(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 EFGH. Therefore, &c.

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

PROP. 26. (Euc. i. 41.)

E D

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 A B C D shall be double of the triangle E B C.

W

B

Complete the parallelogram E BCF. (14. Cor. 1.) Then the parallelogram EBCF is double of the triangle EBC, because it is bisected by the diagonal EC (22. Cor.); and ABCD 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 EB 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.

E

B

E

D

The first part of the proposition is manifest; for the triangles are the halves of parallelograms (26.) upon the B 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, DB 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 A B C, E B C, are upon the same base, or upon equal bases, and between the same parallels, the triangle E B C is equal to A B C, that is, to D BC; 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 plies equally to parallelograms, as to (or converse) part of the proposition aptriangles.

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.

angle ABE. But the base of the triangle ABE is equal to the sum of A D, 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 A BDC 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

E

M

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.

For

From the theorems of this Section rules are easily deduced for the mensuration of rectilineal figures. 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.

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.

F

D E B

H

Let A B and C be two straight lines, of which A B 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 EB. 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.

figure CD is equal to the square of CB; and the figure AD to the rectangle AB, BC. But A D is equal to A F, together with CD. Therefore, the rectangle AB, B C, is equal to the rectangle A C, C B, together with the square of BC.*

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.

A

CR

Let the straight line A B 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 AB, AC, and AB, B C. But the rectangle under AB, AC (31.) is equal to the rectangle under A C, C B 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, C B, 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 A C, C B, by twice the rectangle A C, C B.

Therefore, &c.

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 propositions.

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

Therefore, &c.

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.

D

Let A B, A C, be any two straight lines, and let BA be produced to D, F

so that AD may be equal to A B, 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 C H.

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 A E; 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 AB, 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 C D, C B. Therefore, &c.

PROP. 35. (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.

lines is greater than the sum of their For (32.) the square of the sum of two 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.

(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:

Prop. 30. Ax (B+C+D) = AB+ AC+AD.

4

Prop. 31. (A+B) × B=AB+B2. Prop. 32. (A+B)2=A2+B2+2 A B. Prop. 33. (A-B)2=A2+B2-2 A B. Prop. 33. Cor.· (A+B)2—(A−B)2= A B.

Prop. 34. (A+B)×(A−B)=A2 — B2. Prop. 35. (A+B)2+(A−B)2= 2 (A2 +B2).

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

requires but a very slight acquaintance with the former science, to perceive at once, that, when the lines contain each of them some common part M a certain number of times exactly, these theorems are but so many examples of its rules of addition, subtraction, and multiplication. The perspicuity alone, however, which is displayed in the above expressions, will enable the uninitiated reader to form some notion of the advantages resulting from a more intimate union of the two sciences.

is equal to the rectangle BK (ax. 1.). And, in like manner, it may be shewn that the square of BC is equal to the rectangle K C. Therefore, the squares of A B, A C, together, are equal to the rectangles B K, KC together, that is, to the square of B C.

Next, let the square of B C be equal to the sum of the squares of BA, AC: the angle B A C shall be a right angle. For, let L M N be another triangle having its sides, LM, LN, equal to the sides AB, AC, respectively, and the angle at La right angle. Then, by the former part

Triangles.

PROP. 36. (EUc. i. 47. and 48.) In every right-angled triangle, the square of the hypotenuse, or side opposite to the right angle, is equal to the sum of the squares of the sides which contain that angle: and conversely, if the square of one side of a triangle be equal to the sum of the squares of the other two sides, the angle contained by those two sides shall be a right angle. Let A B C be a right-angled triangle, having the right angle, BAC: the square of BC shall be equal to the sum of the squares of BA, AC. Upon B C, B A, describe the squares CD, AE; produce

E

T

G

B

D K

H

A

M

DB to meet E F, or E F produced, in G; and, through A, draw HK parallel to

BD.

Then, because the sides E B, B G, of the angle E B G, are perpendicular respectively to the sides A B, BC, of the angle A B C, these two angles (18.) are equal to one another; and the right angle B E G is equal to BAC; there fore the two triangles EBG, ABC, having two angles of the one equal to two angles of the other, each to each, and the interjacent sides, E B, A B, equal to one another, are equal in every respect, and B G is equal to B C or BD (5.). And, because the parallelogram AG, and the square A E, are upon the same base, and between the same parallels (24.), A G is equal to A E. Again, because the parallelogram A G, and the rectangle KB, are upon equal bases, BG, BD, and between the same parallels (25.), A G is equal to K B. Therefore, also, A E, that is, the square of AB,

equal to the squares of L M, L N, that is, to the squares of A B, A C, or to the square of B C, and M N is equal to B C (25. Cor.). Therefore, in the triangles ABC, LMN, the three sides of the other, each to each; and, consequently, one are equal to the three sides of the the angle B A C is (7.) equal to the angle MLN, that is, to a right angle.* Therefore, &c.

square of either of the two sides is equal Cor. 1. In a right-angled triangle, the to the difference of the squares of the hypotenuse and the other side.

Cor. 2. It appears, from the demonstration, that if a perpendicular be drawn from the right angle to the hypotenuse, the square of either side is equal to the rectangle under the hypotenuse and segment adjacent to that side. And conversely, if this be the case, the angle at A must be a right angle. For, if the rectangle BC, BP, be equal to the square of BA; then, taking the square of B P from each (31. and 36.) the rect

angle BP, PC, will be equal to the square of AP; and therefore, adding

the

square of P C to each, the rectangle AC: therefore the two rectangles, BC, B C, C P, will be equal to the square of BP, and B C, C P, together, that is (30. Cor.) the square of BC, will be equal to the squares of B A, A C, and the angle BAC will be a right angle (36.),

BAC be a right angle, the square of Cor. 3. And hence it follows that, if the perpendicular on the hypotenuse

*It is evident that the sides of the squares upon

AB, AC, which are opposite to A B, AC, would meet, if produced, in the point H. The latter part of the demonstration is, accordingly, equally applicable to shew that if upon the two sides A B, AC, of any triangle ABC (right-angled or otherwise) any two parallelograms be described whose sides opposite to A B, AC, meet in a point H, and if, upon the base BC of the triangle, a parallelogram be likewise described, having its sides adjoining to the base equal and parallel to AH, the parallelogram upon the base shall be equal to the sum of the parallelograms upon the two sides.