First, let AD fall within the triangle ABC.

Then because the straight line CB is divided into two parts in D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD, and the square on DC; (II. 7.)

to each of these equals add the square on AD;

therefore the squares on CB, BD, DA, are equal to twice the rectangle CB, BD, and the squares on AD, DC;

but the square on AB is equal to the squares on BD, DA, (1.47.) because the angle BDA is a right angle ;

and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC, and twice the rectangle CB, BD:

that is, the square on AC alone is less than the squares on CB, BA, by twice the rectangle CB, BD.

Secondly, let AD fall without the triangle ABC.

Then, because the angle at D is a right angle,
the angle ACB is greater than a right angle; (I. 16.)

and therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD; (II. 12.)

to each of these equals add the square on BC;

therefore the squares on AB, BC are equal to the square on AC, twice the square on BC, and twice the rectangle BC, CD; but because BD is divided into two parts in C,

therefore the rectangle DB, BC is equal to the rectangle BC, CD, and the square on BC; (II. 3.)

and the doubles of these are equal;

that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC:

therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC:

wherefore the square on AC alone is less than the squares on AB, BC; by twice the rectangle DB, BC.

Lastly, let the side AC be perpendicular to BC.

Then BC is the straight line between the perpendicular and the acute angle at B;

and it is manifest, that the squares on AB, BC, are equal to the square on AC, and twice the square on BC. (1. 47.)

Therefore in any triangle, &c. Q. E.D.

F

PROPOSITION XIV. PROBLEM.

To describe a square that shall be equal to a given rectilineal figure.
Let A be the given rectilineal figure.

It is required to describe a square that shall be equal to A.

Describe the rectangular parallelogram BCDE equal to the rectilineal figure 4. (1. 45.)

Then, if the sides of it, BE, ED, are equal to one another,

it is a square, and what was required is now done.

But if BE, ED, are not equal,

produce one of them BE to F, and make EF equal to ED,
bisect BF in G; (I. 10.)

from the center G, at the distance GB, or GF, describe the semicircle BHF,

and produce DE to meet the circumference in H.

The square described upon EH shall be equal to the given rectilineal figure 4.

Join GH.

Then because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in the point E;

therefore the rectangle BE, EF, together with the square on EG, is equal to the square on GF; (II. 5.)

but GF is equal to GH; (def. 15.)

therefore the rectangle BE, EF, together with the square on EG, is equal to the square on GH;

but the squares on HE, EG are equal to the square on GH; (1. 47.) therefore the rectangle BE, EF, together with the square on EG, is equal to the squares on HE, EG;

take away the square on EG, which is common to both; therefore the rectangle BE, EF is equal to the square on HE. But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED;

therefore BD is equal to the square on EH;

but BD is equal to the rectilineal figure A; (constr.)

therefore the square on EH is equal to the rectilineal figure A. Wherefore a square has been made equal to the given rectilineal figure A, namely, the square described upon EH.

Q. E.F.

NOTES TO BOOK II.

IN Book 1, Geometrical magnitudes of the same kind, lines, angles and surfaces, more particularly triangles and parallelograms, are compared, either as being absolutely equal, or unequal to one another.

In Book 11, the properties of right-angled parallelograms, but without reference to their magnitudes, are demonstrated, and an important extension is made of Euc. 1. 47, to acute-angled and obtuse-angled triangles. Euclid has given no definition of a rectangular parallelogram or rectangle: probably, because the Greek expression Tapаóуpaμ ὀρθογώνιον, οι ὀρθογώνιον simply, is a definition of the figure. In English, the term rectangle, formed from rectus angulus, ought to be defined before its properties are demonstrated. A rectangle may be defined to be a parallelogram having one angle a right-angle, or a right angled parallelogram; and a square is a rectangle having all its sides equal.

As the squares in Euclid's demonstrations are squares described or supposed to be described on straight lines, the expression "the square on AB," is a more appropriate abbreviation for "the square described on the line AB," than "the square of AB." The latter expression more fitly expresses the arithmetical or algebraical equivalent for the square on the line AB.

In Euc. 1. 35, it may be seen that there may be an indefinite number of parallelograms on the same base and between the same parallels whose areas are always equal to one another; but that one of them has all its angles right angles, and the length of its boundary less than the boundary of any other parallelogram upon the same base and between the same parallels. The area of this rectangular parallelogram is therefore determined by the two lines which contain one of its right angles. Hence it is stated in Def. 1, that every right-angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles. No distinction is made in Book II, between equality and identity, as the rectangle may be said to be contained by two lines which are equal respectively to the two which contain one right angle of the figure. It may be remarked that the rectangle itself is bounded by four straight lines.

It is of primary importance to discriminate the Geometrical conception of a rectangle from the Arithmetical or Algebraical representation of it. The subject of Geometry is magnitude not number, and therefore it would be a departure from strict reasoning on space, to substitute in Geometrical demonstrations, the Arithmetical or Algebraical representation of a rectangle for the rectangle itself. It is however, absolutely necessary that the connexion of number and magnitude be clearly understood, as far as regards the representation of lines and areas.

All lines are measured by lines, and all surfaces by surfaces. Some one line of definite length is arbitrarily assumed as the linear unit, and the length of every other line is represented by the number of linear units, contained in it. The square is the figure assumed for the measure of surfaces. The square unit or the unit of area is assumed to be that square, the side of which is one unit in length, and the magnitude of every surface is represented by the number of square units contained in it. But here it may be remarked, that the properties of rectangles and squares in the Second Book of Euclid are proved independently

of the consideration, whether the sides of the rectangles can be represented by any multiples of the same linear unit. If, however, the sides of rectangles are supposed to be divisible into an exact number of linear units, a numerical representation for the area of a rectangle may be deduced.

On two lines at right angles to each other, take AB equal to 4, and AD equal to 3 linear units.

Complete the rectangle ABCD, and through the points of division of AB, AD, draw EL, FM, GN parallel to AD; and HP, KQ parallel to AB respectively.

Then the whole rectangle AC is divided into squares, all equal to each other. And AC is equal to the sum of the rectangles AL, EM, FN, GC; (II. 1.) also these rectangles are equal to one another, (1. 36.) therefore the whole AC is equal to four times one of them AL. Again, the rectangle L is equal to the rectangles EH, HR, RD, and these rectangles, by construction, are squares described upon the equal lines AH, HK, KĎ, and are equal to one another.

Therefore the rectangle AL is equal to 3 times the square on AH, but the whole rectangle AC is equal to 4 times the rectangle AL, therefore the rectangle AC is 4 x 3 times the square on AH, or 12 square units:

that is, the product of the two numbers which express the number of linear units in the two sides, will give the number of square units in the rectangle, and therefore will be an arithmetical representation of its area.

And generally, if AB, AD, instead of 4 and 3, consisted of a and b linear units respectively, it may be shewn in a similar manner, that the area of the rectangle AC would contain ab square units; and therefore the product ab is a proper representation for the area of the rectangle AC.

Hence, it follows, that the term rectangle in Geometry corresponds to the term product in Arithmetic and Algebra, and that a similar comparison may be made between the products of the two numbers which represent the sides of rectangles, as between the areas of the rectangles themselves. This forms the basis of what are called Arithmetical or Algebraical proofs of Geometrical properties.

If the two sides of the rectangle be equal, or if b be equal to a, the figure is a square, and the area is represented by aa or a2.

Also, since a triangle is equal to the half of a parallelogram of the same base and altitude;

Therefore the area of a triangle will be represented by half the rectangle which has the same base and altitude as the triangle: in other words, if the length of the base be a units, and the altitude be b units; Then the area of the triangle is algebraically represented by bab. The demonstrations of the first eight propositions, exemplify the obvious axiom, that, "the whole area of every figure in each case, is equal to all the parts of it taken together.'

Def. 2. The parallelogram EK together with the complements AF

FC, is also a gnomon, as well as the parallelogram HG together with the same complements.

Prop. I. For the sake of brevity of expression, "the rectangle contained by the straight lines AB, BC,” is called "the rectangle AB, BC;" and sometimes "the rectangle ABC."

To this proposition may be added the corollary: If two straight lines be divided into any number of parts, the rectangle contained by the two straight lines, is equal to the rectangles contained by the several parts of one line and the several parts of the other respectively.

The method of reasoning on the properties of rectangles, by means of the products which indicate the number of square units contained in their areas, is foreign to Euclid's ideas of rectangles, as discussed in his Second Book, which have no reference to any particular unit of length or measure of surface.

Prop. 1. The figures BH, BK, DL, EH are rectangles, as may readily be shewn. For, by the parallels, the angle CEL is equal to EDK; and the angle EDK is equal to BDG (Euc. I. 29.). But BDG is a right angle. Hence one of the angles in each of the figures BH, BK, DL, ÈH is a right angle, and therefore (Euc. 1. 46, Cor.) these figures are rectangular.

Prop. I. Algebraically. (fig. Prop. 1.)

Let the line BC contain a linear units, and the line A, b linear units of the same length.

Also suppose the parts BD, DE, EC to contain m, n, p linear units respectively.

Then am + n + P,

multiply these equals by b,

therefore ab= bm + bn + bp.

That is, the product of two numbers, one of which is divided into any number of parts, is equal to the sum of the products of the undivided number, and the several parts of the other;

or, if the Geometrical interpretation of the products be restored,

The number of square units expressed by the product ab, is equal to the number of square units expressed by the sum of the products bm, bn, bp.

Prop. II. Algebraically. (fig. Prop. 11.)

Let AB contain a linear units, and AC, CB, m and n linear units respectively.

multiply these equals by a,

Then m + n = a,

therefore am + an = a2.

That is, if a number be divided into any two parts, the sum of the products of the whole and each of the parts is equal to the square of the

whole number

Prop. I. Algebraically. (fig. Prop. 11.)

Let AB contain a linear units, and let BC contain m, and AC, n linear units.

multiply these equals by m,

Then a = m + n,

therefore ma = m2 + mn.

That is, if a number be divided into any two parts, the product of the whole number and one of the parts, is equal to the square of that part, and the product of the two parts.