BOOK SIXTH.

DEFINITIOΝ Ι.

Όρος α'.-Ομοία σχήματα εὐθύγραμμά ἐστιν, ὅσα τάς τε γωνίας ἴσας ἔχει κατὰ μίαν, καὶ τὰς περὶ τὰς ἴσας γωνίας πλευρὰς ἀνάλογον.

Similar rectilinear figures are those which have their angles equal, one by one, and the sides about the equal angles proportional.

DEFINITION II.

Ὅρος β ́.-Αντιπεπονθότα δὲ σχήματά ἐστιν, ὅταν ἑκατέρῳ τῶν σχημάτων ἡγούμενοί τε και ἑπόμενοι λόγων ὦσιν.

Figures are reciprocal when the antecedents and consequents of ratios are in each of the figures.

Annotation.

That is to say, reciprocal figures are such as have the sides about two of their angles proportional in such a manner that a side of the first figure is to a side of the second as the remaining side of the second is to the remaining side of the first.

DEFINITION III.

Όρος γ ́. Ακρον καὶ μέσον λόγον εὐθεῖα τετμῆσθαι λέγεται, ὅταν ἢ ὡς ἡ ὅλη πρὸς τὸ μεῖζον τμήμα οὕτως τὸ μεῖζον πρὸς

τὸ ἔλασσον.

A straight line is said to be cut in extreme and mean ratio when it happens that the whole line is to the greater segment as the greater segment to the less.

DEFINITION IV.

Ὅρος δ'.—Υψος ἐστὶ πάντος σχήματος ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν βάσιν κάθετος ἀγομένη.

The altitude of any figure is the perpendicular let fall from its vertex to its base.

PROPOSITION I.-THEOREM.

Πρότασις α'.—Τὰ τρίγωνα καὶ τὰ παραλληλόγραμμα, τὰ ὑπὸ τὸ αὐτὸ ὕψος ὄντα, πρὸς ἄλληλὰ ἐστιν ὡς αἱ βάσεις.

Triangles and parallelograms of the same altitude are to one another as their bases.

Statement.-Let the triangles ABC and ACD, and the parallelograms EC and CF, have the same altitude, viz., the perpendicular drawn from the point A to BD or BD produced. As the base BC is to the base CD, so is the triangle ABC to the triangle ACD; and as the base BC is to the base CD, so is

the parallelogram EC to the parallelogram CF.

Construction. -Produce BD both ways to the points H and

E A F

L, and take any number of HGB C D straight lines BG and GH, each

equal to the base BC (I. 3); and DK and KL, any number of straight lines each equal to the base_CD. Join AG, AH, AK, and AL.

Proof. Because CB, BG, and GH, are all equal, the triangles AHG, AGB, and ABC, are all equal (I. 38). Therefore, whatever multiple the base HC is of the base BC, the same multiple is the triangle AHC of the triangle ABC. For the same reason, whatever multiple the base LC is of the base CD, the same multiple is the triangle ALC of the triangle ADC. But if the base HC be equal to the base CL, the triangle AHC is

also equal to the triangle ALC (I. 38); and if the base HC be greater than the base CL, the triangle AHC is likewise greater than the triangle ALC; and if less, less. Because there are four magnitudes, viz., the two bases BC and CD, and the two triangles ABC and ACD; and of the base BC, and the triangle ABC, the first and the third, any equimultiples whatever have been taken, viz., the base HC and the triangle AHC; and of the base CD, and the triangle ACD, the second and the fourth, any equimultiples whatever have been taken, viz., the base CL and the triangle ALC. And it has been shown, that, if the base HC be greater than the base CL, the triangle AHC is greater than the triangle ALC; if equal, equal; and if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD (V. Def. 6).

Because the parallelogram CE is double of the triangle ABC (1.41), and the parallelogram CF double of the triangle ACD, and magnitudes have the same ratio which their equimultiples have (V. 15). Therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF. But it has been shown, that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD. And as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF. Therefore, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF (V. 11).

Wherefore, triangles, &c. Q. E. D.

PROPOSITION II.—THEOREM.

Πρότασις β ́. Εὰν τριγώνου παρὰ μίαν τῶν πλευρῶν ἀχθῇ τις εὐθεῖα, ἀνάλογον τεμεῖ τὰς τοῦ τριγώνου πλευράς· καὶ ἐὰν αἱ τοῦ τριγώνου πλευραὶ ἀνάλογον τμηθῶσιν, ἡ ἐπὶ τὰς τομὰς ἐπιζευγμένη εὐθεῖα παρὰ τὴν λοιπὴν ἔσται τοῦ τριγώνου πλευράν.

If a straight line be drawn parallel to one of the sides of a triangle, it cuts the other two sides proportionally; and if the two sides be cut proportionally, the straight line which joins the points of section is parallel to the remaining side of the triangle.

PART I.

Statement. Let DE be drawn parallel to BC, one of the sides of the triangle ABC. The sides AB and AC are cut proportionally; that is, BD is to DA, as CE is to EA.

Construction.-Join BE and CD.

A

E

Proof. The triangle BDE is equal to the triangle CDE (I. 37), because they are on the same base DE, and between the same parallels DE and BC. But ADE is another triangle; and equal magnitudes have the same ratio to the same magnitude (V. 7). Therefore, as the triangle BDE is to the triangle ADE, so is the triangle CDE to the triangle ADE. But the triangle BDE is to the triangle ADE, as BD is to DA (Prop. I.), because their altitude is the perpendicular drawn from the point E to AB, and they are to one another as their bases. For the same reason, the triangle CDE is to the triangle ADE, as CE to EA. Therefore, as BD is to DA, so is CE to EA (V. 11).

Statement.

PART II.

Next, let the sides AB and AC of the

Join

triangle ABC be cut proportionally in the points D and E, that is, so that BD is to DA as CE to EA. DE, and it is parallel to BC.

Proof. The same construction being made, because BD is to DA as CE is to EA. But BD is to DA, as the triangle BDE is to the triangle ADE (Prop. I.); and CE is to EA, as the triangle CDE is to the triangle ADE. Therefore the triangle BDE is to the triangle ADE, as the triangle CDE is to the triangle ADE (V. 11). Wherefore the triangles BDE and CDE have the same ratio to the triangle ADE. Therefore the triangle BDE is equal to the triangle CDE (V.9); and they are on the same base DE. But equal triangles on the same base and on the same side of it are between the same parallels (I. 39). Therefore DE is parallel to BC. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION III.-THEOREM.

Πρότασις γ ́.—Εὰν τριγώνου γωνία δίχα τμηθῇ, ἡ δὲ τέμν ουσα τὴν γωνίαν εὐθεῖα τέμνῃ καὶ τὴν βάσιν, τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἕξει λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευ ραῖς· καὶ ἐὰν τὰ τῆς βάσεως τμήματα τὸν αὐτὸν ἔχῃ λόγον ταῖς λοιπαῖς τοῦ τριγώνου πλευραῖς, ἡ ἀπὸ τῆς κορυφῆς ἐπὶ τὴν τομὴν ἐπιζευγνυμένη εὐθεῖα δίχα τέμνει τὴν τοῦ τριγώνου γωνίαν.

If any angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio to one another which the adjacent sides of the triangle have; and conversely, if the segments of the base have the same ratio to one another which the adjacent sides of the triangle have, the straight line drawn from the ver tex to the point of section bisects the vertical angle.