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Join BC, EF. And because the circles ABC, DEF are equal, the straight lines drawn from their centers are equal : (111. def. 1.) therefore the two sides BG, GC, are equal to the two EH, HF, each to each:

and the angle at G is equal to the angle at H; (hyp.) therefore the base B C is equal to the base EF. (1. 4.) And because the angle at A is equal to the angle at D, (hyp.) the segment BAC is similar to the segment EDF: (111. def. 11.)

and they are upon equal straight lines BC, EF: but similar segments of circles upon equal straight lines, are equal to

one another, (111. 24.)

therefore the segment BAC is equal to the segment EDF: but the whole circle ABC is equal to the whole DEF; (hyp.) therefore the remaining segment BKC is equal to the remaining segment ELF, (1. ax. 3.)

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

PROPOSITION XXVII. THEOREM. In equal circles, the angles which stand upon equal arcs, are equal to one another, whether they be at the centers or circumferences.

Let ABC, DEF be equal circles,
and let the angles BGC, EIF at their centers,
and the angles BAC, EDF at their circumferences,

stand upon the equal arcs BC, EF.
Then the angle BGC shall be equal to the angle EIIF,

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. (111. 20. and

I. ax. 7.)
But, if not, one of them must be greater than the other :
if possible, let the angle BGC be greater than EHF,

and at the point G, in the straight line BG,
make the angle BGK equal to the angle EHF. (1. 25.)

Then because the anglc BGK is equal to the angle EHF, and that equal angles stand upon equal arcs, when they are at the centers; (111. 26.)

therefore the arc BK is equal to the arc EF:
but the arc EF is equal to the arc BC; (hyp.)

therefore also the arc BK is equal to the arc BC,
the less equal to the greater, which is impossible: (1. ax. 1.)

therefore the angle BGC is not unequal to the angle EHF;

that is, it is equal to it:
but the angle at A is half of the angle BGC, (I11. 20.)

and the angle at D, half of the angle EHF; therefore the angle at A is equal to the angle at D. (1. ax. 7.)

Wherefore, in equal circles, &c. Q.E.D.

PROPOSITION XXVIII. THEOREM. In equal circles, equal straight lines cut off equal arcs, the greater equal is 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

arcs BAC, EDF, and the two less BGC, EHF. Then the greater arc BAC shall be equal to the greater EDF,

and the less arc BGC to the less EHF.

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ake K, L, the centers of the circles, (111. 1.) and join BK, KC, EL, LF.

Because the circles ABC, DEF are equal,
the straight lines from their centers are equal : (111. def. 1.)

therefore BK, KC are equal to EL, LF, each to each : and the base B C is equal to the base EF, in the triangles BCK, EFL;

therefore the angle BKC is equal to the angle ELF: (1. 8.) but equal angles stand upon equal arcs, when they are at the centers : (III. 26.)

therefore the arc BGC is equal to the arc EHF: ut the whole circumference ABC is equal to the whole EDF; (hyp.)

therefore the remaining part of the circumference, viz. the arc BAC, is equal to the remaining part EDF. (I. ax. 3.)

Therefore, in equal circles, &c. Q.E.D.

PROPOSITION XXIX. THEOREM.
In equal circles, equal arcs are subtended by equal straight lines.

Let ABC, DEF be equal circles,
and let the arcs BGC, EHF also be equal,

and joined by the straight lines BC, ÈF. Then the straight line B C shall be equal to the straight line EF.

Take K, L, (111. 1.) the centers of the circles, and join BK, KC, EL, LF.

Because the arc BGC is equal to the arc EHF, therefore the angle BKC is equal to the angle ELF: (III. 27.)

and because the circles ABC, DEF, are equal,
the straight lines from their centers are equal; (111. def. 1,)

therefore BK, KC, are equal to EL, LF, each to each :
and they contain equal angles in the triangles B CK, EFL;
therefore the base BC is equal to the base EF (1. 4.)

Therefore, in equal circles, &c. Q. E. D.

PROPOSITION XXX. PROBLEM.
To bisect a given arc, that is, to divide it into two equal parts.

Let ADB be the given arc:
it is required to bisect it.

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Join AB, and bisect it in C; (1. 10.)
from the point C draw CD at right angles to ÁB. (1. 11.)
Then the arc ADB shall be bisected in the point D.

Join AD, DB.
And 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, each to each;

and the angle ACD) is equal to the angle BCD,

because each of them is a right angle:
therefore the base AD is equal to the base BD. (1. 4.)

But equal straight lines cut off equal arcs, (111. 28.) the greater arc equal to the greater, and the less arc to the less;

and the arcs AD, DB are each of them less than a semicircle ; because DC, if produced, passes through the center: (111. 1. Cor.)

therefore the arc AD is equal to the arc DB. Therefore the given arc ADB is bisected in D. Q.E.F.

PROPOSITION XXXI. THEOREM. 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.

Let ABCD be a circle, of which the diameter is B C, and center E, and let CA be drawn, dividing the circle into the segments ABC, ADC.

Join BA, AD, DC. Then the angle in the semicircle BAC shall be a right angle; and the angle in the segment ABC, which is greater than a semicircle,

shall be less than a right angle; and the angle in the segment ADC, which is less than a semicircie,

shall be greater than a right angle.

Join A E, and produce B A to F.
First, because EB is equal to EA, (I. def. 15.)
the angle EAB is equal to EBA; (1. 5.)

also, because is equal to EC,

the angle ECA is equal to EAC; wherefore the whole angle BAC is equal to the two angles EBA,

ECA; (1. ax. 2.) but FAC, the exterior angle of the triangle ABC, is equal to the two

angles EBA, ECA; (1. 32.)
therefore the angle BAC is equal to the angle FAC; (ax. 1.)

and therefore each of them is a right angle: (1. def. 10.) wherefore the angle BAC in a semicircle is a right angle. Secondly, because the two angles ABC, BAC of the triangle ABC are together less than two right angles, (1. 17.)

and that BAC has been proved to be a right angle;

therefore 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 lastly, because ABCD is a quadrilateral figure in a circle, any two of its opposite angles are equal to two right angles: (111. 22.) therefore the angles ABC, ADC, are equal to two right angles :

and ABC has been proved to be less than a right angle;

wherefore the other ADC is greater than a right angle. Therefore, in a circle the angle in a semicircle is a right angle; &c. Q.E.D.

COR. From this it is manisest, 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; (1. 32.) and when the adjacent angles are equal, they are right angles. (I. def. 10.)

PROPOSITION XXXII. THEOREM. If a straight line touch a circle, and from the point of contact a straight line be drawn meeting the circle; the angles which this line makes 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, meeting the circumference in D, and dividing it into the segments DCB, DAB, of which DCB is less than, and DAB greater than a semcircle.

Then 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 DBF shall be equal to the angle which is in the segment DÅB,

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and the angle DBE shall be equal to the angle in the alternate

segment DCB.

EBF From the point B draw BA at right angles to EF, (1. 11.) meeting the circumference in A;

take any point C in the arc DB, and join AD, DC, CB. Because the straight line EF touches the circle ABCD in the point B,

and BA is drawn at right angles to the touching line from the point of contact B,

the center of the circle is in BA: (111. 19.) therefore the angle ADB in a semicircle is a right angle: (111. 31.) and consequently the other two angles BAD, ABD, are equal to a right angle; (1. 32.)

but ABF is likewise a right angle; (constr.) therefore the angle ABF is equal to the angles BAD, ABD: (I. ax. 1.)

take from these equals the common angle ABD: therefore the remaining angle DBF is equal to the angle BAD, (I.ax.3.)

which is in BDA, 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 right angles: (III. 22.) but the angles DBF, DBE are likewise equal to two right angles;

(1. 13.) therefore the angles DBF, DBE are equal to the angles BAD, BCD, (1. ax. 1.) .

and DBF has been proved equal to BAD; therefore the remaining angle DBE is equal to the angle BCD in BDC, the alternate segment of the circle. (I. ax. 2.)

Wherefore, if a straight line, &c. Q.E.D.

PROPOSITION XXXIII. PROBLEM. Upon a given straight line to describe a segment of a circle, which shall contain an angle equal to a given rectilineal angle.

Let AB be the given straight line,

and the angle C the given rectilineal angle. It is required to describe upon the given straight line AB, a segment of a circle, which shall contain an angle equal to the angle C.

First, let the angle C be a right angle.

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