Euclid's Elements of geometry, the first three books (the fourth, fifth, and sixth books) tr. from the Lat. To which is added, A compendium of algebra (A compendium of trigonometry).

SECOND BOOK.

DEFINITIONS.

1. Every right angled parallelogram, is said to be contained by the two right lines which make the right angle.

2. In any parallelogram either of these c parallelograms, (EK or OF,) which are about the diagonal, together with the two complements, (AG and GD,) is called the gnomon.

If there be two right lines, (A and BC,) the latter of which (BC) is divided into any number of parts, (BD, De, EC,) the rectangle under these lines (A and BC,) shall be equal to the rectangles under the undivided line (A,) and the parts of the divided line, (BD, DE, and EC).

From B erect BH, perpendicular to H BC, and in it take BF equal to A, and through F draw FL, parallel to BC, and draw DG, EK, and CL, parallel to BF.

D E C
A

It is evident the rectangle BL, is equal to the rectangles BG, DK, and EL, but the rectangle BL, is the rectangle under A and BC; for BF is equal to A; but the rectangles BG, DK, and EL, are the rectangles under A and BD, A and DE, and under A and EC, for each of the lines BF, DG and EK, is equal to A. (by Prop. 34, B. 1)

COR.-Hence, and from Prop. 34, B. 1, it is evident the area of a rectangle is found, by multiplying the altitude into the base; and from Prop. 35 and 36, B. 1, the area of any parallelogram is found by multiplying the altitude into the base: and from Prop. 37 and 38, B. 1, the area of a triangle is found by multiplying the altitude into half the base.

PROPOSITION II. THEOREM.

If a right line, (AB,) be divided in any way whatsoever, (in C,) the square of the whole line, (AB,) shall be equal to the rectangles under the whole line, (AB,) and each of the parts, (AC, CB.)

For, on AB describe the square ADEF, (by Prop. 46, B. 1,) and from C, draw CE parallel to AD.

The square AF is equal to the rectangles AE and CF. But the rectangle AE is the rectangle under AB and AC, because AD is equal to AB, (by Const.):

D E F

and the rectangle CF, is the rectangle under AB and CB, because CE is equal to AB, (by Prop. 34, B. 1.)

OTHERWISE.

Assume a right line X equal to AB. The rectangle under X and AB, or the square of AB, (by Hypoth.) is equal to the sum of the rectangles under X and AC, and under X and CB, (by Prop. 1, B. 2,) that is to the sum of the rectangles under AB and AC, and under AB and CB.

(If a right line, (AB,) be divided in any way, (in C,) the rectangle under the whole line, (AB,) and either part (AC,) will be equal to the rectangle under the parts, (AC and CB,) together with the square of the part, (AC).

For on AC, describe the square ADFC, and from B, draw BE parallel to AD, until it meet DF produced to E. The rectangle AE is equal to the square ADFC, together with the rectangle CE. But the A rectangle AE is the rectangle under AC A and AB, for AD is equal to AC, (by Const. and Def.31, B. 1,) and the square ADFC,

is the square of AC, (by Const.,) and the rectangle CE is the rectangle under AC and CB, for CF is equal to AC, (by Const., and Defin. 31, B. 1.)

OTHERWISE.

Assume a right line X, equal to AC. The rectangles under X and AF, is equal to sum of the rectangles under X and AC, and under X and CF, (by Prop. 1, B. 2.) But the rectangle under X and AF, is the rectangle under AC and AF; and the rectangle under X and AC, is the square of AC: and the rectangle under X and CF, is the rectangle under AC and CF.

PROPOSITION IV. THEOREM.

If a right line (AB) be divided into any two parts, (in O) the square of the whole line is equal to the squares of the parts, together with twice the rectangle under the parts.

K D

On AB describe the square ACDB, and draw CB, and from O draw OK parallel to AC, and cutting CB in G, and through G draw EF parallel to AB. The square ACDB, is equal to the squares EK and OF, together with the rectangles AG and GD. But OF is the square of OB (by Const., and Cor. of Prop. 43, B. 1,) and because EG is equal to AO, (by Const., and Prop. 34, B. 1,) EK is the square of AO; and because OG and OB are equal, (by Defin. 31, B. 1,) AG is the rectangle under the parts AO and OB, but GD is equal to AG, (by

Prop. 43, B. 1,) therefore AG and GD together are equal to twice the rectangle under the parts.

OTHERWISE.

The square of AB is equal to the sum of the rectangles under AB and AO, and under AB and BO, (by Prop. 2, B. 2,) but the rectangle under AB and AO, is equal to the sum of the rectangle under AO and OB, and the square of AO, (by Prop. 3, B. 2); and the rectangle under AB and BO, is equal to the sum of the rectangle under AO and OB, and the square of OB (by last ref.); therefore, the sum of the rectangles under AB and OA, and under AB and BO, (or the square of AB,) is equal to the sum of the squares of AO and OB, and twice the rectangle under AO and ов.

COR.-Hence it appears that the square of half the line, is the fourth part of the square of the whole; for, a right line being bisected, the rectangle under the parts is equal square of the half.

If a right line (AB) be divided into equal parts (in C), and into unequal parts (in D), the rectangle under the unequal parts (AD and DB) together with the square of the intermediate part (CD) will be equal to the square of half the line (CB.)

Describe on CB the square C KMB, and draw KB, and from D draw DL, parallel to CK, and cutting KB in G, and through G draw HG E parallel to AB, meeting AE drawn from A parallel to CK.

K L M

ढे

Because AC and CB are equal (by Hypoth.), the rectangles AF and CH are equal (by Prop. 36. B. 1), but the rectangles CG and GM are equal (by Prop. 43. B. 1), therefore the rectangle AG, is equal to the Gnomon CHL (by Ax. 2); to both let

FL be added, and the rectangle AG with the square FL, will be equal to the square CKMB. But AG is the rectangle under AD and DB, for DG is equal to DB (by Cor. Prop. 43. B. 1, and Def. 31. B. 1); and FL is the square of CD, because FG is equal to CD (by Prop. 34. B. 1), and C KMB is the square of CB.

OTHERWISE,

The rectangle under AD and DB is equal to the sum of the rectangles under AC and DB, and under CD and DB, (by Prop. 1. B. 2); but the rectangle under AC and DB is equal to the rectangle under CB and DB (because AC and CB are equal), or to the rectangle under CD and DB, with the square of DB (by Prop. 3. B. 2), therefore the rectangle under AD and DB is equal to twice the rectangle under CD and DB together with the square of DB; add to both the square of CD, and the rectangle under AD and DB together with the square of CD, is equal to twice the rectangle under CD and DB, together with the squares of CD and DB, which is the square of CB (by Prop. 4. B. 2).

COR. 1.-Hence it appears if a line be bisected, the rectangle under the parts is greater than if it be divided unequally, and therefore the sum of the squares of the parts is less (by Prop. 4. B. 2.)

COR. 2.-If two equal right lines be so divided, that the rectangle under the segments of one, be equal to the rectangle under the segments of the other, the segments themselves shall be equal.

A GEB

C AFD

+ +

If right lines be bisected, the thing is evident. But if they be not bisected, let there be AB and CD, and divide them in E and F. Let them be bisected in G and H, and because the right lines themselves are equal, (by Hypoth.), the halves of them will be equal, and therefore the squares of their halves (by Cor. 1. Prop. 46. B. 1), but the rectangles under AE and EB, and under

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Euclid's Elements of geometry, the first three books (the fourth, fifth, and ...