47. 1.

a 47. J. * See the

following theor.

b 8. 1.

6 conftr.

2 34. 1.

b 4. 1. &

6. ax.

a 46. 1.

b 1. part,

c hyp.

d 9. ax.

Now, let the fides AB, BC, be known, the one 6 foot, the other 10, Therefore since BCq- ABq = 100 - 36=64 ACq, whence AC = 64= 8.

Which was to be done.

PROP. XLVIII. Plate I. Fig. 47.

If the square made upon one fide BC of a triangle be equal to the squares made on the other fides of the triangle AB, AC, then the angle BAC comprehended under the two other fides of the triangle AB, AC, is a rightangle. Perpendicular to AC draw AD = AB, and join CD. Now is (a) CDq = ADq+ACq=ABq -+ ACq= BCq. * Therefore is CD = BC. And therefore the triangles CAB, CAD, are mutually equilateral. Wherefore the angle CAB (b) = CAD (c) = a right-angle. Which was to be demonftrated.

Schol.

We assumed in the demonstration of the last Proposition, CD = BC, because CDq was equal to BCq: Our assumption we prove by the following

THEOREM. Fig. 48.

The Squares AF, CG of equal right-lines AB, CD, ore equal one to the other: And the fides AB, CD, of equal Squares AF, CG, are equal one to the other.

1. Hypothesis. Draw the diameters EB, HD. Then it is evident that AF is (a) equal to the triangle EAB twice taken, and (6) equal to the triangle HCD twice taken, and equal to (a) CG. Which was to be demonftrated. 2. Hyp. If it may be, let CD be greater than AB. Make CT = AB, and let CS = CTq. Therefore is CS (b) = EB (c) = CG. Which is abfurd.

Corol.

After the fame manner any rectangles equilateral one to another, are demonstrated to be also equal.

The End of the first Book.

The

!

The SECOND BOOK

OF

EUCLIDE'S
ELEMENTS.

I.

E

Definitions.

Very right-angled Parallelogram ABCD, is
said to be contained under two right-lines
AB, AC, comprehending a right-angle.

Therefore when you meet with such as these, the rectangle under BA, AC, or more briefly the rectangle BAC, or BA X AC (or ZA, for ZXA) the rectangle meant is that which is contained under the right-lines BA, AC, fet at right-angles.

II. In every Pgr. ALGE, any one of those parallelograms which are about the diameter, together with its two complements is called a Gnomon. As the Pgr. FB + DL + FK (BEC) is a Gnomon; and likewise the Pgr. FB, DL - BC (FAK) is a Gnomon.

PROP. I. Fig. 49.

If two right-lines AF, AB, are given, and one of them AB be divided into as many parts or fegments as you please; the rectangle comprehended under the two whole right-lines AB, AF, shall be equal to all the rectangles contained under the whole line AF, and the several fegments, AD, DE, EB.

(a) Set AF perpendicular to AB. Thro' F (a) draw an infinite line FG perpendicular to AF. From the points D, E, B, erect perpendiculars DH, EI, BG. Then is AG a rectangle comprehended under AF, AB, and is (6) equal to the rectangles AH, DI, EG, that is (because DH, EI, AF, (c) are equal) to the rectangles under AF, AD, under AF, DE, under AF, EB. Which was to be demonstrated.

Schol

a 1. 2. b 2. ax.

* 1. 2,

*19, ax.

Schol.

If two right-lines given are both divided into how many parts foever, one whole multiplied into the other shall bring out the same product, as the parts of one multiplied into the parts of the other.

For let Z be = A + B -- C, and Y = D + E; then, because IZ (a) = DA + DB+DC, and EZ (a)= EA + EB+EC, and YZ (a)=DZ+EZ, (6) There fore ZY will be = DA + DB + DC + EA + EB + EC. Which was to be demonstrated.

From hence we have a method of multiplying compound

lines into compound ones. For if the rectangles of all the parts be taken, their sum shall be equal to the rectangle of the wholes.

into

But whensoever in the multiplication of lines into themselves you meet with these signs-intermingled with these +, you must also have particular regard to the figns, For of multiplied into - arifeth -; but ofarifeth + ex. gr. let + A be multiplied into B-C; then because + A is not affirmed of all B, but only of that part of it, whereby it exceeds C, therefore AC must remain denied; so that the product will be AB - AC. Or thus; because B confifts of the parts C and B - C, * thence AB=AC+ AXB-C, take away AC from both, then AB - AC=AXB-C. In like manner, if - A be to be multiplied into B - C, then since by virtue of the sign -, A is not denied of all B, but only of fo much as it exceeds C, therefore AC must remain affirmative, whence the product will be AB AC. Or thus; because AB = AC + A×B-C; take away all from both fides, and there will be -AC-AXB - C; add AC to both, and it will be - AB+ AC=-AX B-С.

AB=

This being fufficiently understood, the nine following propositions, and innumerable others of that kind, arifing from the comparing of lines multiplied into themselves (which you may find done to your hand in Vieta, and other analytical Writers) are demonftrated with great facility, by reducing the matter for the most part to almost a simple work.

Furthermore, * it appears that the product arifing from the multiplication of any magnitude into the parts of any number is equal to the product arifing from the multiplication of the fame into the whole number: As 5 A+ 7 A = 12 A, and 4 A x5 A+ 4A x 7A =

4 A

4 AX 12 A. Wherefore what is here delivered of the multiplying of right-lines into themselves, the same may be understood of the multiplying of numbers into themselves, so that whatsoever is affirmed concerning lines in the nine following Theorems, holds good also in numbers; seeing they all immediately depend on, and are deriv'd from this first.

PROP. II. Plate I. Fig. 50.

If a right-line Z be divided any wife into two parts, the rectangles comprehended under the whole line Z, and each of the segments A, E, are equal to the square made of the whole

line Z.

I say that ZA + ZE=Zq. For take B = Z; (a) then is BABE = BZ, that that is is (because (because B B = Z) ZA +ZE=Zq. Which was to be demonstrated.

PROP. III. Fig. 50.

-
-

If a right-line Z, be divided any wife into two parts, the rectangle comprehended under the whole line Z, and one of the segments E, is equal to the rectangle made of the Jegments A, E, and the square described on the faid fegment E.

1

I fay ZE =AE + Eq. (a) For EZ = EA + Eq.

PROP. IV. Fig. 50.

If a right-line Z be cut any wife into two parts, the Square described on the whole line Z, is equal to the squares described on the fegments A, E, and to twice a rectangle made of the Segments A, E, taken together.

I say that Zq = Aq+Eq--2AE. For ZA (a)=Aq + AE, and ZE (a) = Eq + EA. Therefore whereas ZA + ZE (6)=Zq, (c) thence is Zq = Aq + Eq + 2AE. Which was to be demonstrated.

demonftrated.

Otherwise thus; Upon the right-line AB make the square AD, and draw the diameter EB; thro' C, the point wherein the line AB is divided, draw the perpendicular CF; and thro' the point G draw HI parallel to AB.

Because the angle EHG = A is a right-angle, and AEB is (d) half a right (e) therefore is the remaining angle HGE half a right-angle. Therefore is HE (f) = HG (g) = fo EF (g) = AC, so that HF (b) is the square of the right-line AC. After the fame manner is CE proved

to

a 1. 2.

a 1.2.

a 3. 2. b. 2. 2.

cr.ax.

Fig. 51.

d 4. cor.
32. 1.
e 32. 1.
f 6.1.

g 34. I.
h 29. def.

to be CBq. Therefore AG, GD, are rectangles under k 19. ax. 1. AC, CB, wherefore the whole square AI) (k) = ACq +CBq + 2ACB: Which was to be demonstrated.

a 4.1. b 3. 2.

c hyp.

d 1. 2.

Sch. 1. 2.

a 5. 2. & 3. αx.

Coroll.

1. Hence it appears that the Pgrs. which are about

the diameter of a fquare are also squares themselves.
2. That the diameter of any square bisects it angles.
3. That if a = Z, then is Zq = 4 Aq, and Aq
Zq. As on the contrary, if Zq4 Aq, then is A
Z.

A1

C

D

PROP. V.

_ 11 B If a right-line AB be cut into equal parts AC, CB, and into unequal parts AD, DB, the rectangle comprehending under the unequal parts AD, DB, together with the Square that is made of the difference of the parts CD, is equal to the Square described on the half line CB.

I fay that CBq= ADB + CDq.

For these are

all equal.

CBq.

(a) CDq+CDB CDq+CDB+DBq+ + DBq + CDB. CDq+(b) CBD (c) (ACxBD)+CDB. (CDq+ (d) ADB.

This theorem is somewhat differently exprefs'd and more easily demonftrated thus; A rectangle made of the fum and the difference of two right-lines A, E, is equal to the difference of the squares of those lines.

For if AE be multiplied into A-E, * there arifeth Aq-AE-EA-Eq=Aq-Eq. Which was to be de

If the line AB be divided otherwise, (viz.) nearer to the point of bisection, in E; then is AEB ADB.

For AEB (a) = CBq-CEq, and ADB (a) = CBq CDq. Therefore, whereas CDqCEq, thence is AEB - ADB. Which was to be demonstrated.

Coroll.

1. Hence is ADq+DBq C AEg+EBq. For ADq

b 4. 2.

+DBq+2 ADB (6) = ABq (6)

AEq+EBq+2AEB.

Therefore because 2AEB-2ADB, shall ADq-f-DBq.
AEq+EBq. Which was to be demonstrated.

2. Hence is ADqfDBq-AEq (c) -EBq=2AEB