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Book III. and the translations from it, the angle of the grdater segment

is said to be greater, and the angle of the less segment is said to be less, than a right angle.'

Cor. Froin this it is manifest, that if one angle of a triangle be equal to the other two, it is a right angle, because the angle adjacent to it is equal to the same two; and when the adjacent angles are eqiial, they are right angles.

PROP. XXXII. THEOR.

IF a straight line touch a circle, and from the point of contact a straight line be drawn cutting the circle, the angles made by this line with the line touching the circle, shall be equal to the angles which are in the alternate segments of the circle.

11. 1.

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b 19. 3.

Let the straight line EF touch the circle ABCD in B, and from the point B let the straight line BD be drawn, cutting the circle: the angles which DB makes with the touching line EF shall be equal to the angles in the alternate segments of the circle: that is, the angle FBD is equal to the angle which is in the segment DAB; and the angle DBE to the angle in the segment BCD:

From the point B draw a BA at right angles to EF, and take any point C in the circumference BD, and join AD, DC, CB; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at

A
right angles to the touching line
from the point of contact B, the
centre of the circle is b in BA;
therefore the angle ADB in a
semicircle is a right © angle; and
consequently the other two an-
gles BAD, ABD are equal to
a right angle: but ABF is like-
wise a right angle; therefore the
angle ABF is equal to the angles E

B

F BÅD, ABD : fake from these equals the comihon angle ABD; therefore the remaining angle DBP is equal to the angle BAD, which is in the alternate seg. mept of the circle; and because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal to two

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031.3.

32. 1.

22.3.:

right angles; therefore the angles DBF, DBE, being likewise Book III. equal to two right angles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD: therefore the re-f13.1. maining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c, , Q. E. D.

444

,

PROP. XXXIII. PROB.

UPON a given straight line to describe a segmeni See Note: of a circle, containing ari angle equal to a given rectilineal angle.

a

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Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given straight linc AB a segment of a circle, containing ani angle equal to the angle c. First, Let the angle at C be a

HI right angle, and bisect * AB in F,

a 10.1. and from the centre F, at the distance FB, describe the semicir, cle AHB; therefore the angle AHB in a semicircle is b equal A F to the right angle at C.

But if the angle C be not a right angle, at the point A, in the straight line AB, make the angle BAD equal to the angle c 28. 2. C, apd from the point A

H drawd AE at right angles to AD; bisect a AB in F, and

1 d 11, 1. from F drawd FG at right angles to AB, and join GB:

G and because AF is equal to FB, and FG common to the triangles AFG, BFG, the

A two sides AF, FG are equal to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the

D base AG is equal e to the base GB; and the circle described c 4. Ta, from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: and because from the point A the extremity of the diameter AE, AD is drawn at

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g 32: 3.

Book 111. right angles to AE, therefore AD f touches the circle; and be cause AB drawn from the

H € Cor. 16. 3. point of contact A cuts the

circle, the angle DAB is equal
to the angle in the air
alternate

B
segment AHB&: but the an-
gle DAB is equal to the an-

G

E glc C, therefore also the angle Cis equal to the angle in the segment AHB: wherefore, upon the given straight line AB the segment AHB of a circle is described which cone

tains an angle equal to the given angle at C. Which was to 5 be done!

a

PROP. XXXIV. PROB.

TO cut off a segment from a given circle which shall contain an angle equal to a given rectilineal angle:

a 17. 3.

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b 23. I.

Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a segment from the circle ABC that shall contain an angle equal to the given angle D.

Drawa the straight line EF touching the circle ABC in the point B, and at the point B,

A
in the straight line BF, make
b the angle FBC equal to the
angle D; therefore, because
the straight line EF touches
the circle ABC, and BC is
drawn from the point of con-
tact B, the angle FBC is
equal to the angle in the
'alternate segment BAC of Era В.

F
the circle: but the angle FBC
is equal to the angle D; therefore the angle in the segment
BAC is equal to the angle D: wherefore the segment BAC;
is cut off from the given circle ABC .containing, an angle.
equal to the gifen angle D. Which was to be done.

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€ 32. 3.

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IF (wo straight lines within a circle cut one ano. See Note, ther, the rectangle contained by the segments of one of them is equal to the rectangle contained by the seg. ments of the other.

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с

a 3,3

Let the two straight lines AC, BD, within the circle ABCD,
cut one another in the point E: the rectangle contained by AE,
EC is equal to the rectangle contained by
BE, ED.

А

E D If AC, BD pass each of them through the centre, so that E is the centre; it is evident, that AE, EC, BE, ED, being all

B equal, the rectangle AE, EC is likewise equal to the rectangle BE, ED.

But let one of them BD pass through the centre, and cut the other AC, which does not pass through the centre, at right an. gles, in the point E; then, if BD be bisected in F, F is the centre of the circle ABCD; join AF:and because BD, which passes through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE, EC are equala

D
to one another: and because the
straight line BD is.cut into two equal
parts in the point F, and into two
unequal in the point E, the rectangle

F
BE, ED together with the square of
EF, is equal to the square of FB;

A

b 5. that is, to the square of FA; but the

E

с squares of AE, EF are equale to the

c 47. k. square of FA; therefore the rectangle BE, ED, together with the square

B of EF, is equal to the squares of AE, EF: take away the common square of EF, and the remaining rectangle BE, ED is equal to the remaining square of AE; that is, to the rectangle AE, EC,

Next, Let BD, which passes through the centre, cut the other, AC, which does not pass through the centre, in E, but not at right angles: then, as before, if BD be bisected in F, F Is the centre of the cle. Join AF, and from F drawd FG N

A 12.1

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Book 11 perpendicular to AC; therefore AG is equal to GC; where

fore the rectangle AE, EC, together with the square of EG, is a 3. 3.

equalk to the square of AG: to each of these equals add the b 5. 2. square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to

D
the squares of AG, GF: but the
€ 47. 1.

squares of EG, GF are equal c to the
square of EF, and the squares of

F
AG, GF are equal to the square of
AF: therefore the rectangle AE,

A

C EC, together with the square of EF, is equal to the square of AF; that

B is, to the square of FB: but the... square of FB is equalb to the rectangle BE, ED, together with the square of EF : therefore the rectangle AE, EC, together with the square of EF, is equal to the rectangle BE, ED, together with the square of EF : take.

away

the common square of EE, and the remaining rectangle AE, ÉC is therefore equal to the remaining rectangle BE, ED.

Lastly, I et neither of the straight lines AC,BD pass through the centre : take the centre F, and through E, the intersection of the straight lines AC, DB draw the

D diameter GEFH: and because the rectangle AE, EC is equal, as las

E been shown, to the rectangle GE,

A EH; and, for the same reason, the rectangle BE, ED is equal to the

G same rectangle GE, EH; therefore

B the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two straight lines, &c., Q. E. D.

19

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PROP. XXXVI. THEOR

IF from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle,

and the part of it without the circle, shall be equal to the square of the line which touches it.

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