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to one another ; wherefore d E is the center of the circle. from Book III. the center E, at the distance of any of the three AE, EB, EC describe a circle, this shall pass thro' the other points; and the d. 9. 3. circle of which ABC is a segment is described., and it is evident that if the angle ABD be greater than the angle BAD, the center E falls without the segment ABC, whic', therefore is less than a femicircle. but if the angle ABD be less than BAD, the center E falls within the segment ABC, which is therefore greater than a femicircle. wherefore a segment of a circle being given, the circle is defcribed of which it is a segment. Which was to be done.

PROP. XXVI. THEOR.

IN cqual circles,

N equal circles, equal angles stand upon equal cir:

cumferences, whether they be at the centers or circumferences.

Let ABC, DEF be equal circles, and the equal angles BGC, EHF at their centers, and BAC, EDF at their circumferences, the circumference BKC is equal to the circumference ELF.

Join BC, EF, and because the circles ABC, DEF are equal, the straight lines drawn from their centers are equal; therefore the two sides BG, GC, are equal to the two EH, HF; and the angle

1

А.

D

G

H

B В
CE

F

L K at G is equal to the angle at H; therefore the base BC is equal # a. 4. 1. to the base EF. and because the angle at A is equal to the angle at D, the fegment BAC is similar to the segment EDF; and they b.11.Def. 3, are upon equal straight lines BC, EF; but similar segments of circles upon equal straight lines are equal to one another ; c. 24. 3. therefore the segment BAC is equal to the segment EDF. but the whole circle ABC is equal to the whole DEF, therefore the

Book III. remaining fegment EKC is equal to the remaining segment ELF,

and the circumference BKC to the circumference ELF. Wherefore, in equal circles, &c. Q. E. D.

PRO P. XXVII. THEOR.

IN N equal circles, the angles which stand upon equal

circumferences, are equal to one another, whether they be at the centers, or circumferences.

Let the angles BGC, EHF at the centers, and BAC, EDF at the circumferences of the equal circles ABC, DEF stand upon she equal circumferences BC, EF. the angle BGC is equal to the angle EHF, and the angle BAC to the angle EDF.

If the angle BGC be equal to the angle EHF, it is manifest • that the angle BAC is also equal to EDF. but if not, one of them

a. 20. 3.

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b. 23. 1. 4. 26. 3:

is the greater. let BGC be the greater, and at the point G, in the straight line BG, make b the 'angle BGK equal to the angle EHF; but equal angles sand upon equal circumferences", when they are at the center; therefore the circumference BK is equal to the circumference EF. but EF is equal to BC, therefore also BK is equal to BC, the less to the greater, which is impossible. therefore the angle BGC is not unequal to the angle EHF; that is, it is equal to it. and the angle at A is half of the angle BGC, and the angle at D half of the angle EHF. therefore the angle at A is equal to the angle at D. Wherefore, in equal circles, &c. Q. E. D.

Book III.

PROP. XXVIII. THEOR.
IN
N equal circles, equal straight lines cut off equal

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

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 less EHF.

Take K, L the centers of the circles, and join BK, KC, EL, a. 1. 3
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 o to the angle ELF. but equal angles stand upon 4.8. 1.
equal · circumferences, when they are at the centers ; therefore c. 26. 3.
the circumference BGC is equal to ene circumference EHF.
but the whole circle ABC is equal to the whole EDF; the re-
maining part therefore of the circumference, viz. BAC, is equal
to the remaining part EDF. Therefore, in equal circles, &c.
Q. E. D.

PRO P. XXIX. THE O R.
N equal circles equal circumferences are subtended

by equal straight lines.

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IN

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.

Book III.

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

a. 1. 3.

A

D

K

L

b. 27. 3.

BH

С EN
G

H
ference EHF, the angle BKC is equal 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, F,
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 biscct a given circumference, that is, to divide it

into two equal parts.

2. 10. 1.

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

Join AB, and bifect ' 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 sides AC, CD
are equal to the two BC, CD; and

D
the angle ACD is equal to the angle
BCD, because each of them is a right
angle; therefore the base AD is equal

A C B b. 4. I.

to the base BD. but equal straight c. 28. 3.

lines cut off equal o circumferences, the greater equal to the greater,

and the less to the lefs, 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 bifected in D. Which was to be done.

Book III,

PRO P. XXXI. THEOR.
IN Na 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.

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b. 32. I.

Let ABCD be a circle, of which the diameter is BC, and cen-
ter E; and draw CA dividing the circle into the fegments 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 femicircle, 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
EA, the angle EAB is equal - to

a. 5.
EBA ; also, because AE is equal to
EC, the angle EAC is equal to

A
ECA; wherefore the whole angle
BAC is equal to the two angles

D
ABC, ACB. but FAC, the exterior
angle of the triangle ABC, is equal

B
6 to the two angles ABC, ACB ;

E
therefore the angle BAC is equal
to the angle FAC, and each of them

c. 10. Def.i,
is therefore a right angle. where-
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 d 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 o to two right angles; there-e. 2 313 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 manifest, that the circumference of the greater fegment ABC falls without the right angle CAB, bụt the circum,

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