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Book 6 Proposition 8

97v-98r

97v-98r

98v-99r

98v-99r

Ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις. Ἔστω τρίγωνον ὀρθογώνιον τὸ ΑΒΓ ὀρθὴν ἔχον τὴν ὑπὸ ΒΑΓ γωνίαν, καὶ ἤχθω ἀπὸ τοῦ Α ἐπὶ τὴν ΒΓ κάθετος ἡ ΑΔ: λέγω, ὅτι ὅμοιόν ἐστιν ἑκάτερον τῶν ΑΒΔ, ΑΔΓ τριγώνων ὅλῳ τῷ ΑΒΓ καὶ ἔτι ἀλλήλοις. Ἐπεὶ γὰρ ἴση ἐστὶν ἡ ὑπὸ ΒΑΓ τῇ ὑπὸ ΑΔΒ: ὀρθὴ γὰρ ἑκατέρα: καὶ κοινὴ τῶν δύο τριγώνων τοῦ τε ΑΒΓ καὶ τοῦ ΑΒΔ ἡ πρὸς τῷ Β, λοιπὴ ἄρα ἡ ὑπὸ ΑΓΒ λοιπῇ τῇ ὑπὸ ΒΑΔ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΓ ὑποτείνουσα τὴν ὀρθὴν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΑ ὑποτείνουσαν τὴν ὀρθὴν τοῦ ΑΒΔ τριγώνου, οὕτως αὐτὴ ἡ ΑΒ ὑποτείνουσα τὴν πρὸς τῷ Γ γωνίαν τοῦ ΑΒΓ τριγώνου πρὸς τὴν ΒΔ ὑποτείνουσαν τὴν ἴσην τὴν ὑπὸ ΒΑΔ τοῦ ΑΒΔ τριγώνου, καὶ ἔτι ἡ ΑΓ πρὸς τὴν ΑΔ ὑποτείνουσαν τὴν πρὸς τῷ Β γωνίαν κοινὴν τῶν δύο τριγώνων. τὸ ΑΒΓ ἄρα τρίγωνον τῷ ΑΒΔ τριγώνῳ ἰσογώνιόν τέ ἐστι καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον ἔχει. ὅμοιον ἄρα [ἐστὶ] τὸ ΑΒΓ τρίγωνον τῷ ΑΒΔ τριγώνῳ. ὁμοίως δὴ δείξομεν, ὅτι καὶ τῷ ΑΔΓ τριγώνῳ ὅμοιόν ἐστι τὸ ΑΒΓ τρίγωνον: ἑκάτερον ἄρα τῶν ΑΒΔ, ΑΔΓ [τριγώνων] ὅμοιόν ἐστιν ὅλῳ τῷ ΑΒΓ. Λέγω δή, ὅτι καὶ ἀλλήλοις ἐστὶν ὅμοια τὰ ΑΒΔ, ΑΔΓ τρίγωνα. Ἐπεὶ γὰρ ὀρθὴ ἡ ὑπὸ ΒΔΑ ὀρθῇ τῇ ὑπὸ ΑΔΓ ἐστιν ἴση, ἀλλὰ μὴν καὶ ἡ ὑπὸ ΒΑΔ τῇ πρὸς τῷ Γ ἐδείχθη ἴση, καὶ λοιπὴ ἄρα ἡ πρὸς τῷ Β λοιπῇ τῇ ὑπὸ ΔΑΓ ἐστιν ἴση: ἰσογώνιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. ἔστιν ἄρα ὡς ἡ ΒΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν ὑπὸ ΒΑΔ πρὸς τὴν ΔΑ τοῦ ΑΔΓ τριγώνου ὑποτείνουσαν τὴν πρὸς τῷ Γ ἴσην τῇ ὑπὸ ΒΑΔ, οὕτως αὐτὴ ἡ ΑΔ τοῦ ΑΒΔ τριγώνου ὑποτείνουσα τὴν πρὸς τῷ Β γωνίαν πρὸς τὴν ΔΓ ὑποτείνουσαν τὴν ὑπὸ ΔΑΓ τοῦ ΑΔΓ τριγώνου ἴσην τῇ πρὸς τῷ Β, καὶ ἔτι ἡ ΒΑ πρὸς τὴν ΑΓ ὑποτείνουσαι τὰς ὀρθάς: ὅμοιον ἄρα ἐστὶ τὸ ΑΒΔ τρίγωνον τῷ ΑΔΓ τριγώνῳ. Ἐὰν ἄρα ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς γωνίας ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, τὰ πρὸς τῇ καθέτῳ τρίγωνα ὅμοιά ἐστι τῷ τε ὅλῳ καὶ ἀλλήλοις [ὅπερ ἔδει δεῖξαι].

Πόρισμα. Ἐκ δὴ τούτου φανερόν, ὅτι ἐὰν ἐν ὀρθογωνίῳ τριγώνῳ ἀπὸ τῆς ὀρθῆς ἐπὶ τὴν βάσιν κάθετος ἀχθῇ, ἡ ἀχθεῖσα τῶν τῆς βάσεως τμημάτων μέση ἀνάλογόν ἐστιν: ὅπερ ἔδει δεῖξαι [καὶ ἔτι τῆς βάσεως καὶ ἑνὸς ὁποιουοῦν τῶν τμημάτων ἡ πρὸς τῷ τμήματι πλευρὰ μέση ἀνάλογόν ἐστιν].

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another. Let ABC be a right-angled triangle having the angle BAC right, and let AD be drawn from A perpendicular to BC; I say that each of the triangles ABD, ADC is similar to the whole ABC and, further, they are similar to one another. For, since the angle BAC is equal to the angle ADB, for each is right, and the angle at B is common to the two triangles ABC and ABD, therefore the remaining angle ACB is equal to the remaining angle BAD; [I. 32] therefore the triangle ABC is equiangular with the triangle ABD. Therefore, as BC which subtends the right angle in the triangle ABC is to BA which subtends the right angle in the triangle ABD, so is AB itself which subtends the angle at C in the triangle ABC to BD which subtends the equal angle BAD in the triangle ABD, and so also is AC to AD which subtends the angle at B common to the two triangles. [VI. 4] Therefore the triangle ABC is both equiangular to the triangle ABD and has the sides about the equal angles proportional. Therefore the triangle ABC is similar to the triangle ABD. [VI. Def. 1] Similarly we can prove that the triangle ABC is also similar to the triangle ADC; therefore each of the triangles ABD, ADC is similar to the whole ABC. I say next that the triangles ABD, ADC are also similar to one another. For, since the right angle BDA is equal to the right angle ADC, and moreover the angle BAD was also proved equal to the angle at C, therefore the remaining angle at B is also equal to the remaining angle DAC; [I. 32] therefore the triangle ABD is equiangular with the triangle ADC. Therefore, as BD which subtends the angle BAD in the triangle ABD is to DA which subtends the angle at C in the triangle ADC equal to the angle BAD, so is AD itself which subtends the angle at B in the triangle ABD to DC which subtends the angle DAC in the triangle ADC equal to the angle at B, and so also is BA to AC, these sides subtending the right angles; [VI. 4] therefore the triangle ABD is similar to the triangle ADC. [VI. Def. 1] Therefore etc.

Porism. From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base [and further that between the base and any one of the segments the side adjacent to the segment is a mean proportional].