(15. Cor. 3.) AD is parallel to BC. And, for the like reason, A B is parallel to CD.

Or, let the diagonals A C, BD, bisect one another in E: then, because the triangles E A D, E C B, have two sides of the one equal to two sides of the other, each to each, and the included angles AED, CE B, (3.) equal to one another, the angle E A D is (4.) equal to EC B, and, therefore, (15.) AD is parallel to B C. And, for the like reason, A B is parallel to CD.

Therefore, in each of the three cases, the figure is a parallelogram. Therefore, &c.

Cor. 1. (Euc. i. 34. second part of.) A parallelogram is bisected by each of its diagonals; for the triangles into which it is divided are equal to one another.

Cor. 2. The diagonals of a rhombus bisect one another at right angles. For the triangles into which it is divided by either of its diagonals are isosceles triangles, of which that diagonal is the base (6. Cor. 4.)

Cor. 3. (Euc. i. 46. Cor.) If one angle of a parallelogram be a right angle, all its angles will be right angles.

Cor. 4. By help of this Proposition more complete notions may be acquired of the rhombus, rectangle, and square: for, hence it appears, that a rhombus has all its sides equal to one another; that a rectangle has all its angles right angles; and that a square has all its sides equal, and all its angles right angles.

PROP. 23. (EUc. i. 43.)

The complements of the parallelograms, which are about the diagonals of any parallelogram, are equal to one another.

F

B

it D

Let A B C D be A a parallelogram, and through any point E in the diagonal BD let there be drawn the straight lines FG, HK parallel to the sides B C, DC respectively: the complement A E shall be equal to E C. Because ABCD is a parallelogram, of which BD is a diagonal, the triangle ABD is (22. Cor. 1.) equal to CD B. In like manner, because F K, and HG (17. Cor.) are parallelograms, the triangles FBE, HED are equal to the triangles K EB, G D E: therefore, taking these equals from the former, there remains (ax. 3.) the complement AE equal to E C. Therefore, &c.

PROP. 24. (EUc. i. 35.) Parallelograms upon the same base and between the same parallels, ar equal to one another.

WW

Let the parallelograms A B C D, EBCF, be upon the same base B C and between the same parallels AF, BC; the parallelogram A B C D shall be equal to the parallelogram EBCF.

Because AD and EF are each of them (22.) equal to BC, they are (ax. 1.) equal to one another. Therefore, the whole or the remainder AE is equal to the whole or the remainder DF (ax. 2. or 3.). Therefore, the two triangles E A B, FDC, having two sides of the one equal to two sides of the other, each to each, and the included angles E AB, FDC equal, are equal to one another (4. Cor.). Therefore, taking each of these equals from the whole figure A B C F, there remains (ax. 3.) the parallelogram EBCF equal to the parallelogram A B CD.

If the points D, E coincide, AE is the same with AD, and DF the same with EF; therefore, AE and DF, being each of them equal to BC, are equal to one another; and, hence, the triangle E AB is equal to the triangle FD C, and the parallelogram EBCF to the parallelogram ABCD, as before.

(21.), and EBCH is a parallelogram. But the parallelogram E BCH is equal to A B CD, because it is upon the same base BC, and between the same parallels (24.): and for the like reason EB CH is equal to E F G H. Therefore (ax. 1.) ABCD is equal to EFGH. Therefore, &c.

Cor. The squares of equal straight lines are equal to one another: and conversely.

PROP. 26. (Euc. i. 41.) If a parallelogram and a triangle be upon the same base and between the same parallels, the parallelogram shall be double of the triangle. Let the parallelogram ABCD and the triangle E B C be upon the same base B C, and between the same parallels AD, BC. The parallelogram ABCD shall be double of the triangle EBC.

W

Complete the parallelogram E B CF.* (14. Cor. 1.) Then the parallelogram EBCF is double of the triangle EBC, because it is bisected by the diagonal E C (22. Cor.); and A B C D is equal to E B CF, because it is upon the same base, and between the same parallels (24.). Therefore, the parallelogram ABCD is also double of the triangle E B C. Therefore, &c.

Cor. Every triangle is equal to the half of a rectangle of the same base and altitude.

PROP. 27. (Euc. i. 37, 38, 39, & 40.) Triangles upon the same base, or upon equal bases, and between the same parallels, are equal to one another: and conversely, equal triangles, upon the same base, or upon equal bases in the same straight line, and towards the same parts, are between the same parallels.

The first part of the proposition is manifest; for the triangles are the halves of parallelograms (26.) upon the

E

D

same base, or upon equal bases, and between the same parallels; and because these parallelograms are equal to one another (24. or 25.), the triangles, which are their halves, are also equal (ax. 5.).

To complete the parallelogram, in this case, it is only requisite that CF should be drawn through the point parallel to B E, to meet AD produced in F. The word complete, indeed, almost explains itself: in future constructions it will be introduced without further notice.

In the second place, therefore, let the triangles A B C, D B C, standing upon the same base B C, or upon equal bases B C, B C, in the same straight line, and towards the same parts, be equal to one another; and let AD be joined: AD shall be parallel to B C.

For, if not, let A E (14. Cor. 1.) be parallel to BC; and let it meet D B in E. Join E C. Then, by the first part of the proposition, because ABC, E B C, are upon the same base, or upon equal bases, and between the same parallels, the triangle EBC is equal to ABC, that is, to DBC; the less to the greater, which is impossible. Therefore A E is not parallel to BC; and in the same manner it may be shown that no other straight line which passes through A, except A D only, can be parallel to BC; that is, (14. Cor. 1.) AD is parallel to B C.

Therefore, &c.

Cor. 1. It is evident that the second (or converse) part of the proposition applies equally to parallelograms, as to triangles.

Cor. 2. If a quadrilateral be bisected by each of its diagonals, it must be a parallelogram for the two triangles ABC, DBC (see the figure of Prop. 21.) which stand upon any one of its sides BC for a base, and which have their vertices in the side opposite, being equal, each of them, to half the quadrilateral, are equal to one another; and therefore AD is parallel to BC; and, for a like reason, A B is parallel to D C.

PROP. 28.

A trapezoid is equal to the half of a rectangle having the same altitude, and a base equal to the sum of its parallel sides.

D

Let ABCD be a trapezoid, having the side AD parallel to the side B C. The trapezoid ABCD shall be equal to the half of a rectangle having the same altitude, and a base equal to the sum of AD, BC.

Join AC; through D draw DE parallel to A C, and let it meet the base B C produced in E, and join A E.

Then, because D E is parallel to A C, the triangle A CD is (27.) equal to ACE, Therefore, the triangles ABC, ACD, together, are equal to the triangles AB C, ACE, together (ax. 2.); or, the trapezoid ABCD is equal to the tri

C

angle ABE. But the base of the triangle ABE is equal to the sum of AD, BC, because (22.) CE is equal to AD; and (26. Cor.) every triangle is equal to the half of a rectangle of the same base and altitude. Therefore, the trapezoid ABCD is equal to the half of a rectangle of the same altitude, and upon a base which is equal to the sum of A D, BC.

Therefore, &c.

PROP. 29.

If the adjoining sides of a rectangle contain, each of them, the same straight line, a certain number of times exactly, the rectangle shall contain the square of that straight line, as often as is denoted by the product of the two numbers, which denote how often the line itself is contained in the two sides.

Let ABDC be a

rectangle, and let its adjacent sides AB, AC, contain each of them the straight line M a certain number of times exactly, viz., AB 6 times, and AC 4

times the rectangle ABDC shall contain the square of M, 6 x 4, or 24 times. Divide A B, A C, each of them, into parts equal to M; and, through the division-points of each, draw straight lines parallel to the other, thereby dividing the rectangle into six upright rows of four parallelograms each, that is, upon the whole, into twenty-four parallelograms. Now these parallelograms are all of them rectangular, because their containing sides are parallel to AB, AC, the sides of the right angle A (18.). They are also equilateral: for any one of them, as E, has its upright sides each of them (22,) equal to a division of A C, that is, to M; and its other two sides each of them equal to a division of A B, that is, to M. Therefore, they are squares (def. 20.), equal, each of them, to the square of M. And they are twenty-four in number. Therefore, the square of M is contained twenty-four times in the rectangle A B D C.

The same may be said, if, instead of 6 and 4, any other two numbers be taken. Therefore, &c.

Cor. 1. In like manner, it may be shown, that, if there be two straight lines, one of which is contained an exact number of times in one side of a rectangle, and the other an exact number of times in the side adjoining to it; the rectangle

under those two straight lines shall be contained as often in the given rectangle, as is denoted by the product of the two numbers which denote how often the lines themselves are contained in the two sides

Cor. 2. The square of twice M is equal to 4 times M square, because it is a rectangle, in which each of the sides contains M twice. In like manner, the square of 3 times M is equal to 9 times M square-of 4 times M to 16 times M square of 5 times M to 25 times M square, &c.

Cor. 3. The square of 5, or 25, is equal to the sum of 16 and 9. Consequently the square of 5 times M is equal to the square of 4 times M, together with the square of 3 times M.

Scholium.

From the theorems of this Section rules are easily deduced for the mensuration of rectilineal figures. For every rectilineal figure may be divided into triangles; and every triangle, being equal (25. Cor.) to half the rectangle under its base and altitude, contains half as many square units as is denoted by the product of the numbers which express how often the corresponding linear unit is contained in its base and in its altitude. Let this linear unit be, for example, a foot; and let it be required to find how many square feet there are in a triangle whose altitude is 10 feet, and its base 9 feet. The rectangle, of which these are the sides, contains 10 × 9, or 90 square feet (by Prop. 29.); and, therefore, the triangle contains 45 square feet.

Hence, a rectangle is sometimes said to be equal to the product of its base and altitude, a triangle to half the product of its base and altitude, and the like; expressions which must be understood as above, the words rectangle, &c. base, &c. being briefly put for the number of square units in the rectangle, &c. the number of linear units in the base, &c. By the length of a line is commonly understood the number of linear units which it contains; and the term superficial area, or area, is similarly applied to denote the number of square units in a surface. With regard to the measuring unit, a less or a greater is convenient, according to the subject of measurement: a glazier measuring his glass by square inches, a carpenter his planks by square feet, a proprietor his land by acres, and a geographer the extent of countries by square miles.