Having placed the two rectangles so that the angles at A may be vertical and opposite, produce the sides CD, GF until they meet in K; the two rectangles ABCD, AFKD, having the same altitude AD, are to each other as their bases AB, AF. In like manner, the two rectangles AFKD, AFGH, having the same altitude AF, are to each other as their bases, AD, AF. Hence we have these two proportions:

ABCD AFKD: AB AF;

AFKD

:

AFGH:: AD: AH.

Multiplying the corresponding terms of these proportions together, and observing to omit AFKD, since it will occur in an antecedent consequent, we shall have

ABCD AFGH: ABX AD AH× AF.

Scholium. Hence we are at liberty to take for the measure of a rectangle, the product of its base by its altitude, provided we understand by this product the same as the product of two numbers, which numbers denote the linear units in the base and altitude respectively.

This measure, however, is not absolute, but only relative. It supposes that the area of any other rectangle is estimated in a similar manner, by measuring its sides by the same linear unit; we shall thus obtain a second product, and the ratio of these two products is the same as that of the two rectangles, in accordance with this proposition.

For example, if the base of the rectangle A is ten

units, and its height

A

three, the rectangle will be represented by the number 10x3=30, a number which, thus is

olated, has no signification; but if we have a second rectangle B, whose base is twelve units and height seven, this second rectangle will be represented by the number 12x7=84. From which we conclude that the two rectangles A and B, are to each other as 30 to 84. If we take the rectangle A as the unit of measure of surfaces, the rectangle B will have for its measure, that is, it will be of our superficial units.

It is more common and more simple to take a square for the unit of surface, and we choose a square whose side is a unit of length. In this case, the measure which we have regarded as relative becomes absolute; for example, the number 30, which measured the rectangle A, represents 30 superficial units, or 30 squares, each side of which is a unit long.

In geometry, we frequently confound the product of two lines with that of their rectangle, and this expression is even employed in arithmetic to denote the product of two unequal numbers; but we use the term square to denote the product of a number multiplied by itself. The squares of the numbers 1, 2,

3, &c., are 1, 4, 9, &c. From which we see that the square formed on a line of double length is quadruple; on a line of triple length it

is nine times as great, and so of other squares.

PROPOSITION III.

THEOREM. The area of any parallelogram is equal to the product of its base by its height.

G D

F

For the parallelogram ABCD is equivalent to the rectangle ABFG, which has the same base AB, and same altitude BF, (B. II, Prop. II.) But the rectangle ABFG is measured by AB× BF, consequently the parallelogram is measured by AB× BF.

A

B

Cor. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; and in general, parallelograms are to each other as the products of their bases multiplied by their altitudes.

PROPOSITION IV.

THEOREM. The area of a triangle is equal to the product of its base by half its altitude.

For, the triangle ABC is half the. parallelogram ABCF, (B. II, Prop.

III.)

B

The parallelogram is measured by the base BC multiplied into the altitude AD; therefore, the triangle is measured by the base BC into half the altitude AD.

Cor. Triangles of the same altitude are to each other as their bases; and triangles of the same or equal bases are to each other as their altitudes.

PROPOSITION V.

THEOREM. If a line be drawn parallel to one side of a triangle, cutting the other two sides, these sides will be divided into proportional parts.

Let DF be parallel to the side BC of the triangle ABC; then will AD : DB :: AF : FC.

For, draw BF and CD; then, the triangles DBF, FCD are equal to each other, be

B

D

F

cause they have the same base DF, and are between the same parallels DF, BC, (B. II, Prop. II, Cor. 1.) But the two triangles ADF, BDF, on the bases AD, DB, have the same altitude; and the two triangles ADF, CDF, on the bases AF, FC, have also the same altitude; and because triangles of the same altitude are to each other as their bases, (B. IV, Prop. iv, Cor.) therefore,

ADF BDF :: AD : DB;

ADF: CDF :: AF: FC.

But BDF=CDF; consequently, by equality of ratios, we have AD: DB:: AF: FC.

In a similar manner, the theorem is proved when the sides of the triangle are cut in prolongation beyond either the vertex or the base.

Cor. Hence, also, the whole lines AB, AC are proportional to their corresponding proportional segments. Thus, since AD: DB:: AF: FC, we have by composition AD+DB: AD :: AF+FC: AF, or AB AD :: AC: AF: and

:

AD+DB : DB :: AF+FC: FC, or

THEOREM. If two sides of a triangle are cut by a straight line, so that the corresponding parts shall be proportional, this line will be parallel to the third side.

In the triangle ABC, let the line DF be drawn, so that

then will DF be parallel to BC.

For if DF is not parallel to BC, suppose that from the point D, the line. DG be drawn parallel to BC. Then we have (B. IV, Prop. v.)

B

D

AD: DB:: AG: GC. But, by hypothesis, we have AD: DB:: AF: FC; therefore, we must have AG GC AF: FC, or

AG : AF :: GC : FC, which is an impossible result, since the antecedent of the first couplet is less than its consequent, while the antecedent of the second couplet is greater than its consequent. Hence the line drawn from D parallel to BC cannot differ from the line DF; that is, DF is parallel to BC.