Euclid · Elements §10.prop3.91

Side of Area by Rational Line and First Apotome

Passage 220 of 316 · Greek

Summary

It is proved that if an area is contained by a rational straight line and a first apotome, the straight line producing the area (its square root) is an apotome.

§10.prop3.91ἐὰν χωρίον περιέχηται ὑπὸ ῥητῆς καὶ ἀποτομῆς πρώτης, ἡ τὸ χωρίον δυναμένη ἀποτομή ἐστιν.

If an area be contained by a rational straight line and a first apotome, the straight line producing the area is an apotome.

περιεχέσθω γὰρ χωρίον τὸ ΑΒ ὑπὸ ῥητῆς τῆς ΑΓ καὶ ἀποτομῆς πρώτης τῆς ΑΔ· λέγω, ὅτι ἡ τὸ ΑΒ χωρίον δυναμένη ἀποτομή ἐστιν.

For let the area AB be contained by the rational straight line AC and the first apotome AD; I say that the straight line producing the area AB is an apotome.

ἐπεὶ γὰρ ἀποτομή ἐστι πρώτη ἡ ΑΔ, ἔστω αὐτῇ προσαρμόζουσα ἡ ΔΗ· αἱ ΑΗ, ΗΔ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι.

For since AD is a first apotome, let DH be its annex; therefore AH, HD are rational straight lines commensurable in square only.

καὶ ὅλη ἡ ΑΗ σύμμετρός ἐστι τῇ ἐκκειμένῃ ῥητῇ τῇ ΑΓ, καὶ ἡ ΑΗ τῆς ΗΔ μεῖζον δύναται τῷ ἀπὸ συμμέτρου ἑαυτῇ μήκει· ἐὰν ἄρα τῷ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ΔΗ ἴσον παρὰ τὴν ΑΗ παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς σύμμετρα αὐτὴν διαιρεῖ.

And the whole AH is commensurable in length with the set-out rational straight line AC, and AH is greater in square than HD by the square on a straight line commensurable in length with itself; if therefore a rectangle equal to the fourth part of the square on DH be applied to AH, deficient by a square figure, it divides it into parts commensurable in length.

τετμήσθω ἡ ΔΗ δίχα κατὰ τὸ Ε, καὶ τῷ ἀπὸ τῆς ΕΗ ἴσον παρὰ τὴν ΑΗ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΑΖ, ΖΗ· σύμμετρος ἄρα ἐστὶν ἡ ΑΖ τῇ ΖΗ. καὶ διὰ τῶν Ε, Ζ, Η σημείων τῇ ΑΓ παράλληλοι ἤχθωσαν αἱ ΕΘ, ΖΙ, ΗΚ. καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΑΖ τῇ ΖΗ μήκει, καὶ ἡ ΑΗ ἄρα ἑκατέρᾳ τῶν ΑΖ, ΖΗ σύμμετρός ἐστι μήκει.

Let DH be bisected at E, and let there be applied to AH a rectangle equal to the square on EH and deficient by a square figure, and let it be the rectangle contained by AZ, ZH; therefore AZ is commensurable in length with ZH. And through the points E, Z, H let EG, ZI, HK be drawn parallel to AC. And since AZ is commensurable in length with ZH, therefore AH is also commensurable in length with each of AZ, ZH.

ἀλλὰ ἡ ΑΗ σύμμετρός ἐστι τῇ ΑΓ· καὶ ἑκατέρα ἄρα τῶν ΑΖ, ΖΗ σύμμετρός ἐστι τῇ ΑΓ μήκει.

But AH is commensurable with AC; therefore each of AZ, ZH is also commensurable in length with AC.

καί ἐστι ῥητὴ ἡ ΑΓ· ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΑΖ, ΖΗ· ὥστε καὶ ἑκάτερον τῶν ΑΙ, ΖΚ ῥητόν ἐστιν.

And AC is rational; therefore each of AZ, ZH is also rational; so that each of the areas AI, ZK is also rational.

καὶ ἐπεὶ σύμμετρός ἐστιν ἡ ΔΕ τῇ ΕΗ μήκει, καὶ ἡ ΔΗ ἄρα ἑκατέρᾳ τῶν ΔΕ, ΕΗ σύμμετρός ἐστι μήκει.

And since DE is commensurable in length with EH, therefore DH is also commensurable in length with each of DE, EH.

ῥητὴ δὲ ἡ ΔΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ῥητὴ ἄρα καὶ ἑκατέρα τῶν ΔΕ, ΕΗ καὶ ἀσύμμετρος τῇ ΑΓ μήκει· ἑκάτερον ἄρα τῶν ΔΘ, ΕΚ μέσον ἐστίν.

And DH is rational and incommensurable in length with AC; therefore each of DE, EH is also rational and incommensurable in length with AC; therefore each of the areas DG, EK is medial.

κείσθω δὴ τῷ μὲν ΑΙ ἴσον τετράγωνον τὸ ΛΜ, τῷ δὲ ΖΚ ἴσον τετράγωνον ἀφῃρήσθω κοινὴν γωνίαν ἔχον αὐτῷ τὴν ὑπὸ ΛΟΜ τὸ ΝΞ· περὶ τὴν αὐτὴν ἄρα διάμετρόν ἐστι τὰ ΛΜ, ΝΞ τετράγωνα.

Let then the square LM be laid down equal to AI, and let the square NX, equal to ZK, be subtracted, having the angle LOM common with it; therefore the squares LM, NX are about the same diagonal.

ἔστω αὐτῶν διάμετρος ἡ ΟΡ, καὶ καταγεγράφθω τὸ σχῆμα.

Let OP be their diagonal, and let the figure be described.

ἐπεὶ οὖν ἴσον ἐστὶ τὸ ὑπὸ τῶν ΑΖ, ΖΗ περιεχόμενον ὀρθογώνιον τῷ ἀπὸ τῆς ΕΗ τετραγώνῳ, ἔστιν ἄρα ὡς ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως ἡ ΕΗ πρὸς τὴν ΖΗ. ἀλλʼ ὡς μὲν ἡ ΑΖ πρὸς τὴν ΕΗ, οὕτως τὸ ΑΙ πρὸς τὸ ΕΚ, ὡς δὲ ἡ ΕΗ πρὸς τὴν ΖΗ, οὕτως ἐστὶ τὸ ΕΚ πρὸς τὸ ΚΖ·

Since, then, the rectangle contained by AZ, ZH is equal to the square on EH, therefore, as AZ is to EH, so is EH to ZH. But as AZ is to EH, so is AI to EK, and as EH is to ZH, so is EK to KZ; therefore EK is a mean proportional between AI, KZ.

τῶν ἄρα ΑΙ, ΚΖ μέσον ἀνάλογόν ἐστι τὸ ΕΚ. ἔστι δὲ καὶ τῶν ΛΜ, ΝΞ μέσον ἀνάλογον τὸ ΜΝ, ὡς ἐν τοῖς ἔμπροσθεν ἐδείχθη, καί ἐστι τὸ ΑΙ τῷ ΛΜ τετραγώνῳ ἴσον, τὸ δὲ ΚΖ τῷ ΝΞ· καὶ τὸ ΜΝ ἄρα τῷ ΕΚ ἴσον ἐστίν.

But MN is also a mean proportional between LM, NX, as was proved before, and AI is equal to the square LM, and KZ to NX; therefore MN is also equal to EK.

ἀλλὰ τὸ μὲν ΕΚ τῷ ΔΘ ἐστιν ἴσον, τὸ δὲ ΜΝ τῷ ΛΞ· τὸ ἄρα ΔΚ ἴσον ἐστὶ τῷ ΥΦΧ γνώμονι καὶ τῷ ΝΞ. ἔστι δὲ καὶ τὸ ΑΚ ἴσον τοῖς ΛΜ, ΝΞ τετραγώνοις· λοιπὸν ἄρα τὸ ΑΒ ἴσον ἐστὶ τῷ ΣΤ. τὸ δὲ ΣΤ τὸ ἀπὸ τῆς ΛΝ ἐστι τετράγωνον· τὸ ἄρα ἀπὸ τῆς ΛΝ τετράγωνον ἴσον ἐστὶ τῷ ΑΒ· ἡ ΛΝ ἄρα δύναται τὸ ΑΒ. λέγω δή, ὅτι ἡ ΛΝ ἀποτομή ἐστιν.

But EK is equal to DG, and MN to LX; therefore DK is equal to the gnomon UFX and NX. And AK is also equal to the squares LM, NX; therefore the remainder AB is equal to ST. But ST is the square on LN; therefore the square on LN is equal to AB; therefore LN produces AB. I say then, that LN is an apotome.

ἐπεὶ γὰρ ῥητόν ἐστιν ἑκάτερον τῶν ΑΙ, ΖΚ, καί ἐστιν ἴσον τοῖς ΛΜ, ΝΞ, καὶ ἑκάτερον ἄρα τῶν ΛΜ, ΝΞ ῥητόν ἐστιν, τουτέστι τὸ ἀπὸ ἑκατέρας τῶν ΛΟ, ΟΝ· καὶ ἑκατέρα ἄρα τῶν ΛΟ, ΟΝ ῥητή ἐστιν.

For since each of the areas AI, ZK is rational, and they are equal to LM, NX, therefore each of the squares LM, NX is rational, that is, the square on each of LO, ON; therefore each of LO, ON is rational.

πάλιν, ἐπεὶ μέσον ἐστὶ τὸ ΔΘ καί ἐστιν ἴσον τῷ ΛΞ, μέσον ἄρα ἐστὶ καὶ τὸ ΛΞ. ἐπεὶ οὖν τὸ μὲν ΛΞ μέσον ἐστίν, τὸ δὲ ΝΞ ῥητόν, ἀσύμμετρον ἄρα ἐστὶ τὸ ΛΞ τῷ ΝΞ· ὡς δὲ τὸ ΛΞ πρὸς τὸ ΝΞ, οὕτως ἐστὶν ἡ ΛΟ πρὸς τὴν ΟΝ· ἀσύμμετρος ἄρα ἐστὶν ἡ ΛΟ τῇ ΟΝ μήκει.

Again, since DG is medial and is equal to LX, therefore LX is also medial. Since, then, LX is medial, and NX is rational, therefore LX is incommensurable with NX; and as LX is to NX, so is LO to ON; therefore LO is incommensurable in length with ON.

καί εἰσιν ἀμφότεραι ῥηταί· αἱ ΛΟ, ΟΝ ἄρα ῥηταί εἰσι δυνάμει μόνον σύμμετροι· ἀποτομὴ ἄρα ἐστὶν ἡ ΛΝ. καὶ δύναται τὸ ΑΒ χωρίον· ἡ ἄρα τὸ ΑΒ χωρίον δυναμένη ἀποτομή ἐστιν.

And both are rational; therefore LO, ON are rational straight lines commensurable in square only; therefore LN is an apotome. And it produces the area AB; therefore the straight line producing the area AB is an apotome.

ἐὰν ἄρα χωρίον περιέχηται ὑπὸ ῥητῆς, καὶ τὰ ἑξῆς.

Therefore, if an area be contained by a rational straight line, and so forth.

Notes

¦5¦ἡ τὸ ΑΒ χωρίον δυναμένη — Meaning "the [straight line] producing the area AB" (literally "the [straight line] which is equivalent in square to the area AB"). This is the feminine singular present participle of "δύναμαι" (to be equivalent in square to), modifying an omitted feminine noun "εὐθεῖα" (straight line), meaning the line whose square equals the area AB.
¦15¦παραβληθῇ — "παραβληθῇ" is a passive subjunctive (forming a conditional clause with ἐάν) of παραβάλλω, a technical term in the Elements for "application of areas". Together with "παρὰ τὴν ΑΗ" (along AH) and "ἐλλεῖπον εἴδει τετραγώνῳ" (deficient by a square figure), it describes the geometric equivalent of solving a quadratic equation of the form $x(a-x) = b$.
¦55¦τουτέστι τὸ ἀπὸ ἑκατέρας τῶν ΛΟ, ΟΝ — Meaning "that is, the [square] on each of LO, ON." The noun "τετράγωνον" (square) is omitted after "τὸ ἀπὸ." In Greek geometry, the square on a straight line is standardly expressed by the neuter singular article with the preposition ἀπό + genitive ("τὸ ἀπὸ τῆς..."), where here the genitive is "ἑκατέρας τῶν ΛΟ, ΟΝ."

Cite this passage

Euclid, Elements §10.prop3.91. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:10.prop3.91

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