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ADB, the exterior to the interior-which is impossible : therefore the similar segments ACB, ADB, upon the same side of the same straight line, AB, cannot but coincide. Wherefore, Upon the same &c. Q.E.D.

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Similar segments of circles upon equal straight lines are equal to one another.

Let AEB, CFD be similar segments of circles upon equal straight lines AB, CD: the segment AEB is equal to the segment CFD.

For if the segment AEB be applied to the segment CFD, so that the point A

may be on C, and the straight line AB upon CD, the point A

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B

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B shall coincide with the point D, because AB is equal to CD: Therefore, the straight line AB coinciding with CD, the segment AEB must coincide with the segment CFD (3. 23), and must therefore be equal to it.

Wherefore, Similar segments &c.

Q.E.D.

PROP. XXV. PROB.

A segment of a circle being given, to describe the circle of which it is the segment.

Let ABC be the given segment of a circle: it is required to describe the circle of which it is the segment. Bisect AC in D, and from the point D draw DB at right angles to AC, and join AB.

E

First, let the angles DAB, DBA be equal to one another: Then, because the angle DBA is equal to the angle DAB, therefore DB is equal to DA, and

B

B

D

A

E

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A D

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D

therefore also to DC: And because the three straight lines DA, DB, DC are all equal, D is the centre of the circle (3. 9): From the centre D, at the distance of any of the three DA, DB, DC, describe a circle; this shall pass through the extremities of the others, and be the circle, of which ABC is a segment: Also, because the centre D is in AC, the segment ABC is a semicircle. Next, if the angles DAB, DBA be not equal to one another: At the point A, in the straight line BA, make the angle BAE equal to the angle ABD, and produce BD, if necessary, to E, and join EC: Then, because the angle EAB is equal to the angle EBA, the straight line EA is equal to EB: And, because AD is equal to CD, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each, and the angle ADE is equal to the angle CDE, for each of them is a right angle-therefore the base AE is equal to the base CE:

But EA was shewn to be equal to EB; therefore also EB is equal to EC, and the three straight lines EA, EB, EC are equal to one another, and therefore E is the centre of the circle: From the centre E, at the distance of any of the three EA, EB, EC, describe a circle; this shall pass through the other points, and be the circle, of which ABC is a segment: Also, it is evident, that if the angle DAB be greater than the angle DBA, the centre E falls without the segment ABC, which therefore is less than a semicircle; but if the angle DAB be less than the angle DBA, the centre E falls within the segment ABC, which is therefore greater than a semicircle: Wherefore, a segment of a circle being given, the circle is described of which it is a segment. Q. E.D.

PROP. XXVI. THEOR.

In equal circles, equal angles stand upon equal circumferences, whether they be at the centres or circumferences.

Let ABC, DEF be equal circles, and BGC, EHF equal angles at their centres, and BAC, EDF, at their circumferences, which stand on the circumferences BKC, ELF: the circumference BKC shall be equal to the circumference ELF.

Join BC, EF: Then, because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: Therefore the two sides BG, GC are equal to the two EH, HF, and the angle at G is equal to the angle at H-there

H

B

CE

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fore the base BC is equal to the base EF: And, because the angle at A is equal to the angle at D, the segment BAC is similar to the segment EDF, and they are upon equal straight lines BC, EF: But similar segments of circles upon equal straight lines are equal to one another (3. 24); therefore the segment BAC is equal to the segment EDF: But the whole circle ABC is equal to the whole circle DEF; therefore the remaining segment BKC is equal to the remaining segment ELF, and the circumference BKC to the circumference ELF, Wherefore, In equal circles &c. Q. E. D.

PROP. XXVII. THEOR.

In equal circles, the angles which stand upon equal circumferences are equal to one another, whether they be at the centres or circumferences.

Let ABC, DEF be equal circles, and let the angles

BGC, EHF at the centres, and BAC, EDF at the circumferences, stand upon equal circumferences BC, EF: the angle BGC shall be 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 (3. 20) that the angle BAC is also equal to the angle EDF: But, if not, one of them must be the greater: Let BGC be the greater, and at the point G, in the straight line BG, make the angle BGK equal to the angle EHF: Then, because equal angles at the centre stand upon equal cir- B cumferences (3.26), there

fore 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 absurd: 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 also the angle at A is equal to the angle at D.

Wherefore, In equal circles &c.

Q.E.D.

PROP. XXVIII. THEOR.

In 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 circumference BAC is equal to the greater EDF, and the less BGC to the less EHF.

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K

C E

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G

H

Take K, L, the centres of the circles, and join BK, KC, EL, LF: Then, because the circles are equal, the straight lines from their centres are equal; therefore BK, KC are equal to EL, LF, each B to each, and the base BC is equal to the base EF-therefore the angle BKC is equal to the angle ELF: But equal angles, at the centres, stand upon equal circumferences (3. 26); therefore the circumference BGC is equal to the circumference EHF: But the whole circle ABC is equal to the whole circle DEF; therefore the remaining circumference BAC is equal to the remaining circumference EDF.

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

PROP. XXIX. THEOR.

In equal circles, equal circumferences are subtended by equal straight lines.

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

A

Take K, L, the centres of the circles, and join BK, KC, EL, LF: Then, because the circumference BGC is equal to the circumference EHF, the angle BKC is equal to the angle ELF (3.27): And because the circles ABC, DEF are equal, the straight lines from their centres are

B

K

C E

F

equal; therefore BK, KC are equal to LF, EL, each to each, and they contain equal angles-therefore the base BC is equal to the base EF.

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

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