Page 168 - Euclid's Elements of Geometry

ST EW þ.

ELEMENTS BOOK 6

to AC. Thus, as AB is to AC, so AC (is) to CE.
Α A

Β B

Γ C

∆ D
ibþ

Ε
E
Δύο ἄρα δοθεισῶν εὐθειῶν τῶν ΑΒ, ΑΓ τρίτη ἀνάλογον Thus, a third (straight-line), CE, has been found
αὐταῖς προσεύρηται ἡ ΓΕ· ὅπερ ἔδει ποιῆσαι. (which is) proportional to the two given straight-lines,
AB and AC. (Which is) the very thing it was required to
do.

.
Proposition 12
Τριῶν δοθεισῶν εὐθειῶν τετάρτην ἀνάλογον προ- To find a fourth (straight-line) proportional to three
σευρεῖν. given straight-lines.
Α A
Β B

Γ Ε C E

Η G

∆ Θ Ζ D H F
῎Εστωσαν αἱ δοθεῖσαι τρεῖς εὐθεῖαι αἱ Α, Β, Γ· δεῖ δὴ Let A, B, and C be the three given straight-lines. So
τῶν Α, Β, Γ τετράτην ἀνάλογον προσευρεῖν. it is required to find a fourth (straight-line) proportional
᾿Εκκείσθωσαν δύο εὐθεῖαι αἱ ΔΕ, ΔΖ γωνίαν περιέχους- to A, B, and C.
αι [τυχοῦσαν] τὴν ὑπὸ ΕΔΖ· καὶ κείσθω τῇ μὲν Α ἴση ἡ ΔΗ, Let the two straight-lines DE and DF be set out en-
τῇ δὲ Β ἴση ἡ ΗΕ, καὶ ἔτι τῇ Γ ἴση ἡ ΔΘ· καὶ ἐπιζευχθείσης compassing the [random] angle EDF. And let DG be
τῆς ΗΘ παράλληλος αὐτῇ ἤχθω διὰ τοῦ Ε ἡ ΕΖ. made equal to A, and GE to B, and, further, DH to C
᾿Επεὶ οὖν τριγώνου τοῦ ΔΕΖ παρὰ μίαν τὴν ΕΖ ἦκται ἡ [Prop. 1.3]. And GH being joined, let EF have been
ΗΘ, ἔστιν ἄρα ὡς ἡ ΔΗ πρὸς τὴν ΗΕ, οὕτως ἡ ΔΘ πρὸς drawn through (point) E parallel to it [Prop. 1.31].
τὴν ΘΖ. ἴση δὲ ἡ μὲν ΔΗ τῇ Α, ἡ δὲ ΗΕ τῇ Β, ἡ δὲ ΔΘ τῇ Therefore, since GH has been drawn parallel to one
Γ· ἔστιν ἄρα ὡς ἡ Α πρὸς τὴν Β, οὕτως ἡ Γ πρὸς τὴν ΘΖ. of the sides EF of triangle DEF, thus as DG is to GE,
Τριῶν ἄρα δοθεισῶν εὐθειῶν τῶν Α, Β, Γ τετάρτη so DH (is) to HF [Prop. 6.2]. And DG (is) equal to A,
ἀνάλογον προσεύρηται ἡ ΘΖ· ὅπερ ἔδει ποιῆσαι. and GE to B, and DH to C. Thus, as A is to B, so C (is)

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