Triangles, parallels, and area: Book 1 Proposition 46
Translations
Ἀπὸ τῆς δοθείσης εὐθείας τετράγωνον ἀναγράψαι. Ἔστω ἡ δοθεῖσα εὐθεῖα ἡ ΑΒ: δεῖ δὴ ἀπὸ τῆς ΑΒ εὐθείας τετράγωνον ἀναγράψαι. Ἤχθω τῇ ΑΒ εὐθείᾳ ἀπὸ τοῦ πρὸς αὐτῇ σημείου τοῦ Α πρὸς ὀρθὰς ἡ ΑΓ, καὶ κείσθω τῇ ΑΒ ἴση ἡ ΑΔ: καὶ διὰ μὲν τοῦ Δ σημείου τῇ ΑΒ παράλληλος ἤχθω ἡ ΔΕ, διὰ δὲ τοῦ Β σημείου τῇ ΑΔ παράλληλος ἤχθω ἡ ΒΕ. Παραλληλόγραμμον ἄρα ἐστὶ τὸ ΑΔΕΒ: ἴση ἄρα ἐστὶν ἡ μὲν ΑΒ τῇ ΔΕ, ἡ δὲ ΑΔ τῇ ΒΕ. ἀλλὰ ἡ ΑΒ τῇ ΑΔ ἐστιν ἴση: αἱ τέσσαρες ἄρα αἱ ΒΑ, ΑΔ, ΔΕ, ΕΒ ἴσαι ἀλλήλαις εἰσίν: ἰσόπλευρον ἄρα ἐστὶ τὸ ΑΔΕΒ παραλληλόγραμμον. λέγω δή, ὅτι καὶ ὀρθογώνιον. ἐπεὶ γὰρ εἰς παραλλήλους τὰς ΑΒ, ΔΕ εὐθεῖα ἐνέπεσεν ἡ ΑΔ, αἱ ἄρα ὑπὸ ΒΑΔ, ΑΔΕ γωνίαι δύο ὀρθαῖς ἴσαι εἰσίν. ὀρθὴ δὲ ἡ ὑπὸ ΒΑΔ: ὀρθὴ ἄρα καὶ ἡ ὑπὸ ΑΔΕ. τῶν δὲ παραλληλογράμμων χωρίων αἱ ἀπεναντίον πλευραί τε καὶ γωνίαι ἴσαι ἀλλήλαις εἰσίν: ὀρθὴ ἄρα καὶ ἑκατέρα τῶν ἀπεναντίον τῶν ὑπὸ ΑΒΕ, ΒΕΔ γωνιῶν: ὀρθογώνιον ἄρα ἐστὶ τὸ ΑΔΕΒ. ἐδείχθη δὲ καὶ ἰσόπλευρον. Τετράγωνον ἄρα ἐστίν: καί ἐστιν ἀπὸ τῆς ΑΒ εὐθείας ἀναγεγραμμένον: ὅπερ ἔδει ποιῆσαι.
On a given straight line to describe a square. Let AB be the given straight line; thus it is required to describe a square on the straight line AB. Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31] Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34] But AB is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral. I say next that it is also right-angled. For, since the straight line AD falls upon the parallels AB, DE, the angles BAD, ADE are equal to two right angles. [I. 29] But the angle BAD is right; therefore the angle ADE is also right. And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled. And it was also proved equilateral. Therefore it is a square; and it is described on the straight line AB.