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Book III.

PROP. XXVHI.

THEOR.

N equal circles, equal straight lines cut off equal

circumferences, the greater equal to the greater, and the less to the less.

IN

Let ABC, DEF be equal circles, and BC, EF equal straight lines in them, which cut off the two greater circumferences BAC, EDF, and the two less BGC, EHF. the greater BAC is equal to the greater EDF, and the less BGC to the lefs EHF.

Take a K, L the centers of the circles, and join BK, KC, EL, a. 1. 3i LF. and because the circles are equal, the straight lines from A

D

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H their centers are equal, therefore BK, KC, are equal to EL, LF; and the base BC is equal to the base EF; therefore the angle BKC is equal to the angle ELF. but equal angles stand upon b. 8. 1. equal o circumferences, when they are at the centers ; therefore c. 26. 3: the circumference BGC is equal to the circumference EHF. but the whole circle ABC is equal to the whole EDF; the remaining part therefore of the circumference, viz. BAC, is equal to the remaining part EDF. Therefore, in equal circles, &c. Q. E. D.

PROP. XXIX. THEOR.

IN

'N equal circles equal circumferences are subtended

by equal straight lines. .

Let ABC, DEF be equal circles, and let the circumferences BGC, EHF also be equal; and join BC, EF. the straight line BC is equal to the straight line EF.

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Book III.

Take a K, L the centers of the circles, and join BK, KC, EL, LF. and because the circumference BGC is equal to the circumА.

D

a. I. 3.

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H B. 27. 3. ference EHF, the angle BKC is equal b to, the angle -ELF. and

because the circles ABC, DEF are equal, the straight lines from their centers are equal; therefore BK, KC are equal to EL, LF, and they contain equal angles. therefore the base BC is equal to the base EF. Therefore, in equal circles, &c. Q. E. D.

C. 4. I.

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To

10 bifect a given circumference, that is, to divide it

into two equal parts.

2. 10. I.

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Let ADB be the given circumference; it is required to bifect it.

Join AB, and bisect a it in C; from the point C draw CD at right angles to AB, and join AD, DB. the circumference ADB is bifected in the point D.

Because AC is equal to CB, and CD common to the triangles ACD, BCD, the two fides AC, CD are equal to the two BC, CD; and the angle ACD is equal to the angle BCD, because each of them is a right angle; therefore the base AD is equal

А. o to the base BD. but equal straight

C B b. 4. I. C. 28. 3. lines cut off equal circumferences, the greater equal to the greater,

and tle less to the less, and AD, DB are each of them less than a d. Cor. 1. 3. semicircle; because DC passes through the center d. wherefore the

circumference AD is equal to the circumference DB. therefore the given circumference is bisected in D. Which was to be done.

Book III.

PROP. XXXI.

THEOR.

IN

D

N 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 femicircle is greater than a right angle.

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 semicircle 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 semicircle is greater than a right angle. Join AE, and produce BA to F; and because BE is equal to

a. 5. So EA, the angle EAB is equal a to EBA; also, because AE is equal to

F
EC, the angle EAC is equal to
ECÀ; wherefore the whole angle

Α.
BAC is equal to the two angles
ABC, ACB. but FAC, the exterior
angle of the triangle ABC, is equal
b to the two angles ABC, ACB;

B
E

b. 32. I. therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right © angle. where

C. 10. Def.I. fore the angle BAC in a semicircle is a right angle.

And because the two angles ABC, BAC of the triangle ABC are together less than two right angles, and that BAC is a right d. 17. I. angle, ABC must be less than a right angle; and therefore 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- c. 22. 3a fore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle, wherefore the other ADC is greater than a right angle.

Besides, it is manifeft, that the circumference of the greater fegment ABC falls without the right angle CAB, but the circum

Book III. ference of the less segment ADC falls within the right angle CAF. Premyer And this is all that is meant, when in the Greek text, and the

• translations from it, the angle of the greater fegment is said to « be greater, and the angle of the less segment is faid to be less than a right angle.'

Cor. From this it is manifeft, 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 angleš.

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B, II. I.

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 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 fhall 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 3 and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at right angles

A to the touching line from the point

D of contact B, the center of the circle B. 19. 3. is bin BA; therefore the angle ADB

in a semicircle is a right angle, and

consequently the other two angles # 32. 1. BAD, ABD are equal d to a right

angle. but ABF is likewise a right
angle; therefore the angle ABF is E B F
equal to the angles BAD, ABD. take
from these equals the common angle
ABD ; therefore the remaining angle DBF is equal to the angle

C. 31. 3.

BAD, which is in the alternate segment of the circle; and because Book I'l.
ABCD is a quadrilateral figure in a circle, the opposite angles
BAD, BCD are equal to two right angles; therefore the angles e. 22. 3.
DBF, DBE, being likewise equal to two right angles, are equal f. 13. 1.
to the angles BAD, BCD; and DBF has been proved equal to
BAD; therefore the remaining angle DBE is equal to the angle
BCD in the alternate fegment of the circle. Wherefore, if a
straight line, &c. 0. E. D.

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UP a

PON a given straight line to describe a segment See N.

of , a rectilineal angle.

2. 10. 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 equal to the angle C.

First, Let the angle at C be a right angle, and bisect : AB in F, and from the center F, at the distance FB, describe the femicircle AHB; therefore the angle AHB in a semicircle is equal to the right angle A 'F B. b. 31. 3. at C.

But if the angle C be not a right angle, at the point A in the straight line AB, make c the angle BAD equal to the angle C, c. 23. 1. and from the point A draw d AE at right angles to AD; bi

H sect a AB in F, and from F draw

E dFG at right angles to AB, and join GB. and because AF is equal to FB, and FG common to the triangles AFG, BFG, the A two fides AF, FG are equal to

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

D therefore the base AG is equal

C. 4. I. to the base GB; and the circle described from the center G, at

d. II. I.

e

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