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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 cb 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 (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 m1; 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 a g And it was demonstrated that the four ck, bn, gr, and rn are quadruple of ck. Therefore the eight rectangles which contain the gnomon a oh are quadruple of a k; and because a k is the rectangle contained by ab, bc, for bk is equal to bc, four times the rectangle a b, bc is quadruple of ak: but the gnomon a oh was demonstrated to be quadruple of ak; therefore four times the rectangle ab, bc, is equal to the gnomon a oh. To each of these add xh, which is equal (ii. 4. cor.) to the square of a c: therefore four times the rectangle a b, bc, together with the square of a c, is equal to the gnomon a oh and the square xh but the gnomon a oh and xh make up the figure a efd, which is the square of a d: therefore four times the rectangle ab, 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.

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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 a c, 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 e ac is equal (i. 5) to the angle a ec; and because the angle a ce is a right angle, the two others a c, e ac together 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, ebc is half a right angle; and therefore the whole a eb is a right angle: and because the angle gef is half a right angle, and egf a 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 fdb a right angle, for it is equal (i. 29) to the interior and opposite angle e cb, 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 db: and because ac 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 ace is a right angle; therefore the square of ea 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 ae 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 af is equal (i. 47) to the squares of a e, ef, because a ef is a right angle; therefore the square of af is double of the squares of a c, cd: but the squares of a d, df, are equal to the square of af, because the angle a df is a right angle; therefore the squares of a d, df are double of the squares of a c, 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.-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 c and produced to the point d; the squares of a d, db are 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 a c or cb, and join a e, 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 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 a g: then,

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because ac is equal to ce, the angle cea is equal (i. 5) to the angle eac; and the angle ace is a right angle; therefore each of the angles cea, eac is half a right angle (i. 32). For the same reason, each of the angles ceb, ebc is half a right angle; therefore aeb is a right angle and because ebc is half a right angle, dbg is also (i. 15) half a right angle, for they are vertically opposite; but bdg is a right angle, because it is equal (i. 29) to the alternate angle dce; therefore the remaining angle dgb is half a right angle, and is therefore equal to the angle dbg; wherefore also the side bd is equal (i. 6) to the side dg. Again, because egf is half a right angle, and that the angle at f is a right angle, because it is equal (i. 34) to the opposite angle ecd, the remaining angle feg is half a right angle, and equal to the angle egf; wherefore also the side gf is equal (i. 6) to the side fe. And because e c is equal to ca, the square of ec is equal to the square of ca; therefore the squares of ec, ca are double of the square of ca But the square of ea is equal (i. 47) to the squares of ec, ca; therefore the square of ea is double of the square of a c. Again, because gf is equal to fe, the square of gf is equal to the square of fe; and therefore the squares of gf, fe are double of the square of ef: but the square of eg is equal (i. 47) to the squares of gf, fe; therefore the square of eg is double of the square of ef: and ef is equal to cd; wherefore the square of eg is double of the square of cd. But it was demonstrated, that the square of ea is double of the square of a c: therefore the squares of a e, eg, are double of the squares of ac, cd: and the square of ag is equal (i. 47) to the squares of ae, eg; therefore the square of ag is double of the squares of ac, cd: but the squares of a d, dg are equal (i. 47) to the square of ag; therefore the squares of a d, dg are double of the squares of ac, cd: but dg is equal to db; therefore the squares of ad, db are double of the squares of a c, cd. Wherefore, if a straight line, &c. Q. E. D.

PROPOSITION XI.—PROBLEM.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

LET ab be the given straight line; it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts shall be equal to the square of the other part.

Upon ab describe (i. 46) the square abdc; bisect (i. 10) ac in e, and join be; produce ca to f, and make (i. 3) ef equal to e b, and upon af describe (i. 46) the square fgha; ab is divided in h, so that the rectangle a b, bh, is equal to the square of a h

Produce gh to k; because the straight line ac is bisected in e, and produced to the point f, the rectangle cf, fa, together with the square of a e, is equal (ii. 6) to the square of ef: but ef is equal to eb; therefore the rectangle cf, fa, together with the square of a e, is equal to the square of eb: and the squares of ba, ae are equal (i. 47) to the square

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of eb, because the angle eab is a right angle; therefore the rectangle cf, fa, together with the square of a e, is equal to the squares of ba, ae: take away the square of a e, which is common to both, therefore the remaining rectangle cf, fa, is equal to the square of ab; and the figure fk is the rectangle contained by cf fa, for af is equal to fg; and ad is the of square ab; therefore fk is equal to ad: take away the common part ak, and the remainder fh is equal to the remainder hd: and hd is the rectangle contained by ab, bh, for ab is equal to bd; e and fh is the square of a h. Therefore the rectangle a b, bh is equal to the square of a h: wherefore the straight line a b is divided in h, so that the rectangle ab, bh, is equal to the square of a h Which was to be done.

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PROPOSITION XII.—THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from any of the acute angles to the opposite side produced, the square of the side subtending the obtuse angle is greater than the squares of the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse angle.

LET abc be an obtuse-angled triangle, having the obtuse angle acb, and from the point a let ad be drawn (i. 12) perpendicular to be produced. The square of ab is greater than the squares of ac, cb, by twice the rectangle bc, cd.

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Because the straight line bd is divided into two parts in the point c, square of bd is equal (ii. 4) to the squares

of bc, cd, and twice the rectangle bc, cd. To each of these equals add the square of da; and the squares of bd, da are equal to the squares of bc, cd, da, and twice the rectangle bc, cd. But the square of ba is equal (i. 47) to the squares of bd, da, because the angle at d is a right angle; and the square of ca is equal (i. 47) to the squares of cd, da. Therefore the square of ba is equal to the of bc, ca, and twice the rectangle bc, cd; that is, the square of ba is greater than the squares of bc, ca, by twice the rectangle bc, cd. Therefore, in obtuse-angled triangles, &c. Q. E. D.

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PROPOSITION XIII.-THEOREM.

In every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle and the acute angle.

LET abc be any triangle, and the angle at b one of its acute angles, and upon bc, one of the sides containing it, let fall the perpendicular (i. 12) ad from the opposite angle: the square of a c, opposite to the angle b, is less than the squares of cb, b,a, by twice the rectangle cb, bd.

First, let ad fall within the triangle abc; and because the straight line cb is divided into two parts in the point d, the squares of cb, bd, are equal (ii. 7) to twice the rectangle contained by cb, bd, and the square of d c. To each of these equals add the square of ad; therefore the squares of cb, bd, da, are equal to twice the rectangle cb, bd, and the squares of a d, dc. But the square of a b is equal (i. 47) to the squares of bd, da, because the angle bda is a right angle; and the square of ac is equal to the squares of ad, dc. Therefore the squares of square of a c, and twice the rectangle cb, bd; alone is less than the squares of cb, ba by twice

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cb, ba are equal to the that is, the square of a the rectangle cb, bd. Secondly, let ad fall

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without the triangle abc. Then, because the angle at d is a right angle, the angle acb is greater (i. 16) than a right angle; and therefore the square of a b is equal (ii. 12) to the squares of ac, cb, and twice the rectangle bc, cd. To these equals add the square of bc, and the squares of a b, bc are equal to the square of a c, and twice the square of bc, and twice the rectangle bc, cd. But because bd is divided into two parts in c, the rectd angle db, bc is equal (ii. 3) to the rectangle bc, cd and the square of bc. And the doubles of these are equal. Therefore the squares of ab, bc are equal to the square of a c, and twice the rectangle db, bc. Therefore the square of a c alone is less than the squares of ab, bc, by twice the rectangle db, bc.

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Lastly, let the side a c be perpendicular to bc; then is bc the straight line between the perpendicular and the acute angle at b; and it is manifest, that the squares of a b, bc, are equal (i. 47) to the square of ac and twice the square of b c. Therefore, in every triangle, &c. Q. E. D.

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