PROPOSITION XXIV.-THEOREM.

Πρότασις κδ'.—Εὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, ἔχῃ δὲ καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν λόγον καὶ ἕκτον πρὸς τέταρτον· καὶ συντεθὲν πρῶτον καὶ πέμπτον πρὸς δεύτερον τὸν αὐτὸν ἕξει λόγον καὶ τρίτον καὶ ἕκτον πρὸς τέταρτον.

If the first has to the second the same ratio which the third has to the fourth; and the fifth to the second the same ratio which the sixth has to the fourth; the first and fifth together have to the second the same ratio which the third and sixth together have to the fourth.

Statement.-Let AB the first, have to C the second, the same ratio which DE the third, has to F the fourth; and let BG the fifth, have to C the second, the same ratio which EH the sixth has to F the fourth. AG, the first and fifth together, have to C the second, the same ratio which DH, the third and sixth together, have to F the fourth.

Proof. Because BG is to C, as EH is to F, fore, by inversion, C

there

and as C is to BG, so is F to EH. Therefore, by equality, AB is to BG, as DE to EH (Prop. XXII.). Because AG is the sum of AB and BG, and DH the sum of DE and EH. Therefore as AG is to GB, so is DH to HE (Prop. XVIII.). But as GB is to C, so is HE to F; therefore, by equality, as AG is to C, so is DH to F (Prop. XXII.).

Wherefore, if the first, &c. Q. E. D.

PROPOSITION XXV.-THEOREM.

Πρότασις κε'.—Εὰν τέσσαρα μεγέθη ἀνάλογον ᾖ, τὸ μέγιστον καὶ τὸ ἐλάχιστον δύο τῶν λοιπῶν μείζονά ἐστιν.

If four magnitudes be proportionals, the greatest and least of them together are greater than the other two together.

Statement. Let the four magnitudes AB, CD, E, and F be proportionals; that is, AB is to CD, as E to F. And let AB be the greatest of them, and consequently F the least (Prop. XIV.). I say that AB and F are together greater than CD and E together.

Construction.-Take AG equal to E, and CH equal to F.

Proof,-Because AB

A

E

G B C HD

F

is to CD, as E is to F, and AG is equal to E, and CH equal to F, therefore, AB is to CD, as AG is to CH. Because AB is to CD, as AG is to CH, the remainder GB is to the remainder HD, as AB is to CD (Prop. XIX.). But AB is greater than CD; therefore GB is greater than HD. Because AG is equal to E, and CH to F. Therefore, AG and F together are equal to CH and E together. If to the unequal magnitudes GB and HD, of which GB is the greater, there be added equal magnitudes, viz., to GB, the two AG and F, and to HD, the two CH and E. Therefore AB and F together are greater than CD and E together.

Therefore, if four magnitudes, &c. Q. E. D.

F

BOOK SIXTH.

DEFINITION I.

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.

Όρος 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 ᾖ 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 K 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 (I.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.