Again, because the triangles are similar, the triangle ABE has to the triangle FGL the duplicate ratio of that which the side BE has to the side GL (VI. 19). For the same reason, the triangle BEC has to the triangle GLH the duplicate ratio of that which BE has to GL. Therefore the triangle ABE is to the triangle FGL as the triangle BEC is to the triangle GLH (V. 11). Because the triangles are similar, the triangle EBC has to the triangle LGH the duplicate ratio of that which the side EC has to the side LH. For the same reason, the triangle ECD has to the triangle LHK the duplicate ratio of that which EC has to LH. Therefore the triangle EBC is to the triangle LGH as the triangle ECD is to the triangle LHK (V. 11). But it has been proved that the triangle EBC is to the triangle LGH as the triangle ABE is to the triangle FGL. Therefore the triangle ABE is to the triangle FGL as the triangle EBC is to the triangle LGH, and as the triangle ECD is to the triangle LHK. But one of the antecedents is to one of the consequents as all the antecedents are to all the consequents (V. 12). Therefore the triangle ABE is to the triangle FGL as the polygon ABCDE is to the polygon FGHKL. But the triangle ABE has to the triangle FGL the duplicate ratio of that which the side AB has to the homologous side FG (VI. 19). Therefore also the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which AB has to the homologous side FG. Wherefore, similar polygons, &c. Q. E. D.

Cor. 1.—In like manner it may be proved that similar four-sided figures, or figures of any number of sides, are one to another in the duplicate ratio of their homologous sides; and this has been proved of triangles (VI. 19); therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

Cor. 2.-If to AB and FG, two of the homologous sides of the polygon, a third proportional M be taken (VI. 11), AB has to M the duplicate ratio of that which AB has to FG (V. Def. 10). But the four-sided figure or polygon upon AB has to the four-sided figure or polygon upon FG likewise the duplicate ratio of that which AB has to FG (VI. 20, Cor. 1). Therefore AB is to M as the figure upon AB to the figure upon FG (V. 11); and this has been proved in triangles (VI. 19, Cor.) Therefore, universally, it is manifest that if three straight lines be proportionals, the first is to the third as any rectilineal figure upon the first is to a similar and similarly described rectilineal figure upon the second.

PROP. XXI. (THEOREM.)—Rectilineal figures (A and B) which are similar to the same rectilineal figure (C), are similar to one another (that is, A is similar to B).

Because the figure A is similar to the figure C, they are equiangular, and have their sides about the equal angles proportional (VI. Def. 1). Because the figure B

is similar to the figure C, they are equiangular, and have their sides about the equal angles proportionals (VI. Def. 1). Therefore the figures

A

A and B are each of them equiangular to the figure C, and have the sides about their equal angles proportionals. Wherefore the rectilineal figures A and B are equiangular. (I. Ax. 1), and have their sides about the equal angles proportionals (V. 11). Therefore the figure A is similar to the figure B (VI. Def.1). Therefore, rectilineal figures, &c. Q.E.D. PROP. XXII. (THEOREM.)—If four straight lines (AB, CD, EF, and G II) be proportionals (that is, AB is to CD as EF is to GH), the similar rectilineal figures (KAB, LCD, MF, NH) similarly described upon them, are proportionals; and, conversely, if the similar rectilineal figures similarly described upon four straight lines be proportionals (that is, KAB is to LCD as MF is to NH), these straight lines are proportionals. To the straight lines A B and CD find a third proportional

X (VI. 11); and to the straight lines EF and GH a third proportional O.

Because AB is to CD as EF to GH, and CD is to X as GH is to O (V. 11); therefore, ex æquali, AB is to X as EF is to O (V. 22). But AB is to X as the rectilineal figure KAB is to the rectilineal figure LCD (VI. 20, Cor. 2); and EI is to O as the rectilineal figure MF is to the rectilineal figure NH. Therefore the rectilineal figure KAB is to the rectilineal figure LCD as the rectilineal figure MF is to the rectilineal figure N H (V. 11).

Secondly, let the rectilineal figures KAB, LCD, MF, and NH be proportionals. The straight lines AB, CD, EF, and GH are also proportionals.

To the straight lines A B, CD, and EF find a fourth proportional PR; that is, make as AB is to CD so EF to PR (VI. 12). Upon PR describe the rectilineal figure SR, similar and similarly situated to either of the figures MF or NH (VI. 18).

Because AB is to CD as EF is to PR (Const.), and on AB and CD are described similar and similarly situated rectilineal figures KAB and LCD (Hyp.), and on EF and PR the similar and similarly situated rectilineal figures MF and SR (Const.); therefore the rectilineal figures KAB, LCD, MF, and SR are proportionals; that is, KA B is to LCD as MF is to SR. But KAB is to LCD as MF is to NH (Hyp.) Therefore, MF has the same ratio to each of the rectilineal figures NH and SR. Wherefore they are equal to each other (V. 9), and they are similar and similarly situated (Const.) Therefore the straight line GH is equal to the straight line PR (VI. 20, Cor. 7). Because AB is to CD as EF is to PR, and PR is equal to GH; therefore AB is to CD as EF is to GH (V. 7). Q, E. D.

PROP. XXIII. (THEOREM.)-Equiangular parall-lograms (AC and CF) have to one another the ratio which is compounded of the ratios of their sides. Let the angle BCD be equal to the angle ECG. Place BC and

CG in a straight line; and DC and CE shall also be in a straight line (I. 14). Complete the parallelogram DG (I. 31). Take any straight line K. and find La fourth proportional to BC, CG, and K (VI. 12); and M a fourth proportional to DC, CE, and L.

A

D II

B

G

C

KLM E F

Because the ratios of K to L and L to M are the same with the ratios of the sides; that is, of BC to CG, and DC to CE (Const.); but the ratio of K to M is that which is said to be compounded of the ratios of K to L and L to M (V. Def. A). Therefore K has to M the ratio compounded of the ratios of the sides. Because BC is to CG as the parallelogram AC is to the parallelogram CH (VI. 1); but BC is to CG as K is to L (Const.); therefore K is to L as the parallelogram AC is to the parallelogram CH (V. 11). Again, because DC is to CE as the parallelogram CH is to the parallelogram CF (VI. 1); but DC is to CE as L is to M (Const.); therefore L is to M as the parallelogram CH is to the parallelogram CF (V. 11). Wherefore, ex æquali, K is to M as the parallelogram AC is to the parallelogram CF (V. 22). But K has to M the ratio which is compounded of the ratios of the sides. Therefore, also, the parallelogram AC has to the parallelogramn CF the ratio which is compounded of the ratios of the sides. Wherefore, equiangular parallelograms, &c. Q. E. D.

PROP. XXIV. (THEOREM.)—Parallelograms (EG, HK) about the dayonat (AC) of any parallelogram (BD), are similar to the whole and to one another (that is, EG and HK are similar to the whole parallelogram BD, and to each other).

A

B

G

F

H

D

K

Because DC and GF are parallels, the angle ADC is equal to the angle AGF (I. 29). Because BC and EF are parallels, the angle ABC is equal to the angle AEF. Because each of the angles BCD and EFG is equal to the opposite angle DAB (I. 34); therefore they are equal to one another. Wherefore the parallelograms BD and EG are equiangular. Because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC and EAF, they are equiangular to one another. Therefore AB is to BC as AE is to EF (VI. 4). Because the opposite sides of parallelograms are equal to one another (I. 34); therefore AB is to AD as AE is to AG (V. 7); DC to CB as GF to FE, and CD to DA as FG to GA. Therefore the sides of the parallelograms BD and EG about the equal angles are proportionals; and they are similar to one another. (VI. Def. 1). For the same reason the parallelogram BD is similar to the parallelogram HK. Because each of the parallelograms GE and KH is similar to the parallelogram DB; and rectilineal figures which are similar to the same rectilineal figure are similar to one another (VI. 21); therefore the parallelogram GE is similar to the parallelogram KH. Wherefore, parallelograms, &c. Q. E. D.

PROP. XXV. (PROBLEM.)-To describe a rectilineal figure similar to one given rectilineal figure (ABC), and equal to another given rectilinea! figure (D).

A

Upon the straight line BC describe the parallelogram BE equal to the figure ABC (I. 45, Cor.) To the straight line CE apply the parallelogram CM equal to D (I. 45, Cor.), and having the angle FCE equal to the angle CBL. Because these angles are equal, BC and CF are in a straight line, as also LE and EM (I. 29, and I. 14).

Between BC

and CF find a mean proportional

K

дра

B

L

E

M G

GH (VI. 13). Upon GH describe the rectilineal figure KGH similar and similarly situated to the figure ABC (VI. 18).

Because BC is to GH as GH is to CF; and if three straight lines be proportionals, as the first is to the third so is the figure described upon the first to the similar and similarly described figure upon the second (VI. 20, Cor. 2). Therefore BC is to CF as the rectilineal figure ABC is to the rectilineal figure KGH. But BC is to CF as the parallelogram BE is to the parallelogram EF (VI. 1). Therefore the rectilineal figure ABC is to the rectilineal figure KGH as the parallelogram BE is to the parallelogram EF (V. 11). But the rectilineal figure ABC is equal to the parallelogram BE (Const.) Therefore the rectilineal figure KGH is equal to the parallelogram EF (V. 14). But the parallelogram EF is equal to the rectilineal figure D (Const.) Therefore also the rectilineal figure KGH is equal to the rectilineal figure D; and it is similar to the rectilineal figure ABC. Therefore the rectilineal figure KGH has been described similar to the rectilineal figure ABC, and equal to the rectilineal figure D. Q. E. F.

PROP. XXVI. (THEOREM.)—If two similar parallelograms (BD and EG) have a common angle (DAB), and be similarly situated, they are about the same diagonal.

For, if not, let, if possible, the parallelogram BD have its diagonal ALC in a different straight line from AF, the diagonal of the parallelogram EG. Let GF meet ALC in L. Through L draw LK parallel to AD or BC.

A G

B

D

Because the parallelograms BD and KG are about the same diagonal, they are similar to one K another (VI. 24). Therefore DA is to AB as GA E is to AK (VI. Def. 1). Because BD and EG are similar parallelograms (Hyp.), DA is to AB as GA is to AE. Therefore, GA is to AE as GA is tɔ AK (V. 11). Because GA has the same ratio to each of the straight lines AE and AK; therefore AK is equal to AE (V. 9); the less to the greater, which is impossible. Therefore the parallelogram BD cannot have its diagonal in ALC, a different straight line from AF. Wherefore BD and EG are about the same diameter AFC. Therefore, if two similar, &c. Q. E. D.

PROP. XXVII. (THEOREM.)-Of all the parallelograms (BE, BH) inscribed in a triangle (ABC) so as to have one of the angles at the base (BC) common to them all, the greatest is (BE) that described on half the base (BD).

F

A

EM

HN

0

Complete the parallelogram BL, and produce G H to M, and KII to N. Because BD is equal to DC (Hup.), the parallelogram BE is equal to the parallelogram DL, and the parallelogram BO to the parallelogram DN (I. 36). Because DL is a parallelogram and EC its diagonal, the complement DH is equal to the complement HL (I. 43). Adding equals to equals, the parallelogram BH is equal to the gnomon DNM (1. Ax. 2). But the parallelogram DL is greater than the gnomon DNM (I. Ax. 9). K Therefore the parallelogram BE is greater than the gnomon DN M. But the parallelogram BH has been proved equal to the gnomon DNM. Therefore the parallelogram BE is greater than the parallelogram BH. In the same manner, it may be shown that the parallelogram BE, described on half the base BC, is greater than any other parallelogram described on a segment greater or less than half the base. Therefore, of all the parallelograms, &c. Q. E. D.

B

DGC

PROP. XXVIII. (PROBLEM.)—To divide a given straight line (AB) into two parts, such that of the two parallelograms having the same altitude described upon them, one may be similar to a given parallelogram (D), and the other equal to a given rectilineal figure (C) not greater than a parallelogram, similar to the given parallelogram described on half of the given straight line (VI. 27, Cor.)

H G OF

L M

Bisect AB at E (I. 10), and upon EB describe the parallelogram EF similar and similarly situated to D (VI. 18). Complete the parallelogram AF. Then the parallelogram AG which is equal to the parallelogram EF (I. 36), is either equal to the rectilineal figure C, or greater than it (Hyp.) If the parallelogram A G is

T

R

A E SB

AÏ

K N

equal to the rectilineal figure C, what was required is done; for the straight line AB is divided into two parts at E, such that of the two parallelograms AG and EF of the same altitude, described upon them, one EF is similar to the given parallelogram D, and the other A G is equal to the given rectilineal figure C, which is not greater than the parallelogram EF similar to the given parallelogram D, and described on EB, half the given straight line.

But if the parallelogram AG is greater than the rectilineal figure C, the parallelogram EF is also greater than it. Describe the parallelogram KM equal to the excess of EF above C, and similar and similarly situated to the parallelogram D (VI. 25). Because EF and KM are each similar to D (Const.); therefore EF is similar to KM (VI. 21), and greater than it (Const.) Wherefore the sides EG and GF of the parallelogram EF are each greater than their homologous sides KL