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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, ea c is half a right angle (i. 32). For the same reason, each of the angles ceb, eb c is half a right angle; therefore a eb is a right angle: and because e bc 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 dg bis 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

са. But the square of ea is equal (i. 47) to the squares of ec, ca; therefore the square of e a 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 g f, fe are double of the square of ef: but the square of eg

is equal (i. 47) to the squares of g f, fe; therefore the square of eg is double of the square of ef: and ef is equal to cd; wherefore the square of is double of the square of cd. But it was demonstrated, that the square of e a is double of the square of ac: therefore the squares of a e, eg, are double of the squares of a c, cd: and the square of ag is equal (i. 47) to the squares of a e, 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 a g; therefore the

eg

squares of a d, dg are double of the squares of a c, cd: but dg is equal to db; therefore the squares of ad, db are double of the squares of ac, 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 con

tained by the whole and one of the parts shall be equal to the square of

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

Produce g h 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, a e are equal (i. 47) to the square

of eb, because the angle ea b is a right angle; therefore the rectangle cf, fa, together with the square of a e, is equal to

f

8 the squares

of ba, a e: take away the square of a e, which is common to both, therefore the remaining rectangle cf, fa, is equal to the square of a b; and the figure fk is the rectangle contained by cf, fa, for af is equal to fg; and a d is the square of

h b 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 a b, bh, for a b is equal to bd; e and fh is the square of a h. Therefore the rectangle a b, b h is equal to the square of a h: wherefore the straight line a b is divided in h, so that the rectangle a b, bh, is equal to the square of a h.

с

k d Which was to be done.

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

Because the straight line bd is divided into two parts in the point c, the

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 b c 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 squares of bc, ca, and twice the rectangle bc, cd;

b

d 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.

с

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 a bc 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 a bc; and because the straight

line cb is divided into two parts in the point a

d, the squares of cb, bd, are equal (ii. 7) to twice the rectangle contained by cb, bd, and the

square of dc. 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 b d

c angle; and the square of ac is equal to the

squares of a d, dc. Therefore the squares of cb, ba are equal to the square of a c, and twice the rectangle cb, bd; that is, the square of a c alone is less than the squares of c b, ba by twice the rectangle cb, bd. Secondly, let ad fall without the triangle a bc. Then, because the

angle at d is a right angle, the angle a cb is a

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 b c, cd. To these equals add the square of bc, and the squares of a b, bc are equal to

of ac,

and twice the square of bc, and twice the rectangle bc, cd. But because

bd is divided into two parts in c, the rectb

d angle d b, bc is equal (ii. 3) to the rectangle

bc, cd and the square of b c. And the doubles of these are equal. Therefore the squares of a b, bc are equal to the

a square of a c, and twice the rectangle db, b c. Therefore
the
square

alone is less than the squares of a b, bc, by twice the rectangle db, bc.

Lastly, let the side ac 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 a c and twice the

square of b c. Therefore, in every triangle, &c. Q. E. D. b

C

the square

of ac

PROPOSITION XIV.-PROBLEM. To describe a square that shall be equal to a given rectilineal figrire. LET a be the given rectilineal figure; it is required to describe a square that shall be equal to a.

Describe (i. 45) the rectangular parallelogram bcde equal to the rectilineal figure a. If then the sides of it, be, ed, are equal to one another, it is a square, and what was required is now done. But if they are not equal, produce one of them be to f, and make ef

h equal to ed, and bisect bf in and from the centre at

8 the distance gb or gf, describe the semicircle b ħ f, and produce de to h, and join

ъ gh: therefore because the

g straight line bf is divided into two equal parts in the point g, and into two unequal at e, the rectangle be, ef, together with the square of eg, is equal (ii. 5) to the square of gf: but gf is equal to gh: therefore the rectangle bé, ef, together with the square of eg, is equal to the square of gh: but the squares of he, eg are equal (i. 47) to the square of gh: therefore the rectangle be, ef, together with the square of eg, is equal to the squares of

the
square

of

eg which is common to both; and the remaining rectangle be, ef is equal to the square of eh: but the rectangle contained by be, ef is the parallelogram bd, because ef is equal to ed; therefore bd is equal to the square of eh; but bd is equal to the rectilineal figure a; therefore the rectilineal figure a is equal to the square of eh. Wherefore a square has been made equal to the given rectilineal figure a, viz. the square described upon eh. Which was to be done.

he, eg;

take away

EXERCISES ON BOOK II.

SECT. I. PROBLEMS. 1. Given the difference between two squares, to find a line whose square shall be equal to that difference:

2. Given two lines, to produce one of them, so that the rectangle contained by the two given lines shall be equal to the square of the part produced.

3. Given a straight line, to describe on it, as the hypotenuse, a rightangled triangle such that the sum of the hypotenuse and the lesser of the other two sides shall be double of the remaining side of the triangle. 4. Given the sum of the squares of any number of lines, to find a line

shall be equal to the given sum,

whose square

SECT. II.—THEOREMS. 5. If a straight line be drawn from the vertex of an isosceles triangle to any point in the base, the square described on this line, together with the rectangle contained by the segments of the base, is equal to the square described upon either of the equal sides.

6. The difference between the squares of two unequal straight lines is equal to the rectangle contained by their sum and difference.

7. If a straight line be divided into five equal parts, the square of the whole line is equal to the square of the straight line which is made up of four of those parts, together with the square of the straight line which is made up of three of those parts.

8. If a perpendicular be drawn from the right angle of a right-angled triangle to the hypotenuse, the square of the perpendicular is equal to the rectangle contained by the segments of the hypotenuse.

9. If the base of a triangle be bisected, the sum of the squares of the other two sides is equal to twice the square of half the base and twice the square of the line drawn from the point of bisection to the vertical angle.

10. If the middle points of the opposite sides of a quadrilateral figure be joined, the sum of the squares of the joining lines is equal to one-half the sum of the squares of the diagonals of the quadrilateral.

11. If the sides of a triangle be bisected and lines drawn from the points of bisection to the opposite angles, four times the sum of the squares of these lines shall be equal to three times the sum of the squares of the three sides of the triangle.

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