6. SIMILAR ARCS, SECTORS, or, SEGMENTS, are those, which in different circles, correspond to equal angles at the

centre.

Thus, if the angles A and O are equal, the arc BFC will be similar to DGE, the sector BAC to the sector DOE, and the segment BCF, to the segment DEG.

7. The ALTITUDE of a triangle is the perpendicular let fall from the vertex of an angle on the opposite side: this side is then called a base.

8. The altitude of a parallelogram is the perpendicular distance between two opposite sides. These sides are called bases.

9. The altitude of a trapezoid is the perpendicular distance between its two parallel sides.

PROPOSITION I. THEOREM.

Parallelograms which have equal bases and equal altitudes, are

equivalent.

Since the two parallelograms have equal bases, those bases may be placed the one on the other. Therefore, let AB be the common base of the two parallelograms ABCD, ABEF, which have the same altitude: then will they be equivalent.

For, in the parallelogram D CF. EDF CE ABCD, we have

AB=DC, and AD=BC (B. I., P. 28);

and in the parallelogram ABEF,

we have,

hence,

A B

A

B

AB-EF, and AF-BE:

DC-EF (A. 1).

Now, if from the line DE, we take away DC, there will

remain CE; and if from the same line we take away EF, there will remain DF;

therefore, the triangles ADF and BCE are mutually equilateral, and consequently, equal (B. I., P. 10).

But if from the quadrilateral ABED, we take away the triangle ADF, there will remain the parallelogram ABEF; and if from the same quadrilateral, we take away the equal triangle BCE, there will remain the parallelogram ABCD. Hence, any two parallelograms, which have equal bases and equal altitudes, are equivalent.

Scholium. Since the rectangle and square are parallelograms (B. I., D. 25), it follows that either is equivalent to any parallelogram having an equal base and an equal altitude. And generally, whatever property is proved as belonging to a parallelogram, belongs equally to every variety of parallelogram.

PROPOSITION II. THEOREM.

If a triangle and a parallelogram have equal bases and equal altitudes, the triangle will be half the parallelogram.

Place the base of the triangle on that of the parallelogram, so that the triangle ACB, and the parallelogram ABFD shall have the common base AB: then will the triangle be half the parallelogram.

For, since the triangle and the D FE parallelogram have equal altitudes, the vertex C, of the triangle, will

C

be in the upper base of the parallelogram (B. I., P. 23). Through A,

A

B

draw AE parallel to BC, forming the parallelogram ABCE. Now, the parallelograms ABFD, ABCE, are equivalent, having the same base and altitude (P. 1). But the triangle ABC is half the parallelogram BE (B. I., P. 28, c. 1): therefore, it is half the equivalent parallelogram BD (a. 7).

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

PROPOSITION III. THEOREM.

Two rectangles having equal altitudes are to each other as their bases.

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

D

A

F

C

E

B

First. Suppose that the bases are commensurable, and are to each other, for example, as the numbers 7 and 4. If AB be divided into 7 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 each has the same base and altitude (P. 1, s). The rectangle ABCD will contain seven partial rectangles, while AEFD will contain four: hence, the rectangle

ABCD : AEFD :: 7: 4, or as AB : AE. The same reasoning may be applied to any other ratio equally with that of 7 to 4: hence, whatever be the ratio, we have, when its terms are commensurable,

ABCD : AEFD :: AB : AE.

Second. Suppose that the bases AB, AE, are incommensurable: we shall still have

AB: AE.

D

FK C

A

EIOB

ABCD AEFD :: AB : : For, if the rectangles are not to each other in the ratio of AB to AE, they are to each other in a ratio greater or less: that is, the fourth term must be greater or less than AE. Suppose it to be greater, and that we have

ABCD : AEFD :: AB : AO.

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,

forming the new rectangle AK: then, since the bases AB, AI, are commensurable, we have,

ABCD : AIKD :: AB : AI.

But by hypothesis we have

In these two proportions the antecedents are equal; hence, the consequents are proportional (B. II., P. 4), that is, AIKD AEFD :: AI : :

AO.

But AO is greater than AI; which requires that the rectangle AEFD be greater than AIKD: on the contrary, however, it is less (A. 8); hence, the proportion is not true; therefore ABCD cannot be to AEFD, as AB is to a line greater than AE.

In the same manner, it may be shown that the fourth term of the proportion cannot be less than AE; therefore, being neither greater nor less, it is equal to AE. Hence, any two rectangles having equal altitudes, are to each other as their bases.

PROPOSITION IV. THEOREM.

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

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

ABCD : AEGF ABXAD

::

Having placed the two rectangles, so that the angles at A are opposite, produce the sides GE, CD, till they meet in H. Then, the two rectangles ABCD, AEHD, having the same altitude

AEXAF

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,

Multiplying the corresponding terms of these proportions together (B. II., P. 13), and omitting the term AEHD, since it is common to both the antecedent and consequent (B. II., P. 7), we have

ABCD : AEGF :: ABXAD AEXAF

:

Scholium 1. If we take a line of a given length, as one inch, one foot, one yard, &c., and regard it as the linear unit of measure, and find how many times this unit is contained in the base of any rectangle, and also, how many times it is contained in the altitude: then, the product of these two ratios may be assumed as the measure of the rectangle.

For example, if the base of the rectangle A contains ten units and its altitude three, the rectangle will be represented by the number 10x3

A

=30; a number which is entirely abstract, so long as we regard the numbers 10 and 3 as ratios.

But if we assume the square constructed on the linear unit, as the unit of surface, then, the 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.

In this case, the measurement which before was merely relative, becomes absolute: the number 30, for example, by which the rectangle was measured, now represents 30 superficial units, or 30 of those equal squares described on the unit of linear measure: this is called the Area of the rectangle.

Scholium 2. In geometry, the product of two lines frequently means the same thing as their rectangle, and this expression has passed into arithmetic, where it serves to designate the product of two unequal numbers. The term square is employed to designate the product of a number multiplied by itself.