Then since AD is equal to AB,

(Const.)

the square on AD is equal to the square on AB. Add to each the square on AC,

(I. 36, cor. 2.)

then the squares on DA and AC,

are equal to the squares on BA and AC.

But since the angle DAC is a right angle,

(Ax. 2.) (Const.)

the squares on DA and AC, are equal to the square on DC. (I. 37.) Hence the square on DC will be less than, equal to, or greater than the square on BC,

according as the square on BC is greater than, equal to, or less than the sum of the squares on BA and AC;

hence the line DC is less than equal to, or greater than the line BC, according as the square on BC is greater than, equal to, or less than the squares on BA and AC;

therefore the angle DAC is less than, equal to, or greater than the angle BAC,

(I. 4, 6, 18.)

according as the square on BC is greater than, equal to, or less than the squares on BA and AC. But DAC is a right angle;

(Const.)

therefore the angle BAC is greater than, equal to, or less than a right angle,

according as the square on BC is greater than, equal to, or less than the squares on BA and AC. Q. E. D.

SCHOLIA.

1. If each of two adjacent sides, AB ana AD, of a rectangle AC, be divided into any whole number of equal parts, and parallels to both be drawn through the points of division, the rectangle will be divided into a number of equal squares:—

For AL will be a square,

C

since the angle A is right, and AE and AH equal; also FL and LK are squares, for their sides are equal, and the external angles E and H each equal to A.

(Const.)

(Const.)

(I. 20, conv.)

And in the same manner all the others may be shewn to be squares; and they are all equal.

The figure is therefore divided into equal squares.

(1. 36, cor. 2.)

2. Now a line may be represented by a number expressing how many lineal units the line contains; that is, how many times some given magnitude of the same kind, used as a standard or unit of measure for the comparison of all like magnitudes, is contained in that line. Thus, the line AB may be represented by the number 4; meaning that the line contains the standard or lineal unit, RO, four times; and the line AD by the number 3, as containing the same unit three times, the measuring unit being a millimetre, an inch, a foot, or other unit agreed upon as a standard. In the same manner the lines may be represented by any letters, as a and b, each letter being understood to denote a number of lineal units.

3. If the lengths of any two sides of a right-angled

triangle be given in numbers, the third side can be found. Thus, if we suppose the base AB (fig. of cor. 2, Prop. 37) to be represented by the number 8, and the perpendicular BC by the number 6, the sum of the squares of these two lines will be 100, and this is the area of the square whose side is 10; so that AC will be represented by 10.

4. One of the squares in the figure, as AL, being taken as the superficial unit, or unit of area, the measure of the area of the rectangle AC will be the number of these squares.

Now there are evidently as many vertical columns of squares as there are parts in AB or DC, and as many squares in each column as there are parts in AD or BC. Also, there are as many horizontal rows as there are parts in AD, and as many squares in each row as there are parts in AB. Hence the entire number will be found by repeating the number in each column or row as many times as there are columns or rows; in the figure there are four vertical columns, each containing three squares; and three horizontal rows, each containing four squares; the entire number is therefore twelve, reckoning the number either way.

The number of superficial units is then the same as is obtained by multiplying the measure of AB by that of AD; that is, the number of lineal units in AB by the number in AD. Hence the area of a rectangle is briefly expressed by BA × AD, or BA.AD, which means that the two are multiplied together. Also, if a and b be the measures of the adjacent sides, that is, denote the number of lineal units in them respectively, the area, or number of superficial units, will be represented by axbor ab, or in this case 4 into 3, giving 12. Hence, if a be the side of a square its area will be represented by aa or a2. The practical rule then is: To find the area of a rectangle, multiply the adjacent sides together; but this is only an abbreviation of the rule, which ought to be stated at length as follows: Take as many superficial units as there are lineal units in one side, and repeat this number of superficial units as often as there are lineal units in the other side. Hence, of the factors a and b in the above product, one denotes superficial units, and the other an abstract number, indicating how often the other is to be repeated. This method of expressing the areas of figures establishes a connection

between geometry on the one hand, and arithmetic and algebra on the other; and shews the origin of the expressions "square of a number," "rectangle of two numbers," and "product of two lines." The first of these expressions is objectionable, but has got established; the "second power" of a number is a preferable expression; the second and third expressions are both improper and wholly unnecessary.-A rectangle may, however, be represented by one letter, as a; and if one side of such rectangle be represented by x, the other will be Also, if a represent the area of a square, ↓a will represent the side.

5. We often meet in geometry with lines such that no lineal unit can be found which is contained in, or is a measure of, both. Thus, if the side of a square is 1, the diagonal is 2 (schol. 3); if the side is 2, the diagonal is √8; if the side is 3, the diagonal is 18, etc. Now these numbers

2,8,18, etc., consist of a whole number and an infinite decimal, neither terminating, repeating, nor circulating, and therefore incapable of being measured by any lineal unit. Similarly, no lineal unit or number can be found expressing how many times the diameter of a circle is contained in the circumference. Such numbers are called incommensurable. Hence it is that, in applying arithmetic and algebra to the solution of geometrical problems, we meet with difficulties which have no place in the methods followed by Euclid, which apply alike to all magnitudes whether commensurable or incommensurable.

6. Since a parallelogram is equal to a rectangle on the same base and of the same altitude, and since the altitude is equal to the side of the rectangle, it follows that the area of any parallelogram will be found by multiplying the base by the altitude; that is, b being the base and a the altitude of any parallelogram, the area is ab.

7. Since a triangle is half the rectangle on the same base and of the same altitude (I. 32), the area will be found by multiplying the base by the altitude and taking half the product; the area therefore is ab.

8. The area of a trapezium will be found by dividing it into two triangles by a diagonal, and finding the sum of their areas. This is most readily done by multiplying the

diagonal by half the sum of the perpendiculars; but if one of the angles is reflex, so that the perpendiculars lie on the same side of the diagonal, the difference of the perpendiculars must be employed. The expression for the area therefore is dx, in which d is the diagonal, and p and p' the perpendiculars.

2

9. Every polygon may be divided into triangles or trapeziums by drawing diagonals, and its area will be found by finding that of these figures. In the case of a regular polygon of n sides, the area will be n times that of one of the component triangles formed by drawing two lines from the centre of the figure to the extremities of any side, since these triangles are all equal. Thus, if a be the side and p the perpendicular from the centre upon it, the expression for the area is n x orp; that is, the area is equal to the product of half the perimeter by the perpendicular from the centre.

αρ

2

na

PROPOSITIONS.

PROPOSITION A.-THEOREM.

The diagonals of a parallelogram bisect one another.

Let ABCD be a parallelogram;

it is required to prove that its diagonals AC, BD bisect one another in the point O.

D

Because AB is parallel to DC, and DB meets them, the angle ABO is equal to the alternate angle CDO;

for the same reason the angle BAO

is equal to the alternate angle DCO,

(I. 20, conv.)

and the sides AB and CD, between the equal angles, are equal; therefore the sides BO and OA are equal to the sides DO and OC, each to each;

that is, the diagonals bisect one another.

(I. 4, Pt. 2.)

Conversely. If the diagonals of a quadrilateral bisect one another, the figure is a parallelogram.

Let the diagonals AC, BD of the quadrilateral ABCD,

E