BOOK IV.

OF THE PROPORTIONS OF FIGURES, AND THE MEASUREMENT OF AREAS.

Definitions.

1. Similar figures are those which have the angles of the one equal to the angles of the other, each to each, and the sides about the equal angles proportional.

2. Any two sides, or any two angles, which have like positions in two similar figures, are called homologous sides or angles.

A

3. In two different circles, similar arcs, sectors, or segments, are those which correspond to equal angles at the centre. Thus, if the angles A and O are equal, the arc BC will be similar to DE, the sector BAC to the sector DOE, and the segment whose chord is BC, to the segment whose chord is DE.

B

A

D
C

E

4. The base of any rectilineal figure, is the side on which the figure is supposed to stand.

5. The altitude of a triangle is the perpendicular let fall from the vertex of an angle on the opposite side, taken as a base. Thus, AD is the altitude of the triangle BAC

B

6. The altitude of a parallelogram is the perpendicular which measures the distance between two opposite sides taken as bases. Thus, EF is the altitude of the parallelo- Á gram DB.

7. The altitude of a trapezoid is the perpendicular drawn between its two parallel sides. Thus, EF is the altitude of the trapezoid DB.

D

DE

F B

DE C

A F B

8. The area and surface of a figure, are terms very nearly synonymous. The area designates more particularly the superficial content of the figure. The area is expressed numeri

cally by the number of times which the figure contains some other area, that is assumed for its measuring unit.

9. Figures have equal areas, when they contain the same measuring unit an equal number of times.

10. Figures which have equal areas are called equivalent. The term equal, when applied to figures, designates those which are equal in every respect, and which being applied to each other will coincide in all their parts (Ax. 13.): the term equivalent implies an equality in one respect only: namely, an equality between the measures of figures.

We may here premise, that several of the demonstrations are grounded on some of the simpler operations of algebra, which are themselves dependent on admitted axioms. Thus, if we have A=B+C, and if each member is multiplied by the same quantity M, we may infer that A× M=BxM+C×M; in like manner, if we have, A=B+C, and D=E-C, and if the equal quantities are added together, then expunging the +C and -C, which destroy each other, we infer that A+D=B+ E, and so of others. All this is evident enough of itself; but in cases of difficulty, it will be useful to consult some agebraical treatise, and thus to combine the study of the two sciences.

PROPOSITION I. THEOREM.

Parallelograms which have equal bases and equal altitudes, are equivalent.

Let AB be the common base of D CF EDF CE the two parallelograms ABCD, ABEF: and since they are supposed to have the same altitude,

A B

A

B

their
upper bases DC, FE, will be
both situated in one straight line parallel to AB.

Now, from the nature of parallelograms, we have AD=BC, and AF BE; for the same reason, we have DC=AB, and FE=AB; hence DC=FE: hence, if DC and FE be taken away from the same line DE, the remainders CE and DF will be equal: hence it follows that the triangles DAF, CBE, are mutually eqilateral, and consequently equal (Book I. Prop. X.). But if from the quadrilateral ABED, we take away the triangle ADF, there wil! remain the parallelogram ABEF; and if from the same quadrilateral ABED, we take away the equal triangle CBE, there will remain the parallelogram ABĊD.

Hence these two parallelograms ABCD, ABEF, which have the same base and altitude, are equivalent

Cor. Every parallelogram is equivalent to the rectangle which has the same base and the same altitude.

PROPOSITION II. THEOREM.

Every triangle is half the parallelogram which has the same base and the same altitude.

Let ABCD be a parallelogram, and ABE a triangle, having the same base AB, and the same altitude: then will the triangle be half the parallelogram.

For, since the triangle and the parallelogram have the same altitude, the vertex E of the triangle, will be in the line EC, parallel to the base AB. Produce BA, and from E draw EF parallel to AD. The triangle FBE is half the parallelogram FC, and the triangle FAE half the parallelogram FD (Book I. Prop. XXVIII. Cor.).

Now, if from the parallelogram FC, there be taken the parallelogram FD, there will remain the parallelogram AC: and if from the triangle FBE, which is half the first parallelogram, there be taken the triangle FAE, half the second, there will remain the triangle ABE, equal to half the parallelogram AC.

Cor 1. Hence a triangle ABE is half of the rectangle ABGH, which has the same base AB, and the same altitude AH: for the rectangle ABGH is equivalent to the parallelogram ABCD (Prop. I. Cor.).

Cor. 2. All triangles, which have equal bases and altitudes, are equivalent, being halves of equivalent parallelograms.

PROPOSITION III. THEOREM.

Two rectangles having the same altitude, are to each other as their bases.

Let ABCD, AEFD, be two rectan- D gles having the common altitude AD: they are to each other as their bases AB, AE.

E

B

If AB be divided into 7

Suppose, first, that the bases are A commensurable, and are to each other, for example, as the numbers 7 and 4. equal parts, AE will contain 4 of those parts: at each point of division erect a perpendicular to the base; seven partial rectangles will thus be formed, all equal to each other, because all have the same base and altitude. The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four: hence the rectangle ABCD is to AEFD as 7 is to 4, or as AB is to AE. The same reasoning may be applied to any other ratio equally with that of 7 to 4: hence, whatever be that ratio, if its terms be commensurable, we shall have

ABCD AEFD :: AB: AE.

Suppose, in the second place, that the bases D AB, AE, are incommensurable: it is to be shown that we shall still have

ABCD AEFD :: AB : AE.

For if not, the first three terms continuing

the same, the fourth must be greater or less A
than AE. Suppose it to be greater, and that we have

ABCD AEFD :: AB : AO.

:

FK C

EIOB

Divide the line AB into equal parts, each less than EO. There will be at least one point I of division between E and 0: from this point draw IK perpendicular to AI: the bases AB, AI, will be commensurable, and thus, from what is proved above, we shall have

ABCD AIKD :: AB : AI.

:

But by hypothesis we have

ABCD AEFD :: AB: AO.

In these two proportions the antecedents are equal; hence the consequents are proportional (Book II. Prop. IV.); and we find

AIKD AEFD :: AI: AO.

But AO is greater than AI; hence, if this proportion is correct, the rectangle AEFD must be greater than AIKD: on the contrary, however, it is less; hence the proportion is impossible; therefore ABCD cannot be to AEFD, as AB is to a line greater than AE.

Exactly in the same manner, it may be shown that the fourth term of the proportion cannot be less than AE; therefore it is equal to AE.

Hence, whatever be the ratio of the bases, two rectangles ABCD, AEFD, of the same altitude, are to each other as their bases AB, AE.

PROPOSITION IV. THEOREM.

Any two rectangles are to each other as the products of their bases multiplied by their altitudes.

Let ABCD, AEGF, be two rectangles; then will the rectangle,

ABCD AEGF :: AB.AD : AF.AE.

D

B

A

F

Having placed the two rectangles, H so that the angles at A are vertical, produce the sides GE, CD, till they meet in H. The two rectangles ABCD, AEHD, having the same altitude AD, are to each other as their bases AB, AE: in like manner the two rectangles AEHD, AEGF, having the same altitude AE, are to each other as their bases AD, AF: thus we have the two proportions,

G

ABCD AEHD :: AB: AE,

AEHD: AEGF:: AD : AF.

Multiplying the corresponding terms of these proportions together, and observing that the term AEHD may be omitted, since it is a multiplier of both the antecedent and the consequent, we shall have

ABCD AEGF:: ABX AD AEX AF.

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. This product will give the number of superficial units in the surface; because, for one unit in height, there are as many superficial units as there are linear units in the base; for two units in height twice as many; for three units in height, three times as many, &c.

Still this measure is not absolute, but relative: it supposes