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(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 CB, have two sides of the one equal to two sides of the other, each to each, and the included angles A ED, 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.

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Let the parallelograms A B C D, EBCF, be upon the same base B C and between the same parallels A F, BC; the parallelogram A B CD 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 EA B, FDC, having two sides of the one equal to two sides of the other, each to each, and the included angles E A B, F DC 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 EBC F equal to the parallelogram A B C D.

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

Therefore, &c.

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

PROP. 25. (Euc. i. 36.) Parallelograms upon equal bases, and between the same parallels, are equal to one another.

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Let A B C D be a parallelogram, and through any point E in the diagonal BD let there be drawn the straight lines FG, HK parallel to the sides BC, DC respectively: the complement A E shall be equal to E C. Because ABCD is a parallelogram, of which B D 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.

Let the parallelograms ABCD, EFGH, be upon equal bases BC, FG, and between the same parallels

AH, BG. The parallelogram A B C D shall be equal to the parallelogram EFG H.

Join E B, HC. Then because EH (22.) and B C are each of them equal to FG, they are equal to one another; and they are likewise parallel; therefore, EB and HC are also equal and parallel

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(21.), and EB CH is a parallelogram. But the parallelogram E B C H 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 E F G H. 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 A D, BC. The parallelogram ABCD shall be double of the triangle E B C.

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Complete the parallelogram EBC F. (14. Cor. 1.) Then the parallelogram EBCF is double of the triangle EBC, because it is bisected by the diagonal EC (22. Cor.); and A B CD is equal to EB CF, because it is upon the same base, and between the same parallels (24.). Therefore, the parallelogram ABCD is also double of the triangle EBC. 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

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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 C F 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, DB C, standing upon the same base B C, or upon equal bases BC, 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 А В С, 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.

by each of its diagonals, it must be a Cor. 2. If a quadrilateral be bisected 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, AD is parallel to BC; and, for a like are equal to one another; and therefore reason, A B is parallel to D C.

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angle ABE. But the base of the triangle ABE is equal to the sum of AD, BC, because (22.) C E 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 AD, 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

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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 (26. 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 x 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.

SECTION 5. Rectangles under the parts is equal to BD, that is to CB; and the rectangle AC, CB, because (22.) CF

of divided lines.

PROP. 30. (Euc. ii. 1.)

If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two lines shall be equal to the sum of the rectangles contained by the undivided line, and the several parts of the divided line.

Let A B and C be two straight lines, of which AB is divided into the parts AD, DE, EB: the rectangle under C and A B shall be equal to the sum of the rectangles under C and A D, C and DE, C and F EB.

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Draw the straight line AF at right angles to A B, and equal to C (post. 5.): complete the rectangle AG, and through the points D and E draw the straight lines DH and EK parallel to AF (14, Cor.). Then, because D H and EK are each of them (22.) equal to AF, that is, to C, the rectangles AH, DK, EG are equal to the rectangles under C and AD, C and DE, C and E B. But these rectangles make up the whole rectangle AG, which is equal to the rectangle under C and A B. Therefore, the rectangle under C and A B is equal to the rectangles under C and AD, C and D E, C and E B.

Therefore, &c.

Cor. (Euc. ii. 2.) If a straight line be divided into any two parts, the rectangles contained by the whole line and each of the parts, shall be together equal to the square of the whole line.

PROP. 31. (EUc. ii. 3.)

If a straight line be divided into any two parts, the rectangle contained by the whole line and one of the parts, shall be equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

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Let the straight line A B be divided into any two parts AC, CB: the rectangle under AB, B C shall be equal to the rectangle under AC, CB, together with the square of B C.

Draw the straight line BD at right angles to A B (post. 5.), and equal to BC: complete the rectangle ABDE and through C draw C F parallel to B D.

figure CD is equal to the square of
CB; and the figure A D to the rectan-
gle AB, BC. But A D is equal to A F,
together with CD. Therefore, the rec-
tangle A B, B C, is equal to the rectangle
AC, CB, together with the square of
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Therefore, &c.

PROP. 32. (EUc. ii. 4.)

The square of the sum of two lines is greater than the sum of their squares, by twice their rectangle.

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Let the straight line AB be the sum of the two straight lines AC, CB: the square of A B shall be greater than the squares of AC, CB, by twice the rectangle AC, CB.

Because the straight line AB is divided into two parts in the point C, (30. Cor.) the square of A B is equal to the sum of the rectangles under A B, AC, and AB, B C. But the rectangle under AB, AC (31.) is equal to the rectangle under A C, CB together with the square of A C; and, in like manner, the rectangle under AB, BC is equal to the rectangle under A C, C B together with the square of CB. Therefore, the square of A B is equal to twice the rectangle AC, CB, together with the squares of A C, CB; or, which is the same thing, the square of A B is greater than the squares of AC, CB, by twice the rectangle A C, C B.

Therefore, &c.

The figure shews in what manner the square of AB may be divided into two squares equal to those of A C, C B, and two rectangles, each equal to the rectangle A C, C B.

PROP. 33. (Euc. ii. 7.)

The square of the difference of two lines is less than the sum of their squares by twice their rectangle.

Let the straight line,

A B, be the difference

of the two straight lines, A C, CB: the square of A B shall be less than the squares of A C, CB, by twice the rectangle A C, C B.

This proposition, being an obvious consequence of the preceding, might have been added to it as a second corollary: it is of so great importance, however, that it seemed preferable to force it upon the attention of the student, by placing it among the

Then, the figure AF is equal to the propositions,

Because AC is the sum of A B and BC (32.), the square of A C is equal to the squares of AB, BC, together with twice the rectangle, A B, BC: therefore the square of A B is less than the square of AC by the square of B C, together with twice the rectangle AB, BC. Therefore, the same square of A B is less than the squares of A C, B C, by twice the square of BC, together with twice the rectangle AB, BC. But twice the square of BC, together with twice the rectangle A B, BC, is (32.) equal to twice the rectangle A C, CB. fore, the square of AB is less than the squares of A C, B C, by twice the rectangle A C, C B.

Therefore, &c.

There

Cor. (Euc. ii. 8.) The square of the sum of two lines is greater than the square of their difference by four times their rectangle for the former square is greater than the sum of their squares by twice their rectangle (32.), and the latter square is less than the sum of their squares by twice their rectangle.

PROP. 34. (Euc. ii. 5.)

The difference of the squares of two lines is equal to the rectangle under their sum and difference.

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Let A B, A C, be any two straight lines, and let BA be produced to D, so that AD may be equal to AB, and therefore CD equal to the sum, and CB to the difference of A B, AC: the difference of the squares of AB, AC, shall be equal to the rectangle under CD, CB.

Draw the straight line CH perpendicular to CD (post. 5.), and equal to CB; complete the rectangle CHFD; and, through the points A and B, draw AG and B E parallel to CH.

Then, because BE is (22.) equal to CH, that is, to B C, the rectangle A E is equal to the rectangle AB, BC; and, because AD is equal to A B, the rectangle D G is equal (25.) to the rectangle AE; and CE is the square of CB: therefore, the whole rectangle CF is equal to the square of BC, together with twice the rectangle AB, BC. But, because the square of A C is equal (32.) to the squares of A B, B C, together with twice the rectangle A B, BC; if the square of A B be taken from each side, (ax. 3.) the difference of the squares of A C, A B, is equal to the square of B C,

together with twice the rectangle A B, BC. Therefore (ax. 1.) the difference of the squares of A C, A B, is equal to the rectangle CF, that is, to the rectangle under CD, C B. Therefore, &c.

PROP. 33. (EUc. ii. 9. and 10.)

difference of two lines, are together The squares of the sum, and of the double of the squares of the two lines.

For (32.) the square of the sum of two lines is greater than the sum of their squares, by twice their rectangle, and the square of their difference is (33.) as much less than the sum of their squares. Therefore (ax. 9.), the square of the sum, together with the square of the difference, is equal to twice the sum of their squares.

Therefore, &c.

Scholium.

The theorems of this section admit of being enunciated more briefly and perspicuously by the use of certain conventional signs,+, -,×,( ), &c. bor. rowed from Algebra.

That A is equal to B, is thus denoted: A = B, which is read "A is equal to B." The sum of A and B thus: A+B, which is read "A plus B."

The excess of A above B: A- B....... "A minus B."

Twice A, four times A, &c......... 2 A, 4 A, &c.

The rectangle under A and B....A× B, or A B........" A, B."

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The square of A....................A A, or A2......... A square; and,

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(A + B C) or A + B-C signifies that A+B-C (that is, the excess of the sum of A and B above C) is to be be taken as a single quantity.

The theorems of this section may, therefore, be more briefly expressed as follows:

Ax (B+C+D) = AB+

Prop. 30. AC+AD. Prop. 31. (A+B) × B=AB+B2. Prop. 32. (A+B)2=A2+Bo +2 A B. Prop. 33. (A-B)2 = Ao + B2 - 2 A B. Prop. 33. Cor. (A+B)-(A-B)2 = AB.

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Prop. 34. (A+B)×(A−B)= A2 — Bo. Prop. 35. (A+B)2+(A−B)2 = 2 (Ao +B®).

In this borrowing of its notation, may be seen the first glimmerings of the application of Algebra to Geometry. It

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