclid, Elements §2.prop.4. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:2.prop.4

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