PROPOSITION I. THEOREM.

Parallelograms which have equal bases and equal altitudes are equivalent.

Let AB be the common base of P C F E DF CE the two parallelograms ABCD, ABEF: and since they are supposed to have the same altitude, their upbases DC, FE will be situated

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both 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.

It follows (11. I.) that the triangles DAF, CBD, are mutually equilateral, and consequently equal.

But if from the quadrilateral ABED, we take away the triangle ADF, there will remain the parallelogram ABEF: and if from the same quadrilateral ABED, we take away the triangle CBE, there will remain the parallelogram ABCD. 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 ABC is half of the parallelogram ABCD, which has the same base and the same altitude.

For (31. I.), the triangles ABC, F ACD are equal.

Cor. 1. Hence a triangle ABC is half of the rectangle BCEF, which has the same base CB, and the same alti- B tude AO: for the rectangle BCEF is equivalent to the parallelogram ABCD.

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Cor. 2. All triangles, which have equal bases and altitudes, are equivalent.

PROPOSITION III. THEOREM.

Two rectangles having the same altitude 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.

Suppose, first, that the bases are commensurable, and are to each other,

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for example, as the numbers 7 and 4. If AB is 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 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.

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

ABCD: AEFD::AB: AE.

For if not, the first three terms continuing the same, the fourth must be greater or less than AE. Suppose it to be greater, and that we have

ABCD: AEFD::AB: AO.

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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, and we find

AIKD: AEFD :: AI: AO.

But AO is greater than AI; hence, if this proportion is cor

rect, the rectangle AEFD must be greater than AIKD: on the contrary, however, it is less; hence the proportion is impossible; hence ABCD cannot be to AEFD as AB is to a line greater than AE.

Exactly in the same manner, it may be shewn 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 ABCD, AEGF, are to each other as the products of the bases multiplied by the altitudes, so that we have ABCD: AEGF:: AB×AD: AEX AF.

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, E AEHD, having the same altitude, are to each other as their bases AB, AE in like manner the two rect

angles AEHD, AEGF, having the same altitude AE, are to each other as their bases AD, AF: thus we have the two proportions,

ABCD: AEHD::AB: AE,

AEHD: AEGF:: AD: AF.

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

ABCD: AEGF:: ABXAD: 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.

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 rectangles, agreeably to the position just demonstrated.

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For example, if the base of the rectangle A contains 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 second rectangle will 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 3 for its absolute measure, in other words, it would be equal to § of a superficial unit.

It is more common and more simple, to assume the square 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, as the diagram exhibits.

In geometry the product of two lines is frequently confounded with their rectangle, and this expression has passed into arithmetic, where it serves to designate the product of two unequal numbers, the expression square being employed to designate the product of a number multiplied by itself.

The arithmetical squares of 1, 2, 3,

&c. are 1, 4, 9, &c. So likewise the geometrical square constructed on a double line is evidently four times as great as on a single one; on a triple line, is nine times as great, &c.

PROPOSITION V. THEOREM.

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

For the parallelogram ABCD is equivalent FD (1. III. Cor.) to the rectangle ABEF, which has the same base AB, and the same altitude BE: but this rectangle (4. III. Schol.) is measured by ABX BE; therefore ABX BE is A equal to the area of the parallelogram ABCD.

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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: for if A, B, C are any three magnitudes, we have always Ax C: BXC:: A: B.

PROPOSITION VI. THEOREM.

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

For, the triangle ABC (2. III.) is half of the parallelogram ABCE, which has the. same base BC, and the same altitude AD: but the surface of the parallelogram (5. III.) is equal to BCX AD; hence that of the triangle must be BC× AD, or BC×1AD.

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Cor. Two triangles of the same altitude are to each other as their bases, and two triangles of the same base are to each other as their altitudes.

PROPOSITION VII. THEOREM.

The area of the trapezium ABCD is equal to its altitude EF, multiplied by the half sum of its parallel bases, AB, CD.

Through I, the middle point of the side BC, draw KL parallel to the opposite side AD; and produce DC till it meet KL.

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In the triangles IBL, ICK, we have the side IB IC, by construction; the angle LIB=CIK; and (23. I.) since CK and A

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BL are parallel, the angle IBL=ICK; hence the triangles are equal; hence the trapezium ABCD is equivalent to the parallelogram ADKL, and is measured by EFX AL.

But we have AL-DK; and since the triangles IBL and KCI are equal, the side BL-CK: hence AB+CD=AL+ DK=2 AL; hence AL is the half sum of the bases AB, CD; hence the area of the trapezium ABCD is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result AB+CD

which is expressed thus: ABCD=EF× 2

Scholium. If through I, the middle point of BC, the line IH be drawn parallel to the base AB, the point H will also be the middle of AD. For, because the figure AHIL is a parallelogram, as DHIK is likewise, their opposite sides being parallel, we have AH-IL, and DH IK; but since the triangles BIL, CIK are equal, we already have IL-IK; therefore AH-DH.