parallelogram equal to Z and exceeding by a parallelogram similar to X.

Construction.-Bisect AB in E; upon EB construct a parallelogram EFLB, similar to the given parallelogram X, and construct a parallelogram GH similar to the parallelogram EL, and equal to the sum of EL and Z; since GH is greater than EL, its sides GK and KH are greater than the homologous sides of EL, which are FE and FL (Prop. XXII.); on these sides produced take FN and FM equal to GK and KH, and complete the parallelogram NM: this is similar to GH, therefore similar to EL, and it is similarly posited, and therefore they are about the same diagonal; draw the diagonal FBX, through A draw AC parallel to EN, until it meet PN produced.

Proof. Since NM and GH are equal, and GH is equal to the sum of Z and EL, NM is also equal to the sum of Z and EL; take away from both EL, and the gnomon NOL is equal to Z; but since AE and EB are equal, the parallelograms AN and EP are equal (I. 36), and also EP and BM are equal (I. 43); therefore AN is equal to BM; add ON to both, and AX is equal to the gnomon NOL, and therefore equal to the given rectilinear figure Z; and its excess PO is similar to the parallelogram EL, and therefore similar to the given figure X.

Therefore, &c. Q. E. F.

Annotation.

This Proposition is equivalent to the following Problem :

To exscribe to a given triangle a parallelogram equal to a given rectilinear figure, and having an angle equal to one of the angles of the given triangle.

In the accompanying figure, ABC is the given triangle; AX is the inscribed, and AY the exscribed, parallelogram.

A

B

PROPOSITION XXX.-PROBLEM.

Πρότασις λ'.—Τὴν δοθεῖσαν εὐθεῖαν πεπερασμένην ἄκρον καὶ μέσον λόγον τεμεῖν.

To cut a given finite right line in extreme and mean ratio.

Statement.-Let AB be the given straight line. It is required to cut it in extreme and mean ratio.

Construction.—Upon AB describe the square BC (I. 46), and produce CA to G, so that AD described on the part produced may be a square similar

to BC, and CD a rectangle equal to BC (Prop. XXIX.).

A

C

F

E B

Proof. Because BC is equal to CD, and the part CE common to both. From each take CE, and the remainder BF is equal to the remainder AD. Because BF and AD are equiangular and equal; therefore their sides about the equal angles at E are reciprocally proportional (Prop. XXIV.). Wherefore FE is to ED as AE is to EB. But FE is equal to AC (I. 34), that is, to AB; and ED is equal to

AE. Therefore BA is to AE as AE is to EB. But AB is greater than AE. Therefore AE is greater than EB. Wherefore the straight line AB is cut in extreme and mean ratio at the point E. Q. E. F.

OTHERWISE :

Statement.-Let AB be the given straight line. is required to cut it in extreme and mean ratio. Construction.-Divide AB at the

point C, so that the rectangle contained by AB and BC may be equal to A the square of AC (II. 11).

It

C B

Proof. Because the rectangle ABC is equal to the Therefore BA is to AC as AC is to CB.

square of AC.

(Prop. XVII.).

mean ratio at C.

Wherefore AB is cut in extreme and
Q. E. F.

PROPOSITION XXXI.-THEOREM.

Πρότασις λα'. Εν τοῖς ὀρθογωνίοις τριγώνοις, τὸ ἀπὸ τῆς τὴν ὀρθὴν γωνιάν ὑποτεινούσης πλευρᾶς εἶδος ἴσον ἐστὶ τοῖς ἀπὸ τὴν ὀρθὴν γωνιάν περιεχουσῶν πλευρῶν εἴδεσι, τοῖς ὁμοίοις τε καὶ ὁμοίως ἀναγραφομένοις.

In right-angled triangles the figure described upon the side opposite to the right angle is equal to the similar and similarly described figures upon the sides containing the right angle.

Statement.-Let ABC be a right-angled triangle, having the right angle BAC. The rectilinear figure described upon BC is equal to the similar and similarly described figures upon BA and AC.

Construction. Draw the perpendicular AD (I. 12). Proof. Because in the right-angled triangle ABC, AD is drawn from the right angle at A perpendicular to

B D

the base BC, the triangles ABD and ADC are similar to the whole triangle ABC, and to one another (Prop. VIII.). Because the triangle ABC is similar to ADB. Therefore CB is to BA as BA is to BD (Prop. IV.). Because these three straight lines are proportionals, the first is to the third as the figure described upon the first is to the similar and similarly described figure upon the second (Prop. XX., Cor. 2). Therefore CB is to BD as the figure described upon CB is to the similar and similarly described figure upon BA. For the same reason, BC is to CD as the figure de scribed upon BC is to that described upon CA. Therefore, BC is to BD and DC together as the figure described upon BC is to the figures described upon BA and AC together (V. 24). But BD and DC together are equal to BC. Therefore the figure described on BC is equal to the similar and similarly described figures on BA and AC.

Wherefore, in right-angled triangles, &c. Q. E. D.

OTHERWISE :

Since similar figures are to each other in a twofold ratio of their homologous sides (Prop. XXIII.), the figure on BC has to the figure upon BA a ratio twofold that of CB to BA. But the square of BC has to the square of BA a ratio twofold that of CB to BA (Prop. XX., Cor. 1); therefore the figure upon CB is to that on BA as the square of CB is to the square of BA (V. 11). For the same reason the figure upon BC is to the figure upon CA as the square of BC is to the square of CA; therefore the figure upon BC is to the figures upon BA, AC as the square of BC is to the squares of the lines BA, AC (V. 24). But the square of BC is equal to the squares of BA, AC (I. 47); there

fore the figure upon BC is equal to the figures similar and similarly described upon BA, AC. Q. E. D.

PROPOSITION XXXII. THEOREM.

Πρότασις λβ'.—Εὰν δύο τρίγωνα συντεθῇ κατὰ μιάν γωνιάν, τὰς δυό πλευρὰς ταῖς δυσὶ πλευραῖς ἀνάλογον ἔχοντα, ὥστε τὰς ὁμολόγους αὐτῶν πλευρὰς καὶ παραλλήλους εἶναι· αἱ λοιπαὶ τῶν τριγώνων πλευραὶ ἐπ ̓ εὐθείας ἔσονται.

If two triangles have two sides proportional, and be so placed at an angle that the homologous sides are parallel, the remaining sides of the triangles form one right line.

Statement.-Let the triangles be ABC and CDE, having the sides AB, BC proportional to the sides CD, DE, and AB, CD parallel, and also BC, DE parallel; I say that AC, CE are in the same straight line.

Proof. Because AB and CD are parallel, the alternate angles B and BCD are equal (I. A 29); and also, since CB and

B

D

E

ED are parallel, the angles D and BCD are equal; therefore B and D are equal; and since the sides about these angles are proportional, the triangles ABC and CDE are equiangular (Prop. VI.); therefore the angles ACB and CED are equal; but BCD is equal to CDE, and if DCE be added, ACD and DCE are together equal to CED, EDC, and DCE; therefore ACD and DCE are equal to two right angles (I. 32); and therefore AC and CE form one right line (I. 14).

Therefore, if two triangles, &c. Q. E. D.