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* 20. 3. & 7 Ax. 1.

23. 1.

Let ABC, DEF be equal circles, and let the angles BGC, EHF at their centres, and BAC, EDF at their circumferences, stand upon

the 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 *, that the angle BAC is also equal to EDF. But, if not, one of them must be greater than the other : let BGC be the greater, and at the point G, in the straight line BG, make * the angle BGK

G equal to the angle EHF. Then, be- B CE cause the angle BGK is equal to the angle EHF, and that equal angles stand upon equal circumferences *, when they are at the centres, therefore the circumference BK is equal to the circumference EF : but EF is equal + to BC; therefore also BK is equal t 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.

* 26. 3.

+ Hyp. + 1 Ax.

+ 20.3. †7 Ax

PROPOSITION XXVIII.

Theor.-In equal circles, equal straight lines cut off equal

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

• 1. 3.

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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 shall be equal to the greater DF, and the less BGC to the less EHF. Take K, L, the centres of the A

D
circles, and join BK, KC, EL, LF:

K
and because the circles are equal,
the straight lines from their centres B В

CE
are + equal; therefore BK, KC are G

H equal to EL, LF, each to each ; and the base BC is equal t to the base EF; therefore the angle BKC is equal to the angle ELF: but equal angles stand upon equal

* circumferences, when they are at the centres ;

+1 Def. 3.

+ Hyp.

• 8.1. * 26. 3.

therefore the circumference BGC is equal to the circumference EHF: but the whole circle ABC is equal + to the whole EDF; + Hyp. therefore the remaining part of the circumference, viz. BAC, is equal t to the remaining part EDF. Therefore, in equal +3 Ax. circles, &c.

Q. E. D.

PROPOSITION XXIX.

THEOR.-In 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 shall be equal to the straight line EF.
Take * K, L, the centres of the circles, and join BK, KC,

1. 3. EL, LF: and because the circumference BGC is equal to the circumfer

D

. 27. 3. ence EHF, the angle BKC is equal to the angle ELF: and because the

K

L circles ABC, DEF are equal, the

B CE

F straight lines from their centres are t G

H

+1 Def. 3. equal; therefore BK, KC are equal to EL, LF, each to each ; and they contain equal angles; therefore the base BC is equal * to the base EF. Therefore, in equal * 4. 1. circles, &c. Q. E. D.

PROPOSITION XXX.

PROB.—To bisect a given circumference, that is, to divide it

into two equal parts.

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

Join AB, and bisect * it in C; from the point C, draw CD * 10..). at right angles + to AB: the circumference ADB shall be + 11.1. 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 A angle BCD, because each of them is a right angle; therefore the base AD is equal * to the base BD.

But • 4. 1. equal straight lines cut off equal* circumferences, the greater • 28. 3.

equal to the greater, and the less to the less; and AD, DB * Cor. 1. 3. are each of them less than a semicircle, because DC * passes

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

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Theor.-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.

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Let ABCD be a circle, of which the diameter is BC, and the centre E ; and draw CA, dividing the circle into the segments ABC, ADC, and join BA, AD, DC: 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 semicircle, shall be greater than a right angle.

Join AE, and produce BA to F: and be15 Def. 1. cause BE is equal † to EA, the angle EAB is

F * 5. 1.

equal * to EBA ; also, because AE•is equal
to EC, the angle EAC is equal to ECA ;

D + 2 Ax. wherefore the whole angle BAC is equal + to

the two angles ABC, ACB : but FAC, the . 32. 1.

exterior angle of the triangle ABC, is equal

to the two angles ABC, ACB; therefore the +1 Ax. angle BAC is equal + to the angle FAC; and therefore each of * 10 Def. 1. them is a right * angle: wherefore the angle BAC in a semi

circle, is a right angle.

And because the two angles ABC, BAC of the triangle * 17. 1.

ABC, are together less * than two right angles, 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. 23. 3. And because ABCD is a quadrilateral figure in a circle, any

two of its opposite angles are equal to two right angles; 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.

Besides it is manifest, that the circumference of the greater

segment ABC falls without the right angle CAB; but the circumference of the less segment ADC falls within the right angle CAF. “ And this is all that is meant, when in the Greek text, and the translations from it, the angle of the greater segment is said to be greater, and the angle of the less segment is said to be less, than a right angle.” Q. E. D.

Cor. From this it is manifest, 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 + 32. 1. the adjacent angles are equal, they are + right angles.

| 10 Def. I.

PROPOSITION XXXII.

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THEOR.-If a straight line touch a circle, and from the point

of contact a straight line be drawn cutting 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, 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 DBF shall be equal to the angle which is in the segment DAB, and the angle DBE shall be equal to the angle in the segment DCB.

From the point B, draw * BA at right angles to EF, and . 11. 1. take any point c, in the circumference DB, 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 touching line from the point of

19. 3. contact B, the centre of the circle is* in BA; therefore the angle ADB in a semicircle is a right * angle ; and consequently the other two angles BAD, ABD, are equal * to a right

32. 1. E

F angle: but ABF is likewise a + right angle;

+ Constr. therefore the angle ABF is equal + to the angles BAD, ABD: + 1 Ax. take from these equals the common angle ABD ; therefore the remaining angle DBF is equal † to the angle BAD, which is in † 3 Ax. 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: but the angles DBF, DBE are

* 22. 3.

• 31. 3.

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likewise equal * to two right angles; therefore the angles DBF, DBE are equal + 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 segment of the circle. Wherefore, if a straight line, &c.

+ 2 Ax.

Q. E. D.

PROPOSITION XXXIII.

See N.

Prob.—Upon a given straight line to describe a segment of a

circle, which shall contain an angle equal to a given recli

lineal angle.

B

* 23. 1.

11.1. 10. l. 11. 1.

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, which shall contain an angle equal to the angle C.

First, let the angle at c be a right angle ; . 10. 1.

bisect * AB in F, and from the centre F, at

the distance FB, describe the semicircle AHB: 31.3. therefore the angle AHB in a semicircle, is * A

equal to the right angle at C.

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, and from the point A, draw * AE at right angles to AD: bisect * AB in F, and from F, draw * FG at right angles to AB, and join GB. And because AF

H
is equal to FB, and FG common to the
triangles AFG, BFG, the two sides AF,

FG are equal to the two BF, FG, each to + 10 Def. 1. each; and the angle AFG is equal † to the angle BFG ; therefore the base AG is

D equal * to the base GB; and therefore the circle described from the centre G, at the distance GA, shall pass through the point B : let this be the circle AIIB : the segment All B shall contain an angle equal to the given rectilineal angle C. Because from the point A, the extremity

H of the diameter AE, AD is drawn at right

A
F

B * Cor. 16. 3. angles to AE, therefore AD * touches the

G circle: and because AB, drawn from the

E

D point of contact A, cuts the circle, the

angle DAB is equal to the angle in the alternate segment + Constr. AHB: but the angle DAB is equal † to the angle C; therefore

B

# 4. 1.

. 32. 3.

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