and LM of the parallelogram KM. From GE cut off GX equal to LK (I. 3), and from GF cut off GO equal to LM. Complete the parallelogram OX (I. 31), and OX is equal and similar to KM. But KM is similar to EF (Const.) Therefore OX is similar to EF, and it is similarly situated. Therefore OX and EF are about the same diagonal GB (VI. 26). Produce XP to R, and OP to S. The straight line AB is divided at S, as required.

Because EF is equal to C and KM together (Const.); but XO has been proved equal to KM; therefore the remainder, the gnomon ERO, is equal to the remainder C (1. Ax. 3). But it may be shown, as in the last proposition, that the parallelogram TS is equal to the gnomon ERO. Therefore TS is equal to C. Also, SR is similar to D (VI. 24). Therefore the straight line AB is divided into two parts at S, such that of the two parallelograms TS and SR having the same altitude described upon them, one, SR is similar to the given parallelogram D, and the other TS is equal to C, a given rectilineal figure not greater than a parallelogram EF similar to the given parallelogram D, described on EB half of the given straight line AB. Q. E. F.

PROP. XXIX. (PROBLEM.)-To produce a given straight line (AB) so that of the two parallelograms of the same altitude described on the whole line thus produced and on the part produced, the one on the part produced may be similar to a given parallelogram (D). and the other on the whole produced equal to a given rectilineal figure (C).

K

H

Bisect AB at E (I. 10), and upon EB describe the parallelogram EL similar and similarly situated to D (VI. 18). Make the parallelogram GH equal to EL and C together, and similar and similarly situated to D (VI. 25). Because GH is similar to EL (VI. 21), and greater than it (Const.); therefore the sides GK and KH of the parallelogram GH, are greater than the homologous sides EF and FL of the parallelogram EL. Produce FL and FE, and make FM equal to KH, and FN to KG (I. 3). Complete the parallelogram MN (I. 31). MN is equal and similar to GH (Const.)

M

D

A

E

B

N

PX

But GH is similar to EL. Therefore MN is similar to EL. Wherefore EL and MN are about the same diagonal FX. Produce LB to P, and AB to O. The straight line AB is produced to O as required (IV. 26). Because G H is equal to EL and C together, and GH is equal to MN (Const.); therefore MN is equal to EL and C (I. Ax. 1). From these equals take away the common part EL. Therefore the remainder, the gnomon NOL is equal to C (I. Ax. 3). Because AE is equal to EB the parallelogram AN is equal to the parallelogram NB (I. 36); that is. to BM (1.43). To these equals add NO. Therefore the parallelogram AX is equal to the gnomon NOL (I. Ax. 2). But the gnomon NOL has been proved equal to C. Therefore also AX is equal to C, and PO is similar to D (VI. 24). Wherefore the given straight line AB has been produced, so that of the two parallelograms AX and PO, of the same altitude described on

the whole AO thus produced, and on the part BO produced, the one PO on the part produced is similar to the given parallelogram D, and the other AX on the whole produced is equal to the given rectilineal figure C. Q. E. F.

PROP. XXX. (PROBLEM.)-To cut a given straight line (AB) in extreme

and mean ratio.

Upon AB describe the square BC (I. 46), and produce AC to G, so that AD described on the part produced may be a square similar to BC, and CD a rectangle equal to BC (VI. 29).

Be

A

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 (I. Ax. 3). cause BF and AD are equiangular (Const.) and equal. Therefore their sides about the equal angles at E are reciprocally proportional (VI. 14). Wherefore FE is to ED as AE is to EB. But FE is equal to AC (I. 34), that is, to AB (Def 30); and ED is equal to AE. Therefore BA is to AE as AE is to EB.

Ꮐ D

E

B

H

But AB is

greater than AE (Const.) Therefore AE is greater than EB (V. 14). Wherefore the straight line AB is cut in extreme and mean ratio at the point E (VI Def. 3) Q. E. F.

OTHERWISE.-Let AB be the given straight line.

cut it in extreme and mean ratio.

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

It is required to

A.

C B

Because the rectangle AB.BC is equal to the square of AC. Therefore BA is to AC as AC is to CB (VI. 17). Wherefore AB is cut in extreme and mean ratio at C (VI. Def. 3). Q. E. F.

PROP. XXXI. (THEOREM.)-In any right-angled triangle (ABC) the rectilineal figure described upon the side (BC) opposite to the right angle is equal to the similar and similarly described rectilineal figures upon the sides (BA, AC) containing the right angle.

Draw the perpendicular AD (I. 12).

B

D

Because in the right-angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD and ADC are similar to the whole triangle ABC, and to one another (VI. 8). Because the triangle ABC is similar to ADB; therefore CB is to BA as BA is to BD (VI. 4). 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 (VI. 20, Cor. 2). Therefore CB is to BD as the figure described upon CB is to the similar and similarly described figure upon BA. Wherefore, inversely, DB is to BC as the figure described upon BA is to that described upon BC (V. B.) For the same reason, DC is to CB as the figure described upon CA is to that described upon CB. Therefore BD and DC together are to BC as the figures described upon BA and AC together are to that described upon BC (V. 24).

But BD and DC together are equal to BC (I. Ax. 9). Therefore the figure described on BC is equal to the similar and similarly described figures on BA and AC (V. A). Wherefore, in right-angled triangles, &c. QE. D.

PROP. XXXII. (THEOREM.)—If two triangles (ABC, DCE) which have two sides (BA, AC) of the one respectively proportional to two sides (CD, D E) of the other, be joined at one angle, so as to have their homologous sides parallel to one another (AB parallel to DC, and AC to DE), and proceeding from their vertices in the same direction; the remaining sides (BC, CE) shall be in the same straight line

A

B

C F

Because AB is parallel to DC, and the straight line AC meets them, the alternate angles BAC and ACD are equal (I. 29). For the same reason the angle CDE is equal to the angle ACD. Therefore BAC is equal to CDE (I Ax. 1). Because the triangles ABC and DCE have the angle at A equal to the angle at D, and the sides about these angles proportionals, that is, BA to AC,as CD to DE (Hyp.); therefore the triangle ABC is equiangular to DCE (VI. 6); and the angle ABC is equal to the angle DČE. But the angle BAC was proved to be equal to ACD. Therefore the whole angle ACE is equal to the two angles ABC and BAC (I. Ax. 2). To each of these equals add the common angle ACB. Therefore the two angles ACE and ACB are equal to the three angles ABC, BAC, and ACB (I. Ax. 2). But the three angles ABC, BAC, and ACB are equal to two right angles (1. 32). Therefore the two angles ACE and ACB are equal to two right angles (I. Ax. 1). Because at the point C, in the straight line A Č, the two straight lines BC and CE on the opposite sides of it make the adjacent angles ACE and ACB equal to two right angles; therefore BC and CE are in the same straight line (I. 14). Wherefore, if two triangles, which have two sides of the one proportional, &c. Q. E. D.

PROP. XXXIII. (THEOREM.)-In equal circles (ABC, DEF) the angles either at the centres (BGC and EHF) or the circumferences (BẮC, DEF) have the same ratio to one another as the arcs (BC, EF) on which they stand, so also have the sectors (BC, EHF).

Take any number of arcs CK and KL, each equal to BC; and any number FM and MN, each equal to EF. Join GK, GL, HM and HN. Because the arcs BC, CK and

KL are all equal, the angles BGC,
CGK and KGL are also all equal A

(III. 27). Therefore what mul

tiple soever the arc BL is of the arc BC, the same multiple is the angle BGL of the angle BGC. For the same reason, whatever multiple the arc EN is of the

arc EF, the same multiple is the angle EHN of the angle EHF. If the arc BL be equal to the arc EN, the angle BGL is equal to the angle EHN (III. 27); if the arc BL be greater than the arc EN, the angle BGL is greater than the angle EHN; and if less, less. But there are four magnitudes, the two arcs BC and EF, and the two

angles BGC and EHF; and of the arc BC and the angle BGC have been taken any equimultiples whatever, viz., the arc BL and the angle BGL; and of the arc EF, and of the angle EHF, any equimultiples whatever, viz., the arc EN and the angle EHN. And it has been proved that if the arc BL be greater than the arc EN, the angle BGL is greater than the angle EHN; if equal, equal; and if less, less. Therefore the arc BC is to the arc EF as the angle BGC is to the angle EHF (V. Def. 5). But the angle BGC is to the angle EHF as the angle BAC is to the angle EDF (V. 15); each being double of each (III. 20). Therefore the arc BC is to the arc EF as the angle BGC is to the angle EHF, and as the angle BAC is to the angle EDF.-Next, the arc BC is to the arc EF as the sector BGC is to the sector EHF. Join BC and CK, and in the arcs BC and CK take any points X and O. Join BX, XC, CO and OK.

B XC

D

Η

N

M

E

Because in the triangles GBC and GCK the two sides BG and GC are equal to the two sides CG and GK, each to each, and they contain equal angles (Const.); therefore the base BC is equal to the base CK (1.4), and the triangle GBC to the triangle GCK. Because the arc BC is equal to the arc CK, the remaining part of the whole circumference of the circle ABC is equal to the remaining part of the whole circumference of the same circle (I. Ax. 3). Therefore the angle BXC is equal to the angle COK (III. 27). Wherefore the segment BX C is similar to the segment COK (III. Def. 11). But they are upon equal straight lines, BC and CK; and similar segments of circles upon equal straight lines are equal to one another (III. 24). Therefore the segment BXC is equal to the segment CO K. But the triangle BGC was proved to! be equal to the triangle CGK. Therefore the whole, the sector BGC, is equal to the whole, the sector CGK. For the same reason, the sector KGL is equal to each of the sectors BGC and CGK. In the same manner the sectors EHF, FHM, and MHN may be proved equal to one another. Therefore, what multiple soever the arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC. For the same reason, whatever multiple the arc EN is of the arc EF, the same multiple is the sector EHN of the sector EHF. If the arc BL be equal to the arc EN, the sector BGL is equal to the sector EHN; if the arc BL be greater than the arc EN, the sector BGL is greater than the sector ĚHN; and if less, less. But there are four magnitudes, the two arcs BC and EF, and the two sectors BGC and EHF; and of the arc BC and the sector BGC, the arc BL and the sector BGL are any equimultiples whatever; and of the are EF and the sector EHF, the arc EN and the sector EHN are any equimultiples whatever. And it has been proved that if the arc BL be greater than the arc EN, the sector BGL is greater than the sector EHN; if equal, equal; and if less, less. Therefore, the arc BC is to the arc EF as the sector BGC is to the sector EHF (V. Def. 5). Wherefore, in equal circles, &c. Q. E. D.

TAYLOR AND GREENING, PRINTERS, GRAYSTOKE-PLACE, FEITER-LANE, LONDON.