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lelogram CL may be also shown to be equal to the square HC. Therefore the whole square BDEC, is equal to the two squares GB, HC, and BDEC is the square described upon the side BC, and GB, HC, are the squares described on BA, AC. Therefore the square described on the side BC, subtending the right angle, is equal to the squares described on the sides BA, AC, containing the right angle. Therefore in right angled triangles, &c.

Q. E. D.

Deductions.

1. Describe a square which shall be equal to two given squares.

2. Describe a square which shall be equal to the difference between two given squares.

3. The square described upon the diameter of another square is double of that square.

4. If the sides of the square described upon the hypothenuse of a right angled triangle be produced to meet the sides of the squares described upon the legs, they will cut off triangles equiangular, and equal to the given triangle.

The name of Pythagoras is rendered immortal in the annals of geometry by the discovery of this famous, useful, and elegant proposition. Some authors relate that he was so transported with joy, that he offered to the gods a sacrifice of a hundred oxen, as a token of gratitude for their inspiring him with it. This circumstance, however, is doubted by others, as being inconsistent with his religious opinions, which prohibited bloody sacrifices. Be this as it may, never had enthusiasm a better foundation. The problem deservedly ranks amongst the first class of geometrical truths, both from the singularity of its result, and the variety of cases to which it is applicable in every branch of the mathematics.

PROPOSITION XLVIII.

THEOREM.

If the square which is described upon one of the sides of a triangle be equal to the squares which are described on the other two sides, then shall the angle which is contained by these two remaining sides be a right angle.

If the square described on BC, one of the sides of the triangle ABC, be equal to the squares described upon the remaining sides of the triangle B

a

BA, AC, the angle BAC is a right angle. For draw from the point A, AD at right angles to AC, and make AD equal to BA, and join DC. Therefore because DA is equal to AB, the square also described on DA is equal to the square described on AB. Add the square of AC, which is common. Therefore the squares of DA, AC, are equal to the squares of BA, AC. But the square described on DC is equal to the squares described on DA, AC, for DAC is a right angle; also the square 47. 1. of BC is equal to the squares of BA, AC. Therefore the square of DC is equal to the square BC. fore the side DC is also equal to the side CB. because DA is equal to AB, and AC common, the two DA, AC, are equal to the two BA, AC; and the base DC is equal to the base CB. Therefore the angle DAC is equal to the angle BAC. But DAC is a right angle; b 8. 1. therefore BAC will also be right angle. If therefore the square, &c. Q. E. D.

a

Where

And

EUCLID'S ELEMENTS.

BOOK II.

DEFINITIONS.

1. Every right angled parallelogram is said to be contained under two right lines, comprehending a right angle.

A E

D

2. In every parallelogram either of those parallelograms about the diameter, together with the complements, is called a gnomon.* Thus the parallelogram HG, together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the posite angles of the parallelograms, which make the gnomon.

op

PROPOSITION I.

THEOREM.

F

H

K

B G

C

If there be two right lines, one of which is divided into any number of parts, the rectangle comprehended under the whole and divided line, is equal to the rectangles contained under the whole line, and the several segments of the divided line.

Let A and BC be two right lines, and let BC be any how divided in the points D, E ; the rectangle contained under the right lines A and BC, is equal to the rectangles contained under A and BD, A and DE, and A and EC.

B

F

D E C

a 11. 1.

K L H

b 3. 1.

A

From the point в drawa BF, at right G angles to BC, and make BG equal to a : and let GH be drawn through G, parallel to BC; and through D, E, C, draw DK, EL, and CH, parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH, but the rectangle BH is contained under A and BC, for it is contained under GB and BC,

* From the Greek word гpwv, which signifies a carpenter's square.

31. 1.

d 34. 1.

and GB is equal to A; and вк is contained under A and BD, for it is contained under BG and BD, and BG is equal to a ; and the rectangle DL is contained under A and ᎠᎬ ; because DK, that is d BG, is equal to a ; so likewise the rectangle EH is that contained under a and EC. Therefore the rectangle under A and BC is equal to the rectangles under A and BD, A and DE, and A and EC. Therefore if there be two right lines, &c. Q. E. D.

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Put a equal to the line A, and b the line BC, and suppose it be divided into the parts e, f, g; then shall * Ax. 8. 1. ab = ae + af + ag; for b* = e + ƒ+ g, multiply each by a, and we shall have ab = ae + af + ag. Q. E. D.t

PROPOSITION II.

THEOREM.

If a right line be divided into any two parts, the rectangles contained under the whole and each of the parts are together equal to the square of the whole line.

A

C B

Let the right line AB be any how divided in the point c, the rectangle contained under AB, BC, together with the rectangle AB, AC, is equal to the square of AB. For let the square ADEB, be describeda on AB, and through c let cr be drawn parallel to AD or BE. Then AE is © Ax. 8. 1. equal to the rectangles AF, CE; and AE is the square of AB; and AF is the rectangle

a 46. 1.

b 31. 1.

FE

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

=

=

y de; then because dr ad + bd drex, therefore xy will be =

For, let x = a + b + c, and + cd, and ex ae + be + ce, and xy ad + bd cd + ae + be + ce. Q. E. D. From hence we deduce a method of multiplying compound lines into compound lines. For if the rectangles of all the parts be taken, their sum shall be equal to the rectangle of the wholes.

--

arises + .

But whensoever in the multiplication of lines into themselves you meet with the sign + intermingled with particular regard must be paid to them. For of + multiplied into into arises; but of For example, 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.

This being sufficiently understood, the nine following propositions, and innumerable others of that kind, arising from the comparing of lines multiplied into themselves, which the reader may find done in Vieta and other analytical writers, are demonstrated with great facility, by reducing the matter for the most part to almost a simple work.

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