§12.prop.3#1πᾶσα πυραμὶς τρίγωνον ἔχουσα βάσιν διαιρεῖται εἰς δύο πυραμίδας ἴσας τε καὶ ὁμοίας ἀλλήλαις καὶ τῇ ὅλῃ τριγώνους ἐχούσας βάσεις καὶ εἰς δύο πρίσματα ἴσα· καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος.
Every pyramid having a triangular base is divided into two pyramids equal and similar to one another and to the whole, having triangular bases, and into two equal prisms; and the two prisms are greater than half of the whole pyramid.
ἔστω πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΒΓ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον· λέγω, ὅτι ἡ ΑΒΓΔ πυραμὶς διαιρεῖται εἰς δύο πυραμίδας ἴσας ἀλλήλαις τριγώνους βάσεις ἐχούσας καὶ ὁμοίας τῇ ὅλῃ καὶ εἰς δύο πρίσματα ἴσα· καὶ τὰ δύο πρίσματα μείζονά ἐστιν ἢ τὸ ἥμισυ τῆς ὅλης πυραμίδος.
Let there be a pyramid of which the base is the triangle ABG, and the vertex the point D; I say that the pyramid ABGD is divided into two pyramids equal to one another, having triangular bases, and similar to the whole, and into two equal prisms; and the two prisms are greater than half of the whole pyramid.
τετμήσθωσαν γὰρ αἱ ΑΒ, ΒΓ, ΓΑ, ΑΔ, ΔΒ, ΔΓ δίχα κατὰ τὰ Ε, Ζ, Η, Θ, Κ, Λ σημεῖα, καὶ ἐπεζεύχθωσαν αἱ ΘΕ, ΕΗ, ΗΘ, ΘΚ, ΚΛ, ΛΘ, ΚΖ, ΖΗ. ἐπεὶ ἴση ἐστὶν ἡ μὲν ΑΕ τῇ ΕΒ, ἡ δὲ ΑΘ τῇ ΔΘ, παράλληλος ἄρα ἐστὶν ἡ ΕΘ τῇ ΔΒ. διὰ τὰ αὐτὰ δὴ καὶ ἡ ΘΚ τῇ ΑΒ παράλληλός ἐστιν.
For let AB, BG, GA, AD, DB, DG be bisected at the points E, Z, H, Q, K, L, and let QE, EH, HQ, QK, KL, LQ, KZ, ZH be joined. Since AE is equal to EB, and AQ to DQ, therefore EQ is parallel to DB. For the same reasons indeed, QK is also parallel to AB.
παραλληλόγραμμον ἄρα ἐστὶ τὸ ΘΕ ΒΚ· ἴση ἄρα ἐστὶν ἡ ΘΚ τῇ ΕΒ. ἀλλὰ ἡ ΕΒ τῇ ΕΑ ἐστιν ἴση· καὶ ἡ ΑΕ ἄρα τῇ ΘΚ ἐστιν ἴση.
Therefore QEBK is a parallelogram; therefore QK is equal to EB. But EB is equal to EA; therefore AE is also equal to QK.
ἔστι δὲ καὶ ἡ ΑΘ τῇ ΘΔ ἴση· δύο δὴ αἱ ΕΑ, ΑΘ δυσὶ ταῖς ΚΘ, ΘΔ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ· καὶ γωνία ἡ ὑπὸ ΕΑΘ γωνίᾳ τῇ ὑπὸ ΚΘΔ ἴση·
And AQ is also equal to QD; therefore the two EA, AQ are equal to the two KQ, QD, each to each; and the angle EAQ is equal to the angle KQD.
βάσις ἄρα ἡ ΕΘ βάσει τῇ ΚΔ ἐστιν ἴση.
Therefore the base EQ is equal to the base KD.
ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΑΕΘ τρίγωνον τῷ ΘΚΔ τριγώνῳ.
Therefore the triangle AEQ is equal and similar to the triangle QKD.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΘΗ τρίγωνον τῷ ΘΛΔ τριγώνῳ ἴσον τέ ἐστι καὶ ὅμοιον.
For the same reasons indeed, the triangle AQH is also equal and similar to the triangle QLD.
καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΕΘ, ΘΗ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΔ, ΔΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ οὖσαι, ἴσας γωνίας περιέξουσιν.
And since two straight lines meeting one another, EQ, QH, are parallel to two straight lines meeting one another, KD, DL, which are not in the same plane, they will contain equal angles.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΕΘΗ γωνία τῇ ὑπὸ ΚΔΛ γωνίᾳ.
Therefore the angle EQH is equal to the angle KDL.
καὶ ἐπεὶ δύο εὐθεῖαι αἱ ΕΘ, ΘΗ δυσὶ ταῖς ΚΔ, ΔΛ ἴσαι εἰσὶν ἑκατέρα ἑκατέρᾳ, καὶ γωνία ἡ ὑπὸ ΕΘΗ γωνίᾳ τῇ ὑπὸ ΚΔΛ ἐστιν ἴση, βάσις ἄρα ἡ ΕΗ βάσει τῇ ΚΛ ἴση· ἴσον ἄρα καὶ ὅμοιόν ἐστι τὸ ΕΘΗ τρίγωνον τῷ ΚΔΛ τριγώνῳ.
And since two straight lines, EQ, QH, are equal to two straight lines, KD, DL, each to each, and the angle EQH is equal to the angle KDL, therefore the base EH is equal to the base KL; therefore the triangle EQH is equal and similar to the triangle KDL.
διὰ τὰ αὐτὰ δὴ καὶ τὸ ΑΕΗ τρίγωνον τῷ ΘΚΛ τριγώνῳ ἴσον τε καὶ ὅμοιόν ἐστιν.
For the same reasons indeed, the triangle AEH is also equal and similar to the triangle QKL.
ἡ ἄρα πυραμίς, ἧς βάσις μέν ἐστι τὸ ΑΕΗ τρίγωνον, κορυφὴ δὲ τὸ Θ σημεῖον, ἴση καὶ ὁμοία ἐστὶ πυραμίδι, ἧς βάσις μέν ἐστι τὸ ΘΚΛ τρίγωνον, κορυφὴ δὲ τὸ Δ σημεῖον.
Therefore the pyramid, of which the base is the triangle AEH, and the vertex the point Q, is equal and similar to the pyramid of which the base is the triangle QKL, and the vertex the point D.
καὶ ἐπεὶ τριγώνου τοῦ ΑΔΒ παρὰ μίαν τῶν πλευρῶν τὴν ΑΒ ἦκται ἡ ΘΚ, ἰσογώνιόν ἐστι τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ, καὶ τὰς πλευρὰς ἀνάλογον ἔχουσιν· ὅμοιον ἄρα ἐστὶ τὸ ΑΔΒ τρίγωνον τῷ ΔΘΚ τριγώνῳ.
And since, in the triangle ADB, QK has been drawn parallel to one of the sides, AB, the triangle ADB is equiangular to the triangle DQK, and they have their sides proportional; therefore the triangle ADB is similar to the triangle DQK.
διὰ τὰ αὐτὰ δὴ καὶ τὸ μὲν ΔΒΓ τρίγωνον τῷ ΔΚΛ τριγώνῳ ὅμοιόν ἐστιν, τὸ δὲ ΑΔΓ τῷ ΔΛΘ. καὶ ἐπεὶ δύο εὐθεῖαι ἁπτόμεναι ἀλλήλων αἱ ΒΑ, ΑΓ παρὰ δύο εὐθείας ἁπτομένας ἀλλήλων τὰς ΚΘ, ΘΛ εἰσιν οὐκ ἐν τῷ αὐτῷ ἐπιπέδῳ, ἴσας γωνίας περιέξουσιν.
For the same reasons indeed, the triangle DBG is also similar to the triangle DKL, and the triangle ADG to the triangle DLQ. And since two straight lines meeting one another, BA, AG, are parallel to two straight lines meeting one another, KQ, QL, which are not in the same plane, they will contain equal angles.
ἴση ἄρα ἐστὶν ἡ ὑπὸ ΒΑΓ γωνία τῇ ὑπὸ ΚΘΛ.
Therefore the angle BAG is equal to the angle KQL.