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1. Every right-angled parallelogram is called a rectangle, and is said to be contained by any two of the straight lines which contain one of the right angles.
2. In any parallelogram, one of the parallelograms about a diameter, together with the two complements, is called a gnomon.
Thus EK with the complements AF, FC is the gnomon AKG.
[Book ii. treats of the properties and relations of rectangles.
E F G B
Let AB represent four yards, and AC, perpendicular to it, three yards; complete the rectangle ABDC, and draw perpendiculars to AB, AC through E, H, &c.
The rectangle is now divided into twelve squares and each of their sides is a yard: thus it contains 12 sq. yds., which is said to be its area. Similar reasoning will hold whatever be the linear unit taken and whatever the number of units in
AB, AC. Hence, generally, the area of a rectangle whose sides contain a and b linear units is ab square units. Now it is proved in i. 41 that a triangle is half of any parallelogram (and therefore half of the rectangle) on the same base and between the same parallels. Hence the area of a triangle whose altitude (that is, the perpendicular from the vertex on the base) is a units long, and whose base is b units long, is fab.
But it may happen that there is no number, whole or fractional, which will express the number of times which AC contains the unit of length. For instance if AB were the diagonal of a square whose side is equal to AC: then if AB contains a units we should have AC =
(i. 47), a' nonterminating decimal equal to no exact number, whole or fractional, and AB, AC have no common measure, that is, they are incommensurable. In geometry we frequently meet with incommensurable magnitudes, in which case our method of finding the area of the rectangle AD would fail. Thus we cannot satisfactorily demonstrate the propositions of the second book concerning rectangles and their properties by means of algebraical processes, because, (1) such demonstrations would only hold when the sides of the rectangle were commensurable, and, (2) because the subject of Geometry is not number, but magnitude: we must prove the propositions by means of geometrical constructions and demonstrations. It may be useful however to give algebraical proofs—(rather, analogous or corresponding results in algebra)—of the propositions, observing that they do not hold when the lines have no common measure, and they cannot in any case replace the more rigorous, more general, and more elegant geometrical demonstrations given by Euclid.
The only abbreviations admitted for the square on AB' is 'sq. on AB,' and for the rectangle contained by AB and CD,' rect. AB, CD.]
PROPOSITION I. THEOREM.
If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangle contained by the undivided line, and the several parts of the divided line.
Let AB and CD be two st. lines ; and let CD be divided into any parts in E, F.
Then shall rect. AB, CD = rect. AB, CE, together with rect. AB, EF and rect. AB, FD.
Constr. From C draw CG I CD, (i. 11) and make
CH= AB; (i. 3) through H draw HM || CD, and
through E, F, D draw EK, FL, DM || CH. (i. 31) Dem. Then rect. CM = rects. CK, EL, FM.
But CM is contained by AB, CD, for it is contained by CH, CD and CH = AB; and CK is contained by AB, CE, for it is contained by CH, CE, and CH =AB; similarly EL is contained by AB, EF, for EK = CH= AB; and FM is contained by AB, FD;
therefore the rect. AB, CD = rect. AB, CE, together with rect. AB, EF and rect. AB, FD.
Therefore, if there be, &c. Q. E. D. [The corresponding formula in algebra runs thus :Let AB, CD be commensurable and let AB contain a linear units, CD b units, and suppose CE, EF, FD contain respectively x,y,z units. Then b= x + y +z: multiply these equals by a, ... ab = ax + ay + az; that is, if there be two numbers, one of which is divided into any number of parts, the product of the two numbers is equal to the sum of the products of the undivided number and the several parts of the other.']
PROPOSITION II. THEOREM.
If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square on the whole line.
Let the st. line AB be divided into any two parts at C.
Then rect. AB, BC together with rect. AB, AC, = sq. on AB.
Constr. On AB describe sq. AE, (i. 46) and through C
draw CF || AD or BE. Dem. Then AE = rects. AF, CE. But AE is the sq. on AB; AF = the rect. AC, AB, for it is contained by AC, AD, and AD = AB;
and CE = rect. AB, CB, for BE= AB; .. rect. AB, AC together with rect. AB, BC = sq. on AB.
Therefore, if a straight line, &c. Q. E. D. [Corresponding result in algebra :—Let AB contain 6 units, AC, BC, x, y units respectively; then b =
= x + y, ..62 = bx +by. The reader should interpret this formula in terms corresponding to the enunciation of the proposition.]