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Book III.

PRO P. XXXI.

THE O R.

IN a circle, the angle in a semicircle is a right angle;

but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

b 32. I.

Let ABCD be a circle, of which the diameter is BC, and center E; and draw CA dividing the circle into the segments ABC, ADC, and join BA, AD, DC; the angle in the femi. circle BAC is a right angle; and the angle in the segment ABC, which is greater than a semicircle, is less than a right angle; and the angle in the segment ADC, which is less than a femicircle, is greater than a right angle.

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA ; also, because AE a 5. 1. is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB: But FAC, A the exterior angle of the triangle

D
ABC, is equal o to the two angles
ABC, ACB; therefore the angle
BAC is equal to the angle FAC, B
and each of them is therefore a

E

cIo. def. 1. right angle: Wherefore the angle BAC in a semicircle is a right an. gle.

And because the two angics ABC, BAC of the triangle ABC are together less d than two right angles, and that BAC & 17.1. is a right angle, ABC must be less than a right angle; and there, fore the angle in a segment ABC greater than a semicircle, is less than a right angle.

And because ABCD is a quadrilateral-figure in a circle, any two of its opposite angles are equal to two right angles; there. e 22 3. fore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC iş greater than a right angle.

Belides, it is manifeft, that the circumference of the greater segment ABC falls without the right angle CAB, but the circumference of the lefs segment ADC falls within the right angle CAF. And this is all that is meant, when in the

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Book III. · Greek text, and the translations from it, the angle of the mo 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 manifeft, that if one angle of a tri. angle 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 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 segments of the circle.

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 center of the circle is in BA;

D
therefore the angle ADB in a semi-
circle is a right © angle, and con-
sequently the other two angles BAD,
ABD are equal d to a right angle :
But ABF is likewise a right angle ;
therefore the angle ABF is equal to
the angles BAD, ABD: Take from
these equals the common angle

F

B ABD; therefore the remaining an. gle 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

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two right angles; therefore the angles DBF, DBE, being like. Book III. wise equal f to two right angles, are equal to the angles BAD, im 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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PON a given straight line to describe a segment of See N.

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

6 31. 31

Let AB he the given straight line, and the angle at C the given rectilineal angle; it is required to describe upon the given Itraight line AB a segment of a circle, containing an angle ea qual to the angle C.

First, Let the angle at C be a right angle, and bifect ' AB in F,

H Н a 10. and from the center F, at the di.

С. Itance FB, describe the semicircle AHB; therefore the angle AHB in a semicircle is b equal to the right angle at C.

A F B 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. 1. C, and from the point A draw d AE at right angles to

H AD; bisect * AB in 1, and from F draw d FG at right

E angles to AB, and join GB:

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

A

F two lides AF, FG are equal

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

D\ base AG is equal to the base GB; and the circle described from the center 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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e 4. I.

Book III. right angles to AE, therefore ADF touches the circle; and because AB drawn from the point

н f Cor. 16. 3. of contact A cuts the circle,

the angle DAB is aqual to
the angle in the alternate seg A

F 6 32. 3. ment AHB 8 : But the angle

B DAB is equal to the angle C, therefore also the angle C is

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

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a 17. 3.

b 23. 1.

O cut off a segment from a given circle which shall

contain an angle equal to a given rectilineal angle. 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 angle D.

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

A
make the angle FBC e-
qual to the angle D:
Therefore, because the
straight line EF touches the
circle ABC, and BC 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.
cle ABC containing an angle equal to the given angle D:
Which was to be done.

C 32• 3.

PR P.

Book III.

PRO P. XXXV.

THE O R.

If two straight lines within a circle cut one another, the Sec 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.

A

D If AC, BD pass each of them through E the center, so that E is the center; 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 center, and cut the other AC, which does not pass through the center, at right angles, in the point E : Then, if BD be bisected in F, F is the center of the circle ABCD, and join AF: And because BD, which paffes through the center, cuts the straight line AC, which does not pass through the center, at right

D angles in E, AE, EC are equal a to

a 3. 3. 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 BE, ED

F together with the square of EF, is e

b s. 2. qual to the square of FB: that is, A to the square of FA; but the squares

E of AE, EF are equal to the square

C 47. I. of PA, 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 center, cut the other AC, which does not pass through the center, in E, but not at right angles: Then, as before, if BD be bifected in F, F is the center of the circle. Join AF, and from F draw FG perpen-d 12. 1, G

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