§6.36λϚ΄.
36.
Ἡμικύκλιον τὸ ἐπὶ τῆς ΑΒ, καὶ παράλληλος τῇ ΑΒ ἡ Γ∠, καὶ κάθετοι ἤχθωσαν αἱ ΓΕ, ∠Η ὅτι ἴση ἐστὶν ἡ ΑΕ τῇ ΗΒ.
εἰλήφοθω τὸ κέντρον τοῦ κύκλου τὸ Ζ, καὶ ἐπεζεύχθωσαν αἱ ΓΖ, Ζθωσαν ἴση ἄρα ἐστὶν ἡ ΓΖ τῇ Ζ∠· ὥστε καὶ τὸ ἀπὸ τῆς ΓΖ ἴσον τῷ ἀπὸ τῆς Ζ∠ τετραγώνῳ.
Let there be a semicircle on ΑΒ, and let ΓΔ be parallel to ΑΒ, and let the perpendiculars ΓΕ, ΔΗ be drawn; (to prove) that ΑΕ is equal to ΗΒ. Let the center of the circle be taken as Ζ, and let ΓΖ, ΖΔ be joined; therefore ΓΖ is equal to ΖΔ; so that the square on ΓΖ is also equal to the square on ΖΔ.
ἀλλὰ τῷ μὲν ἀπὸ ΓΖ τετραγώνῳ ἴσα ἐστὶν τὰ ἀπὸ τῶν ΓΕ, ΕΖ τετράγωνα, τῷ δὲ ἀπὸ ∠Ζ τετραγώνῳ ἴσα ἐστὶν τὰ ἀπὸ τῶν ∠Η, ΗΖ τετράγωνα· καὶ τὰ ἀπὸ τῶν ΓΕ, ΕΖ ἄρα τετράγωνα ἴσα ἐστὶν τοῖς ἀπὸ τῶν ΖΗ, Η∠ τετραγώνοις.
But to the square on ΓΖ, the squares on ΓΕ, ΕΖ are equal, and to the square on ΔΖ, the squares on ΔΗ, ΗΖ are equal; therefore the squares on ΓΕ, ΕΖ are also equal to the squares on ΖΗ, ΗΔ.
ὧν τὸ ἀπὸ ΓΕ τετράγωνον ἴσον ἐστὶν τῷ ἀπὸ τῆς ∠Η τετραγώνῳ· λοιπὸν ἄρα τὸ ἀπὸ τῆς ΕΖ τετράγωνον λοιπῷ τῷ ἀπὸ ΖΗ τετραγώνῳ ἐστὶν ἴσον· ἴση ἄρα ἐστὶν ἡ ΕΖ τῇ ΖΗ. ἔστιν δὲ καὶ ὅλη ἡ ΑΖ ὅλῃ τῇ ΖΒ ἴση· λοιπὴ ἄρα ἡ ΑΕ λοιπῇ τῇ ΗΒ ἐστιν ἴση· ὅπερ ἔδει δεῖξαι.
Of these, the square on ΓΕ is equal to the square on ΔΗ; therefore the remaining square on ΕΖ is equal to the remaining square on ΖΗ; therefore ΕΖ is equal to ΖΗ. And the whole ΑΖ is also equal to the whole ΖΒ; therefore the remaining ΑΕ is equal to the remaining ΗΒ; which was to be proved.
§6.37λζ΄.
37.
Ἡμικύκλιον τὸ ἐπὶ τῆς ΑΒ, καὶ ἀπὸ τυχόντος τοῦ Γ διήχθω ἡ Γ∠, καὶ κάθετος ἤχθω ἡ ∠Ε ὅτι τὸ ἀπὸ ΑΓ τοῦ ἀπὸ Γ∠ ὑπερέχει τῷ ὑπὸ συναμφοτέρου τῆς ΑΓ, καὶ τῆς ΑΕ.
ὅτι ἄρα τὸ ἀπὸ ΑΓ ἴσον ἐστὶν τῷ τε ἀπὸ ∠Γ, τουτέστιν τοῖς ἀπὸ ∠Ε, ΕΓ καὶ τῷ ὑπὸ συναμφοτέρου τῆς ΑΓΒ καὶ τῆς ΑΕ. ὅτι ἄρα κοινοῦ ἀφαιρεθέντος τοῦ ὑπὸ ΓΑΕ λοιπὸν τὸ ὑπὸ ΑΓE ἴσον ἐστὶν τῷ τε ἀπὸ ∠Ε, τουτέστιν τῷ ὑπὸ ΑΕΒ, καὶ τῷ ἀπὸ ΓΕ καὶ τῷ ὑπὸ ΑΕ, ΓΒ. κοινοῦ ἀφαιρεθέντος τοῦ ἀπὸ ΓΕ, ὅτι λοιπὸν τὸ ὑπὸ ΑΕΓ ἴσον ἐστὶν τῷ τε ὑπὸ ΑΕΒ καὶ τῷ ὑπὸ ΑΕ, ΒΓ ἔστιν δέ.
Let there be a semicircle on ΑΒ, and from any point Γ let ΓΔ be drawn, and let the perpendicular ΔΕ be drawn; (to prove) that the square on ΑΓ exceeds the square on ΓΔ by the rectangle contained by the sum of ΑΓ and ΑΕ, and ΑΕ. Therefore, the square on ΑΓ is equal to the square on ΔΓ, that is, to the squares on ΔΕ, ΕΓ, and the rectangle contained by the sum of ΑΓ, ΓΒ and ΑΕ. Therefore, the common rectangle contained by ΓΑ, ΑΕ being subtracted, the remaining rectangle contained by ΑΓ, ΓΕ is equal to the square on ΔΕ, that is, to the rectangle contained by ΑΕ, ΕΒ, and the square on ΓΕ, and the rectangle contained by ΑΕ, ΓΒ. The common square on ΓΕ being subtracted, the remaining rectangle contained by ΑΕ, ΓΕ is equal to both the rectangle contained by ΑΕ, ΕΒ and the rectangle contained by ΑΕ, ΒΓ; which indeed is the case.
§6.38## Κἰς τὸ πόρισμα τοῦ α΄ βιβλίου. λή.
## On the Porism of the First Book. 38.
Θέσει ὄντος παραλληλογράμμου τοῦ Α∠ ἀπὸ δοθέντος τοῦ Ε διαγαγεῖν τὴν ΕΖ καὶ ποιεῖν ἴσον τὸ ΖΓΗ τρίγωνον τῷ Α∠ παραλληλογράμμῳ.
Given the parallelogram ΑΔ in position, from a given point Ε to draw ΕΖ and make the triangle ΖΓΗ equal to the parallelogram ΑΔ.
γεγονέτω.
Let it be done.
ἐπεὶ οὖν ἴσον ἐστὶν τὸ ΖΓΗ τρίγωνον τῷ Α∠ παραλληλογράμμῳ, τὸ δὲ Α∠ παραλληλόγραμμον διπλάσιόν ἐστιν τοῦ ΑΓ∠ τριγώνου, καὶ τὸ ΖΓΗ ἄρα τρίγωνον διπλάσιόν ἐστιν τοῦ ΑΓΔ τριγώνου.
Since, then, the triangle ΖΓΗ is equal to the parallelogram ΑΔ, and the parallelogram ΑΔ is double the triangle ΑΓΔ, therefore the triangle ΖΓΗ is also double the triangle ΑΓΔ.
ὡς δὲ τὸ τρίγωνον πρὸς τὸ τρέγωνον, διὰ τὸ περὶ τὴν αὐτὴν γωνίαν τὴν Γ οὕτως ἐστὶν τὸ ὑπὸ ΖΓΗ πρὸς τὸ ὑπὸ ΑΓ∠.
And as the triangle is to the triangle, because they have the same angle Γ, so is the rectangle contained by ΖΓ, ΓΗ to the rectangle contained by ΑΓ, ΓΔ.
δοθὲν δὲ τὸ ὑπὸ ΑΓ∠ δοθὲν ἄρα καὶ τὸ ὑπὸ ΖΓΗ. καὶ δοθέντος τοῦ Ε εἰς θέσει τὰς ΑΓ, Γ∠ διῆκται εἰς χωρίου ἀποτομήν· θέσει ἄρα ἐστὶν ἡ ΕΖ.
συντεθήσεται δὲ οὕτως·
And the rectangle contained by ΑΓ, ΓΔ is given; therefore the rectangle contained by ΖΓ, ΓΗ is also given. And from the given point Ε, to the straight lines ΑΓ, ΓΔ which are given in position, a line has been drawn cutting off an area; therefore ΕΖ is given in position.
ἔστω τὸ μὲν τῇ θέσει παραλληλόγραμμον τὸ Α∠, τὸ δὲ δοθὲν τὸ Ε. διήχθω ἀπὸ τοῦ Ε εἰς θέσει τὰς ΖΓΗ εὐθεῖα ἡ ΕΖ ἀποτέμνουσα χωρίον τὸ ΖΓΗ ἴσον δοθέντι χωρίῳ τῷ διπλασίονι τοῦ ΑΓ∠.
And it will be synthesized as follows: let the parallelogram given in position be ΑΔ, and the given point be Ε. Let there be drawn from Ε to the straight lines ΓΖ, ΓΗ given in position, a straight line ΕΖ cutting off the area ΖΓΗ equal to the given area which is double of ΑΓΔ.
καὶ κατὰ τὰ αὐτὰ τῇ ἀναλύσει δείξομεν ἴσον τὸ ΖΓΗ τρίγωνον τῷ Α∠ παραλληλογράμμῳ· ἡ ΕΖ ἄρα ποιεῖ τὸ πρόβλημα.
And in the same way as in the analysis we shall show that the triangle ΖΓΗ is equal to the parallelogram ΑΔ; therefore ΕΖ does the problem.
φανερὸν οὖν, ὅτι μόνη, ἐπεὶ κἀκείνη μόνη.
It is clear, then, that it is unique, since that one is also unique.