have an angle in one equal to an angle in the other, and their sides about these equal angles proportionals, the tri

*6. 6. #4.6.

†3 Ax.

angle ABE is equiangular, to the triangle FGL; and therefore similar to it; wherefore the angle ABE is equal to the angle FGL: and, because the polygons are similar, the whole angle ABC is *1 Def. 6. equal to the whole angle FGH; therefore the remaining angle EBC is equal to the remaining angle LGH: and because the triangles ABE, FGL are similar, EB: BA:: LG: GF; and also because the polygons are similar, AB : BC:: FG : *11. 5. GH; therefore,* EB: BC :: LG: GH; that is, the sides about the equal angles EBC, LGH are proportionals; therefore the triangle EBC is equiangular to the triangle LGH, and similar to it;

*6. 6. *4. 6.

E

D

Ꭺ

for a like reason, the triangle ECD) is similar to the triangle LHK: therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles.

Also these triangles are each to each, as the polygons themselves, and these are to each other as the squares of the homologous sides.

#2.5.

Because the triangle ABE is similar to the triangle FGL, ABE is to FGL as the square of the side BE to the square of the side GL: for a similar reason, the triangle BEC is to GLH as the square of BE to the square of GL; therefore, triangle ABE triangle FGL :: triangle BEC: triangle GLH. Again, because the triangle EBC is similar to the triangle LGH, EBC is to LGH as the square of the side EC to the square of the side LH: for the like reason, the triangle ECD is to the triangle LHK as the square of EC to the square of LH therefore, triangle EBC: triangle LGH :: triangle ECD: triangle LHK: but it has been proved, that triangle EBC triangle LGH ::

triangle ABE : triangle FGL; therefore, triangle ABE: triangle FGL:: triangle EBC : triangle LGH, and triangle

M F

G

ECD: triangle LHK: but as one of the antecedents to one of the consequents,* so are all the antecedents to #5.5. all the consequents: therefore triangle ABE : triangle FGL:: polygon ABCDE: polygon FGHKL but the triangle ABE is to the triangle FGL as the square of the side AB to the square of the homologous side FG; therefore also the polygon ABCDE is to the polygon FGHKL as the square of AB to the square of 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 of any number of sides, are to one another as the squares of their homologous sides: and it has already been proved in triangles: therefore, universally, similar rectilineal figures are to one another as the squares of their homologous sides.

+11. 6.

+2.5.

COR. 2. And if to AB, FG, two of the homologous sides, a thirdt proportional M be taken, AB is to M as #16., 6. Cor. the square of AB to the square of FG:* but the polygon upon AB is to the polygon upon FG as the square of AB to the square of FG; therefore,† as AB is to M, so is the figure upon AB to the figure #19.6. Cor. upon FG: which was also proved in triangles: therefore, universally, it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any rectilineal figure upon the first, to a similar, and similarly described rectilineal figure upon the second.

PROP. XXI. THEOR.

Rectilineal figures which are similar to the same rectilineal figure, are also similar to one another.

Let each of the rectilineal figures A, B be similar to the rectilineal figure C: the figure A shall be similar to the figure B.

#1 Def. 6.

Because A is similar to C, they are equiangular, and also have their sides about the equal angles proportional:* again, because B is similar to C, they are equiangular, and have their sides

about the equal angles pro

#1 Def. 6. portionals: therefore the figures A, B are each of

them equiangular to C, and have the sides about the equal angles of each of them and of C proportionals.

*2.5.

B

Wherefore the rec

tilineal figures A and C are equiangular, and have #1 Ax. 1. their sides about the equal angles proportionals: #1 Def. 6. therefore A is similar to B. Therefore rectilineal figures, &c.

2. E. D.

PROP. XXII. THEOR.

If four straight lines be proportionals, the similar rectilineal figures similarly described upon them shall also be proportionals: and if the similar rectilineal figures similarly described upon four straight lines be proportionals, those straight lines shall be proportionals.

Let the four straight lines AB, CD, EF, GH be proportionals, viz. AB to CD, as EF to GH; and upon AB, CD let the similar rectilineal figures KAB, LCD be similarly

described; and upon EF, GH the similar rectilineal figures MF, NH, in like manner; the rectilineal figure KAB shall be to LCD, as MF to NH.

#11. 6.

#2.5. #11.5.

:

To AB, CD take a third proportional* X; and to EF, GH a third proportional O; then because AB CD EF GH, therefore CD* : X :: GH : 0; wherefore* AB X :: EF: 0; but AB: Cor. 2. X: figure KAB: figure LCD,* and EF: 0 :: figure MF figure NH; therefore, KAB: LCD :: MF:

#20. 6.

NH.

#18. 6.

And if the rectilineal figure KAB be to LCD, as MF to NH; the straight line AB shall be to CD as EF to GH. Make as AB to CD, so EF to PR; and upon #12. 6. PR describe the rectilineal figure SR, similar and similarly situated to either of the figures MF, NH; then, because AB: CD :: EF: PR, and that upon AB, CD are described the similar and similarly situated rectilineal figures KAB, LCD, and upon EF, PR in like manner, the similar rectilineal figures MF, SR; therefore KAB: LCD:: MF: SR; but by the hypothesis KAB: LCD ::

#9. 5.

MF: NH; therefore the figures NH, SR, are equal* Cor. 2. to one another; they are also similar, and similarly situated; therefore GH is equal to PR; and because AB CD :: EF: PR, and that PR is equal to GH; therefore AB: CD :: EF: GH. If, therefore, four straight lines, &c. 2. E. D.

PROP. XXIII. THEOR.

Equiangular parallelograms are to each other as the
rectangles of their containing sides.

Let AC, DF be equiangular parallelograms, having the angle B equal to the angle E; then will AC be to DF as the rectangle AB BC to the rectangle DEEF. Draw AG, DH perpendicular to BC, EF respectively. Because parallelograms upon the same base and between the same parallels are equal, the parallelogram AC is equal to

#1. 6.

the rectangle AG BC and the parallelogram DF to the rectangle DH⚫EF. Also the rectangles AG BC, AB BC having the equal altitudes BC, are to each other as their bases AG, AB. In like manner the rectangles DH EF, DE EF are to each other as the bases DH, DE. But the triangles ABG, DEH having the angles at B and G respectively equal to those at E, H are equiangular; thereforet

+4. 6.

#2. 5.

#10.5.

AG: AB :: DH: DE;

consequently*

AG BC AB·BC :: DH⚫EF : DE-EF,
and alternately*

AG BC DH⚫EF :: AB·BC: DE EF;

that is, the parallelograms AC, DF are to each other as the rectangles AB BC, DE EF. Wherefore equiangular parallelograms, &c. Q. E. D.