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Book III. 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 the angles BGC, EHF at the centres, and BAC, EDF at the circumferences of the equal circles ABC, DEF stand upon the 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 manifesta that the angle BAC is also equal to EDF: but, if not, one of

a 20. 3.

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them is the greater: let BGC be the greater, and at the point b 23. 1. G, in the straight line BG, make the angle BGK equal to the c 26. 3. angle EHF; but equal angles stand upon equal circumferences,

when they are at the centre ; 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 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 a K, L the centres of the circles, and join BK, KC, a 1. 3. EL, LF: and because the circles are equal, the straight lines

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from theit centres 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 b 8. 1. stand upon equal circumferences, when they are at the cen- c 26,3. tres; therefore 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 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 lin BC is equal to the straight line EF.

Book III. Take a K, L, the centres of the circles, and join BK, KC,

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

A

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b 27. 3. circumference EHF, the angle BKC is equal o to the angle

ELF: and because the circles ABC, DEF are equal, the straight lines from their centres are equal: therefore BK, KC are equal

to EL, LF, and they contain equal angles: therefore the base c4. 1. BC is equal to the base EF. Therefore, in equal circles, &c.

Q. E. D.

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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. a 10.1.

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 bisected in the point D.

Because AC is equal to CB, and CD common to the triangles
ACD, BCD, the two sides AC, CD

D
are equal to the two BC, CD; and
the angle ACD is equal to the angle
BCD, because each of them is a rigbt

angle; therefore the base AD is equal 64.1. b to the base BD: but equal straight

A с B c 28. 3. lines cut off equal c circumferences, the greater equal to the

greater, and the less to the less, and AD, DB are each of them d Cor. 1. less than a semicircle ; because DC passes through the centred:

wherefore the circumference AD is equal to the circumference

B: therefore the given circumference is bisected in D. Which was to be done.

Book III.

PROP. XXXI. 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.

Let ABCD be a circle, of which the diameter is BC, and centre 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 EA, the angle EAB is equal a to EBA; also, because AE a 5. 1. is equal to EC, the angle EAC is equal to ECA; wherefore the

F whole angle BAC is equal to the two angles ABC, ACB; but FAC, the exterior angle of the triangle

U ABC, is equal b to the two angles

b 32. 1, ABC, ACB; therefore the angle BAC is equal to the angle FAC,

B and each of them is therefore a

С right e angle : wherefore the angle

E

def. 1. BAC in a semicircle is a right angle.

And because the two angles ABC, BAC of the triangle ABC are together lessd than two d 17. 1. right angles, and that BAC is a right 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 é to two right angles; there- e 22. 3. fore the angles ABC, ADC are equal 10 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 circunference 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

c 10.

Book III. 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.'

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 the adjacent angies are equal, they are right angles.

PROP. XXXII. THEOR.

IF 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.

D

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 FBD is equal to the angle which is in the segment

DAB, and the angle DBE to the angle in the segment BCD. a 11. 1. 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; and because the straight line EF touches the circle ABCD in the point B, and BA is drawn at

А right angles to the touching line

from the point of contact B, the b 19. 3. centre of the circle is b in BA ;

therefore the angle ADB in a c 31. 3. semicircle is a right angle, and

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

angle: but ABF is likewise a
right angle; therefore the angle
ABF is equal to the angles BAD,

E
B

F
ABD: take from these equals the
common angle ABD; therefore the remaining angle DBF is equal
to the angle BAD, which is in the alternate segment of the circle ;

and because ABCD is a quadrilateral figure in a circle, the oppo22. 3. site angles BAD, BCD are equale to two right angles; therefore

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