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Book III. ( Greek'text, and the translations from it, the angle of the

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

Cor. From 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 equal, they are right angles.

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IF
[F a straight line touches 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 fegments of the circle.

a II. I.

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 BD 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 * 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 right angles to the touch.

A
ing line from the point of contact
B, the centre of the circle is bin BA;

D
therefore the angle ADB in a semi-
circle is a right angle, and confe-
quently the other two angles BAD,

с
ABD are equal to a right angle:
But ABF is likewise a right angle;
therefore the angle ABF is equal to
the angles BAD, ABD: Take from

E these equals the common angle

F

B ABD; therefore the remaining angle DBF is equal to the angle BAD, which is in the alternate segment of the circle ; and because ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal to

two

C 31. 3.

d 32. I.

e 22. 3.

two right angles; therefore the angles DBF, DBE, being like: Book III. wife equalf to two right angles, are equal to the angles BAD, W BCD; and DBF has been proved equal to BAD: Therefore the f 13. 1. remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c. Q. E. D.

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UPO

PON a given straight line to describe a segment of sec N.

a circle, containing an angle equal to a given rectilineal angle.

Въ31. 3.

dil. I.

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 line AB a segment of a circle, containing an angle eequal to the angle C. First, Let the angle at C be a

H right angle, and bileetAB in F, ő

a ro, I, and from the centre F, at the diftance FB, describe the semicircle AHB; therefore the angle AHB in a femicircle is equal to the right A F 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 23. I. C, and from the point A draw d AE at right angles to

H
AD; bifect * AB in F, and
from F draw d FG at right
angles to AB, and join GB:

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

B two lides AF, FG are equal

F to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the

DI bale AG is equal to the base GB; and the circle described e 4. 1. from the centre G, at the distance GA, shall pass through the point B; let this be the circle AHB: And because froin the point A the extremity of the diameter AE, AD is drawn at

A

Book III. right angles to AE, therefore AD touches the circle ; and be. mcause AB drawn from the point

H fCor. 16.3. of contact A cuts the circle,

the angle DAB is equal to the
angle in the alternate segment

F
AHB: But the angle DAB

B 8 32. 3.

is equal to the angle C, there.
fore also the angle C is equal

E
to the angle in the segment D
AHB: Wherefore, upon the
given straight line AB the seg-
ment AHB of a circle is described wbich contains an angle e-
qual to the given angle at C. Which was to be done.

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T

O cut off a segment from a given circle which shall

contain an angle equal to a given rectilineal angle.

a 17. 3

1 23. 1

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

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

A
make b the angle FBC e.
qual to the angle D:
Therefore, because the
straight line EF touches
the circle ABC, and BC is D
drawn from the point of
contact B, the angle FBC
is equal to the angle in

E B F
the alternate segment BAC
of 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 cir-
ele ABC containing an angle equal to the given angle Di
Which was to be done.

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© 32. 3.

PROP

Book IIT.

PRO P. XXXV. THE O R.

IF
Ftwo straight lines within a circle cut one another, the See N.

rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other.

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.

D:
If AC, BD pass each of them through E
the centre, so that E is the centre; it is
evident, that AE, EC, BE, ED, being all
equal, the rectangle 'AE, EC is likewise B
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 angles, in the point E: Then, if BD be bisected in F, F is the centre of the circle ABCD; and join AF: And because BD, which pafles through the centre, cuts the straight line AC, which does not pass through the centre, at right angles in E, AE, EC are equal to

D

a 3. 3. one another : And because the fraight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED

F together with the square of EF, is e

b qual to the square of FB; that is, A to the square of FA ; but the squares of AE, EF are equal to the square

c 47. I. of FA ; therefore the rectangle BE,

B ED, together with the square of EF, is equal to the squares of AE, EF : Take away the common {quare 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 bisećted in F, F is the centre of the circle. Join AF, and from F draw a FG perpen- d 12. I. G

dicular

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a 3. 3• bs. 2.

Book III. dicular to AC; therefore AG is equal“ to GC; wherefore

the rectangle AE, EC, together with the square of EG, is equal to the square of AG: To each of these equals add the square of GF; therefore the rectangle AE, EC, together with the squares of EG, GF, is equal to

the squares of AG, GF: But the D € 47. I.

squares of EG, GF are equal to the
square of EF; and the squares of
AG, GF are equal to the square of
AF: Therefore the rectangle AE, A

E

C EC, together with the square of EF,

G is equal to the square of AF; that is,

B. to the square of FB : But the square of FB is equal to the rectangle BE, ED, together with the square of EF ; therefore the rectangle AE, EČ, 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 EF, and the remaining rectangle AE, ÉC is therefore equal to the remaining rectangle BE, ED.

Lastly, Let neither of tảe straight lines AC, BD pass through the centre: Take the centre F, and through E, the intersection of the H ftraight lines AC, DB, draw the diameter GEFH: And because the

D rectangle AE, EC is equal, as has been shown, to the rectangle GE,

E

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

G
same rectangle GE, EH ; therefore
the rectangle AE, EC is equal to

B
the rectangle BE, ED. Wherefore, if two straight lines, &c.
Q. E. D.

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IF
F from any point without a circle owo 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.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the

circle,

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