Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

PROPOSITION IV.-THEOREM. If a straight line be divided into any two parts, the square of the whole line

is equal to the squares of the two parts, together with twice the rectangle

contained by the parts. LET the straight line a b be divided into any two parts in c; the square of a b is equal to the squares of a c, cb, and to twice the rectangle contained by a c, cb.

Upon a b describe (i. 46) the square a deb, and join bd, and through c draw (i. 31) cgf parallel to ad or be, and through g draw hk parallel to a b or de. And because of is parallel to a d, and bd falls upon them, the exterior angle bgc is equal (i. 29) to the interior and opposite angle a db; a

с b but adb is equal (i. 5) to the angle a bd, because ba is equal to a d, being sides of a square;

8 wherefore the angle cgb is equal to the angle h

k gbc; and therefore the side bc is equal (i. 6) to the side

cg: But cb is equal (i. 34) also to gk, and cg, to bk; wherefore the figure cgkb is equilateral. It is likewise rectangular; for is parallel to bk, and cb meets thern; the angles d

f k bc, gcb are therefore equal to two right angles ; and k b c is a right angle; wherefore gcb is a right angle: and therefore also the angles (i. 34) cgk, gk b, opposite to these, are right angles, and c8 kb is rectangular; but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side cb. For the same reason hf also is a square, and it is upon the side hg, which is equal to a c. Therefore hf, ck are the squares of ac, cb; and because the complement ag is equal (i. 43) to the complement ge, and that a g is the rectangle contained by a c, cb, for gc is equal to cb; therefore ge is also equal to the rectangle a c, cb; wherefore ag, ge are equal to twice the rectangle ac, cb. And hf, ck are the squares of ac, cb; wherefore the four figures h f, ck, ag, ge are equal to the squares of ac, cb, and to twice the rectangle a c, cb. But hf, ck, ag, ge make up the whole figure adeb, which is the square of a b. Therefore the square of a b is equal to the squares of a c, cb, and twice the rectangle ac, cb. Wherefore, if a straight line, &c. Q. E. D.

CoR. From the demonstration, it is manifest that the parallelograms about the diameter of a square are likewise squares.

cg

PROPOSITION V.-THEOREM. If a straight line be divided into two equal parts, and also into two unequal

parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the

line. LET the straight line a b be divided into two equal parts in the point c,

a

m

and into two unequal parts at the point d ; the rectangle ad, db, together with the square of c d, is equal to the square of cb.

Upon cb describe (i. 46) the square cefb, join be, and through d draw (i. 31) dhg parallel to ce or bf; and through h draw klm paralleì to c b or ef; and also through a draw a k parallel to cl or bm. And because the complement ch is equal (i. 43) to the complement hf,

to each of these add dm; therefore db

the whole cm is equal to the whole

df; but c m is equal (i. 36) to a l, 1 h! because a c is equal to cb; therefore k

also al is equal to df. To each of these add ch, and the whole a h is equal to d fand ch: but a h is the rect

angle contained by a d, db, for dh is 8 f

equal (ii

. 4.cor.) to db; and df together

with ch is the gnomon cmg; therefore the gnomon cm & is equal to the rectangle ad, db: to each of these

which is equal (ii. 4. cor.) to the square of cd ; therefore the gnomon cmg, together with lg, is equal to the rectangle ad, db, together with the square of cd ; but the gnomon cmg and lg make up the whole figure cefb, which is the square of cb: therefore the rectangle a d, db, together with the square of cd, is equal to the square of c b. Wherefore, if a straight line, &c. Q. E. D.

Cor. From this proposition it is manifest, that the difference of the squares of two unequal lines ac, cd, is equal to the rectangle contained by their sum and difference.

addlg,

PROPOSITION VI.-THEOREM. If a straight line be bisected, and produced to any point, the rectangle con

tained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the

straight line which is made up of the half and the part produced. LET the straight line a b be bisected in c, and produced to the point d; the rectangle ad, db, together with the square of cb, is equal to the square of cd.

Upon cd describe (i. 46) the square cefd, join d e, and through b draw (i. 31) bhg parallel to ce or d f, and through h draw klm parallel to a dor e f, and also through a draw a k parallel to cl or dm; a

b. d and because a c is equal to cb, the

rectangle al is equal (i. 36) to ch;

but ch is equal (i. 43) to hf; there1 k

h

fore also al is equal to hf: to each of these add cm; therefore the whole am is equal to the gnomon cmg: and a m is the rectangle contained by ad, db, for d m is equal (ii. 4. cor.)

to db: therefore the gnomon cmg

8 f is equal to the rectangle a d, db: ada to each of these lg, which is equal to the square of cb; therefore the rect

m

e

and lg

angle a d, db, together with the square of cb, is equal to the gnomon cmg, and the figure 18 ; but the gnomon cmg.

make

up

the whole figure cefd, which is the square of cd; therefore the rectangle ad, db, together with the square of cb, is equal to the square of cd Wherefore, if a straight line, &c. Q. E. D.

a

PROPOSITION VII.—THEOREM. If a straight line be divided into any two parts, the squares of the whole

line, and of one of the parts, are equal to twice the rectangle contained by

the whole and that part, together with the square of the other part. LET the straight line a b be divided into any two parts in the point c; the squares of a b b c are equal to twice the rectangle a b, bc, together with the square of a c.

Upon a b describe (i. 46) the square a deb, and construct the figure as in the preceding propositions ; and because ag is equal (i. 43) to ge, add to each of them ck; the whole a k is therefore equal to the whole ce; therefore ak, ce, are double of a k: but a k, ce are the gnomon a k f, together with the

b square ck; therefore the gnomon a kf, together with the square ck, is double of a k: but twice the rectangle a b, bc is double of a k, for bk is

k equal (ii. 4. cor.) to bc: therefore the gnomon akf, together with the square ck, is equal to twice the rectangle a b, bc: to each of these equals add hf, which is equal to the square of a c; therefore the gnomon a kf, together with the squares d

f ck, h f, is equal to twice the rectangle a b, bc, and the square of a c: but the gnomon a kf, together with the squares ck, hf, make up the whole figure a deb and ck, which are the squares of a b and bc: therefore the squares of a b and b c are equal to twice the rectangle a b, bc, together with the square of a c. Wherefore, if a straight line, &c. Q. E. D.

60

PROPOSITION VIII.-THEOREM.

If a straight line be divided into any two parts, four times the rectangle

contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made

up of the whole and that part. LET the straight line a b be divided into any two parts in the point c; four times the rectangle a b, bc, together with the square of a c, is equal to the square of the straight line made up of a b and b c together.

Produce a b to d, so that bd be equal to cb, and upon a d describe the square a efd; and construct two figures such as in the preceding; Because cb is equal to bd, and that cb is equal (i. 34) to gk, and bd to kn; therefore gk is equal to kn: for the same reason, p r is equal,

m

r

to ro; and because c b is equal to bd, and gk to kn, the rectangle ck is equal (i. 36) to bn, and gr to rn; but ck is equal (i. 43) to rn, because they are the complements of the parallelogram co ; therefore also bn is equal to gr; and the four rectangles bn, ck, gr, rn are therefore equal to one another, and so are quadruple of one of them ck. Again, because c b is equal to bd, and that bd is equal (ii. 4. cor.) to bk, that

is, to cg, and cb equal to gk, that is, to gp a с ъ d

(ii. 4. cor.); therefore cg is equal to gp: and

because cg is equal to gp, and pr to ro, the glk

na rectangle ag is equal to mp, and pl to rf:

but mp is equal (i. 43) to p1 because they are P

o the complements of the parallelogram ml;

wherefore a g is equal also to rf: therefore the four rectangles ag, mp, pl, rf, are equal to one another, and so are quadruple of one of them

And it was demonstrated that the h 1 f four ck bn, gr, and rn are quadruple of ck.

Therefore the eight rectangles which contain the gnomon a o h are quadruple of a k; and because a k is the rectangle contained by a b b c, for bk is equal to bc, four times the rectangle a b, bc is quadruple of a k: but the gnomon a oh was demonstrated to be quadruple of a k; therefore four times the rectangle ab, bc, is equal to the gnomon a oh. To each of these add x h, which is equal (ii. 4. cor.) to the square of a c: therefore four times the rectangle a b, b c, together with the square of a c, is equal to the gnomon a oh and the square x h: but the gnomon a oh and xh make up the figure a efd, which is the square of ad: therefore four times the rectangle a b, bc, together with the square of a c, is equal to the square of a d, that is, of a b and be added together in one straight line. Wherefore, if a straight line, &c. Q. E. D.

a 8;

e

PROPOSITION IX.-THEOREM.

If a straight line be divided into two equal, and also into two unequal

parts, the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of

section. LET the straight line a b be divided at the point c into two equal, and at d into two unequal parts: the squares of ad, db are together double of the squares of a c, cd.

From the point c draw (i. 11) ce at right angles to a b, and make it equal to ac or

cb, and join e a, e b; through d draw (i. 31) f

df parallel to ce, and through f draw fg parallel to a b; and join a f: then, because a c

is equal to ce, the angle e ac is equal (i. 5) с d

to the angle a ec; and because the angle a ce b

is a right angle, the two others a c, ea c together make one right angle (i. 32); and they are equal to one another;

e

each of them therefore is half of a right angle. For the same reason, each of the angles ceb, e bcis half a right angle; and therefore the whole a e b is a right angle: and because the angle gef is half a right angle, and egfa right angle, for it is equal (i. 29) to the interior and opposite angle e cb, the remaining angle efg is half a right angle; therefore the angle gef is equal to the angle efg, and the side eg equal (i. 6) to the side gf: again, because the angle at b is half a right angle, and fd b a right angle, for it is equal (i. 29) to the interior and opposite angle ecb, the remaining angle bfd is half a right angle; therefore the angle at b is equal to the angle bfd, and the side df to (i. 6) the side d b: and because a c is equal to ce, the square of a c is equal to the square of ce; therefore the squares of a c, ce, are double of the square of a c: but the square of e a is equal (i. 47) to the squares of a c, ce, because a ce is a right angle; therefore the square of e a is double of the square of a c: again, because eg is equal to gf, the square of eg is equal to the square of gf; therefore the squares of eg, gf are double of the square of gf; but the square of ef is equal to the squares of eg, gf; therefore the square of ef is double of the square gf; and gf is equal (i. 34) to cd; therefore the square of ef is double of the square of cd: but the square of a e is likewise double of the square of a c; therefore the

squares of a e, ef are double of the squares of ac, cd: and the square of a f is equal (i. 47) to the squares of a e, e f, because a ef is a right angle; therefore the square of a f is double of the squares of a c, cd: but the squares of a d, df, are equal to the square of a f, because the angle a df is a right angle; therefore the squares of a d, df are double of the squares of ac, cd: and df is equal to db; therefore the squares of a d, db are double of the squares of a c, cd. If therefore a straight line, &c. Q. E. D.

PROPOSITION X.—THEOKEM. If a straight line be bisected, and produced to any point, the square of the

whole line thus produced, and the square of the part of it produced, are together double of the square of half the line bisected, and of the square

of the line made up of the half and the part produced. LET the straight line a b be bisected in c and produced to the point d; the squares of a d, db are double of the squares of ac, cd.

From the point c draw (i. 11) ce at right angles to ab: and make it equal to ac or cb, and join a é, eb; through e draw (i. 31) ef parallel to a b, and through d draw df parallel to ce: and because the straight line ef meets the parallels e c, fd, the angles cef, efd are equal (i. 29) to two right angles; and therefore the angles bef, efd are less than

f two right angles; but straight lines which with another straight line make the interior angles upon the same side less than two right angles, do meet (12 ax.) if produced far a

b

d enough : therefore eb, fd shall meet if produced towards b, d: let them meet in g, and join a g: then,

8

« ΠροηγούμενηΣυνέχεια »