AB: CD :: EF: GH, and

CD: X:: GH: (11. 5.) O, ex æquali (22. 5.)
AB: X:: EF: O. But

:

AB X (2. Cor. 20. 6.):: KAB: LCD; and
EF: 0 (2. Cor. 20. 6.) MF: NH; therefore

KAB: LCD (2. Cor. 20. 6.) :: MF : NH.

And if the figure KAB be to the figure LCD, as the figure MF to the figure NH, AB is to CD, as EF to GH.

Make (12. 6.) as AB to CD, so EF to PR, and upon PR describe (18. 6.) the rectilineal figure SR similar, and similarly situated to either of the

figures MF, NH: then, because that as AB to CD, so is EF to PR, and upon AB, CD are described the similar and similarly situated rectilineals KAB, LCD, and upon EF, PR, in like manner, the similar rectilineals MF, SR; KAB is to LCD, as MF to SR; but by the hypothesis, KAB is to LCD, as MF to NH; and therefore the rectilineal MF having the same ratio to each of the two NH, SR, these two are equal (9. 5.) to one another; they are also similar, and similarly situated; therefore GH is equal to PR and because as AB to CD, so is EF to PR, and because PR is equal to GH, AB is to CD, as EF to GH.

PROP. XXIII. THEOR.

Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Let AC, CF be equiangular parallelograms having the angle BCD equal to the angle ECG; the ratio of the parallelogram AC to the parallelogram CF, is the same with the ratio which is compounded of the ratios of their sides.

Let BC, CG be placed in a straight line; therefore DC and CE are also in a straight line (7. 1.); complete the parallelogram DG; and, taking any straight line K, make (12. 6.) as BC to CG, so K to L; and as DC to CE, so make (12. 6.) L to M: therefore the ratios of K to L, and L to M, are the same with the ratios of the sides, viz. of BC to CG, and of DC to.CE. But the ratio of K to M, is that which is said to be compounded (def. 10. 5.) of the ratios of K to L, and L to M; wherefore also K has to

A

B

D H

G

M the ratio compounded of the ratios of the sides of the parallelograms. Now, because as BC to CG, so is the parallelogram AC to the parallelogram CH (1. 6.); and as BC to CG, so is K to L; therefore K is (11. 5.) to L, as the parallelogram AC to the parallelogram CH: again, because as DC to CE, so is the parallelogram CH to the parallelogram CF: and as DC to CE, so is L to M; therefore Lis (11. 5.) to M, as the parallelogram CH to the parallelogram CF: therefore, since it has been proved, that as K to L, so is the parallelogram AC to the parallelogram CH; and as L to M, so the parallelogram CH to the parallelogram CF; ex æquali (22. 5.), K is to M, as the parallelogram AC to the parallelogram CF; but K has to M the ratio which is compounded of the ratios of the sides; therefore also the parallelogram AC has to the parallelogram CF the ratio which is compounded of the ratios of the sides.

K L M

E F

COR. Hence, any two rectangles are to each other as the products of their bases multiplied by their altitudes.

SCHOLIUM.

Hence the product of the base by the altitude may be assumed as the measure of a rectangle, provided we understand by this product the product of two numbers, one of which is the number of linear units contained in the base, the other the number of linear units contained in the altitude.

Still this measure is not absolute but relative: it supposes that the area of any other rectangle is computed in a similar manner, by measuring its sides with the same linear unit; a second product is thus obtained, and the ratio of the two products is the same as that of the two rectangles, agreeably to the proposition just demonstrated.

For example, if the base of the rectangle A contained three units, and its altitude ten, that rectangle will be represented by the number 3×10, or 30, a number which signifies nothing while thus isolated; but if there is a second rectangle B, the base of which contains twelve units, and the altitude seven, this rectangle would be represented by the number 12×7=84; and we shall hence be entitled to conclude that the two rectangles are to each other as 30 is to 84; and therefore, if the rectangle A were to be assumed as the unit of measurement in surfaces, the rectangle B would then have 4 for its absolute measure; or, which amounts to the same thing, it would be equal to 34 of a superficial unit.

84

30

It is more common and more simple to assume the squares as the unit of surface; and to select that square whose side is the unit of length. In this case, the measurement which we have regarded merely as relative, becomes absolute: the number 30, for instance, by which the rectangle A was measured, now represents 30 superficial units, or 30 of those squares, which have each of their sides equal to unity.

COR. 1. Hence, the area of any parallelogram is equal to the product of its base by its altitude.

COR. 2. It likewise follows, that the area of any triangle is equal to the product of its base by half its altitude.

PROP. XXIV. THEOR.

The parallelograms about the diameter of any parallelogram, are similar to the whole, and to one another.

Let ABCD be a parallelogram, of which the diameter is AC; and EG, HK the parallelograms about the diameter: the parallelograms EG, HK are similar, both to the whole parallelogram ABCD, and to one another.

A

E

B

F

H

Because DC, GF are parallels, the angle ADC is equal (21. 1.) to the angle AGF: for the same reason, because BC, EF are parallels, the angle ABC is equal to the angle AEF: and each of the angles BCD, EFG is equal to the opposite angle DAB (28. 1.), and therefore are equal to one another, wherefore the parallelograms ABCD, AEFG are equiangular And because the angle ABC is equal to the angle AEF, and the angle BAC common to the two triangles BAC, EAF, they are equiangular to one another; therefore (4. 6.) as AB to BC, so is AE to EF; and because the opposite sides of parallelograms are equal to one another (28. 1.), AB is (7. 5.) to AD, as AE to AG; and DC to CB, as GF to FE; and also CD to DA. as FG to GA: therefore the sides of the parallelograms ABCD, AEFG about the equal angles are proportionals; and they are therefore similar to one another (def. 1. 6.); for the same reason, the parallelogram ABCD is similar to the parallelogram FHCK. Wherefore each of the parallelograms, GE, KH is similar to DB: but rectilineal figures which are similar to the same rectilineal figure, are also similar to one another (21.6.); therefore the parallelogram GE is similar to KH.

D

PROP. XXV. PROB.

To describe a rectilineal figure which shall be similar to one, and equal to another given rectilineal figure.

Let ABC be the given rectilineal figure, to which the figure to be described is required to be similar, and D that to which it must be equal. _It is required to describe a rectilineal figure similar to ABC, and equal to D.

Upon the straight line BC describe (Cor. Prob. 16. 1.) the parallelogram BE equal to the figure ABC; also upon CE describe (Cor. Prob. 16. 1.) the parallelogram CM equal to D, and having the angle FCE equal to the angle CBL therefore BC and CF are in a straight line (21. 1. 7. 1.), as also LE and EM; between BC and CF find (13. 6.) a mean proportional GH, and upon GH describe (18. 6.) the rectilineal figure KGH similar, and similarly situated, to the figure ABC. And because BC is to GH as

GH to CF, and if three straight lines be proportionals, as the first is to the third, so is (2. Cor. 20. 6.) the figure upon the first to the similar and similarly described figure upon the second; therefore as BC to CF, so is the

figure ABC to the figure KGH but as BC to CF, so is (1. 6.) the parallelogram BE to the parallelogram EF: therefore as the figure ABC is to the figure KGH, so is the parallelogram BE to the parallelogram EF (11. 5.): but the rectilineal figure ABC is equal to the parallelogram BE; therefore the rectilineal figure KGH is equal (14. 5.) to the parallelogram EF : but EF is equal to the figure D; wherefore also KGH is equal to D; and it is similar to ABC. Therefore the rectilineal figure KGH has been described similar to the figure ABC, and equal to D.

PROP. XXVI. THEOR.

If two similar parallelograms have a common angle, and be similarly situated, they are about the same diameter.

Let the parallelograms ABCD, AEFG be similar and similarly situated, and have the angle DAB common; ABCD and AEFG are about the same diameter.

G

D

KH

E

F

For, if not, let, if possible, the parallelogram BD have its diameter AHC in a different straight line from AF, the diameter of the parallelogram EG, and let GF meet AHC in H ; and through H draw HK parallel to AD or BC; therefore the parallelograms ABCD, AKHG being about the same diameter, are 1995 pipe similar to one another (24. 6.): wherefore, as DA to AB, so is (def. 1. 6.) GA to AK; but because ABCD and AEFG are similar parallelograms, as DA is to AB, so is GA to AE; therefore (11. 5.) as GA to AE, so GA to AK; wherefore GA has the same ratio to each of the straight lines AE, AK; and consequently AK is equal (9. 5.) to AE, the less to the greater, which is impossible; therefore ABCD and AKHG are not about the same diameter; wherefore ABCD and AEFG must be about the same diameter.

B

PROP. XXVII. THEOR.

Of all the rectangles contained by the segments of a given straight line, the greatest is the square which is described on half the line.

A

Let AB be a given straight line, which is bisected in C; and let D be any point in it, the square on AC is greater C D B than the rectangle AD, DB. For, since the straight line AB is divided into two equal parts in C, and into two unequal parts in D, the rectangle contained by AD and DB, gether with the square of CD, is equal to the square of AC (5. 2.). The square of AC is therefore greater than the rectangle AD.DB.

PROP. XXVIII. PROB.

to

To divide a given straight line, so that the rectangle contained by its segments may be equal to a given space; but that space must not be greater than the square of half the given line.

Let AB be the given straight line, and let the square upon the given straight line C be the space to which the rectangle contained by the segments of AB must be equal, and this square, by the determination, is not greater than that upon half the straight line AB.

E

C

D

GB

F

Bisect AB in D, and if the square upon AD be equal to the square upon C, the thing required is done: But if it be not equal to it, AD must be greater than C, according to the determination: Draw DE at right angles to AB, and make it equal to C: produce ED to F, so that EF be equal to AD or DB, and from the centre E, at the distance EF, describe a circle meeting AB in G. Join EG; and because AB is divided equally in D, and unequally in G, AG.GB+ DG2=(5. 2.) DB2= EG2. But (37. 1.) ED2+DG2=EG2; therefore, AG. GB+DG2=ED2 +DG2, and taking away DG2, AG.GB=ED2. Now ED=C, therefore the rectangle AG.GB is equal to the square of C: and the given line AB is divided in G, so that the rectangle contained by the segments AG, GB is equal to the square upon the given straight line C.

PROP. XXIX. PROB.

To produce a given straight line, so that the rectangle contained by the segments between the extremities of the given line, and the points to which it is produced, may be equal to a given space.

Let AB be the given straight line, and let the square upon the given straight line C be the space to which the rectangle under the segments of AB produced, must be equal.