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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 ab be divided into any two parts in c; the square of ab is equal to the squares of ac, cb, and to twice the rectangle contained by ac, cb.

Upon ab describe (i. 46) the square adeb, and join bd, and through c draw (i. 31) cgf parallel to ad or be, and through g draw hk parallel to ab orde. And because cf is parallel to ad, and bd falls

upon them, the exterior angle bgc is equal

(i. 29) to the interior and opposite angle adb;
but adb is equal (i. 5) to the angle abd, be-
cause ba is equal to ad, being sides of a square;
wherefore the angle cgb is equal to the angle h
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 cg
is parallel to bk, and cb meets them; the angles d
kbc, gcb are therefore equal to two right

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angles; and kbc is a right angle; wherefore gcb is a right angle: and therefore also the angles (i. 34) cgk, gkb, opposite to these, are right angles, and cgkb 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 ac. 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 ac, cb, for gc is equal to cb; therefore ge is also equal to the rectangle ac, 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 hf, ck, ag, ge are equal to the squares of ac, cb, and to twice the rectangle ac, cb. But hf, ck, ag, ge make up the whole figure adeb, which is the square of ab. Therefore the square of a b is equal to the squares of ac, 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.

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 ab be divided into two equal parts in the point c, and into two unequal parts at the point d; the rectangle ad, db, together with the square of cd, 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 parallel to cbor ef; and also through a draw ak parallel to cl or bm. And because the complement ch is equal (i. 43) to the complement hf,

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to each of these add dm; therefore the whole cm is equal to the whole df; but cm is equal (i. 36) to al, because ac is equal to cb; therefore also al is equal to df. To each of these add ch, and the whole ah is equal to d f and ch: but ah is the rectangle contained by ad, db, for dh is equal (ii. 4. cor.) to db; and df together with ch is the gnomoncmg; therethe rectangle ad, db: to each of these

fore the gnomon cmg is equal to add1g, which is equal (ii. 4. cor.) to the square of cd; therefore the gnomon cmg, together with 1g, is equal to the rectangle ad, db, together with the square of cd; but the gnomon cmg and 1g make up the whole figure cefb, which is the square of cb: therefore the rectangle ad, db, together with the square of cd, is equal to the square of cb. 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.

PROPOSITION VI. - THEOREM.

If a straight line be bisected, and produced to any point, the rectangle contained 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 ab 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 de, and through b draw (i. 31) bhg parallel to ce or df, and through h draw klm parallel to ad or ef, and also through a draw ak parallel to clordm;

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and because ac is equal to cb, the rectangle al is equal (i. 36) to ch; but ch is equal (i. 43) to hf; therefore also al is equal to hf: to each of these add cm; therefore the whole am is equal to the gnomon omg: and am is the rectangle contained by ad, db, for dm is equal (ii. 4. cor.) to db: therefore the gnomon cmg is equal to the rectangle ad, db: add

to each of these 1g, which is equal to the square of cb; therefore the rect

angle ad, db, together with the square of cb, is equal to the gnomon cmg, and the figure 18; but the gnomon cmg and 1g 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.

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 ab be divided into any two parts in the point c; the squares of ab, bc are equal to twice the rectangle ab, bc, together with the square of a c.

Upon a b describe (i. 46) the square adeb, 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 ak is therefore equal to the whole

ce; therefore ak, ce, are double of ak: but

ak, ce are the gnomon akf, together with the
square ck; therefore the gnomon akf, together
with the square ck, is double of ak: but twice
the rectangle ab, bc is double of ak, for bkis h
equal (ii. 4. cor.) to bc: therefore the gnomon akf,

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together with the square ck, is equal to twice the
rectangle ab, bc: to each of these equals add
hf, which is equal to the square of ac; there-
fore the gnomon akf, together with the squares d

ck, hf, is equal to twice the rectangle ab, bc,

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and the square of ac: but the gnomon akf, together with the squares ck, hf, make up the whole figure adeb and ck, which are the squares of ab and bc: therefore the squares of ab and be are equal to twice the rectangle ab, bc, together with the square of a c. Wherefore, if a straight line, &c. Q. E. D.

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

Produce ab to d, so that bd be equal to cb, and upon ad describe the square aefd; 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, pr is equal to ro; and because cbis 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 cb is equal to bd, and that bd is equal (ii. 4. cor.) to bk, that

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is, to cg, and cb equal to gk, that is, to gp (ii. 4. cor.); therefore cg is equal to gp: and because cg is equal to gp, and pr toro, the rectangle ag is equal to mp, and pl to rf: but mp is equal (i. 43) to pl, because they are the complements of the parallelogram ml; wherefore ag 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 ag. And it was demonstrated that the four ck, bn, gr, and rn are quadruple of ck. Therefore the eight rectangles which contain

the gnomon ach are quadruple of ak; and because ak is the rectangle contained by ab, bc, for bk is equal to bc, four times the rectangle ab, be is quadruple of ak: but the gnomon aoh was demonstrated to be quadruple of ak; therefore four times the rectangle ab, bc, is equal to the gnomon aoh. To each of these add xh, which is equal (ii. 4. cor.) to the square of ac: therefore four times the rectangle ab, bc, together with the square of ac, is equal to the gnomon aoh and the square xh: but the gnomon aoh and xh make up the figure aefd, which is the square of ad: therefore four times the rectangle ab, bc, together with the square of ac, is equal to the square of a d, that is, of ab and be added together in one straight line. Wherefore, if a straight line, &c. Q. E. D.

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 ab 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 ac, cd.

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From the point c draw (i. 11) ce at right angles to ab, and make it equal to ac or cb, and join ea, eb; through d draw (i. 31) df parallel to ce, and through f draw fg parallel to ab; and join af: then, because ac is equal to ce, the angle eacis equal (i. 5) to the angle aec; and because the angle ce right angle, the two others aec, eactogether make one right angle (i. 32); and they are equal to one another;

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each of them therefore is half of a right angle. For the same reason, each of the angles ceb, ebcis half a right angle; and therefore the whole aeb is a right angle: and because the angle gef is half a right angle, andegfa right angle, for it is equal (i. 29) to the interior and opposite angle ecb, 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 bis half a right angle, and fdb 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 bis equal to the angle bfd; and the side df to (i. 6) the side db: and because ac is equal to ce, the square of ac is equal to the square of ce; therefore the squares of ac, ce, are double of the square of ac: but the square of ea is equal (i. 47) to the squares of ac, ce, because ace is a right angle; therefore the square of ea is double of the square of ac: 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 ac; therefore the squares of a e, ef are double of the squares of ac, cd: and the square of af is equal (i. 47) to the squares of ae, ef, because aef is a right angle; therefore the square of af is double of the squares of ac, cd: but the squares of ad, df, are equal to the square of af, because the angle adf is a right angle; therefore the squares of ad, df are double of the squares of ac, cd: and dfis equal to db; therefore the squares of ad, db are double of the squares of ac, cd. If therefore a straight line, &c. Q. E. D.

PROPOSITION X. - THEOREM.

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 ab be bisected in cand produced to the point d; the squares of ad, 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 ae, eb; through e draw (i. 31) ef parallel to ab, and through d draw df parallel to ce: and because the straight line ef meets the parallels ec, fd, the angles cef, efd are equal (i. 29)

to two right angles; and therefore
the angles bef, efd are less than
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
enough : therefore eb, fd shall
meet if produced towards b, d: let
them meet in g, and join ag: then,

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