PROP. XXVII.-THEOREM.

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.

B

K

If the angle BGC be equal to the angle EHF, it is manifest (III. 20.) that the angle BAC is also equal to EDF. But if not, one of them is the greater; let BGC be the greater, and at the point G, in the straight line BG, make (I. 23.) the angle BGK equal to the angle EHF; but equal angles stand upon equal circumferences (III. 26.) when they are at the centre; therefore

EF:

1. The circumference BK is equal to the circumference

but EF is equal to BC; therefore also (Ax. 1.)

2. BK is equal to BC,

the less to the greater, which is impossible: therefore

3. 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 (Ax. 7.)

4. The angle at A is equal to the angle at D.

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

PROP. XXVIII.-THEOREM.

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 (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF. And because the circles are equal, the straight lines from their centres (III. Def. 1.) are equal; therefore

1.

BK, KC, are equal to EL, LF, each to each;

and the base BC is equal (Hyp.) to the base EF; therefore (I. 8.)

2.

The angle BKC is equal to the angle ELF:

but equal angles stand (III, 26.) upon equal circumferences, when they are at the centres; therefore

3. The circumference BGC is equal to the circumference

EHF:

but the whole circle ABC is equal (Hyp.) to the whole EDF; the remaining part therefore of the circumference, viz.

4. BAC is equal to the remaining part EDF.

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

PROP. XXIX.- -THEOREM.

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 is equal to the straight line EF.

Take (III. 1.) K, L, the centres of the circles, and join BK, KC, EL, LF, and because the circumference BGC is equal to the circumference EHF, (III. 27.)

1. The angle BKC is equal to the angle ELF:

and because the circles ABC, DEF, are equal, the straight lines from their centres (III. Def. 1.) are equal; therefore

2. BK, KC, are equal to EL, LF, each to each;

and they contain equal angles; therefore (I. 4.)

3. The base BC is equal to the base EF.

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

PROP. XXX. -PROBLEM.

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 (I. 10.) in C; from the point C draw (I. 11.) CD at right angles to AB; the circumference ADB is bisected in the point D.

D

Join AD, DB. And because AC is equal to CB, and CD common to the triangles ACD, BCD,

and

1.

The two sides AC, CD, are equal to the two BC, CD,

each to each;

2. The angle ACD is equal to the angle BCD, because each of them is a right angle; therefore (I. 4.)

3. The base AD is equal to the base BD.

But equal straight lines cut off (III. 28.) equal circumferences, the greater equal to the greater, and the less to the less; and

4. AD, DB, are each of them less than a semicircle, because DC passes through the centre (III. 1. Cor.) Wherefore 5. The circumference AD is equal to the circumference DB. Therefore the given circumference is bisected in D. Which was to be done.

PROP. 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 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.

A

D

B

E

Join AE, and produce BA to F; and because BE is equal (I. Def. 15.)

to EA, (I. 5.)

3. The whole angle BAC is equal to the two angles ABC, ACB:

but FAC, the exterior angle of the triangle ABC, is equal (I. 32.) to the two angles ABC, ACB; therefore (Ax. 1.)

4. The angle BAC is equal to the angle FAC;

and each of them is therefore (I. Def. 10.) a right angle: wherefore 5. The angle BAC in a semicircle is a right angle.

And because the two angles ABC, BAC, of the triangle ABC, are together less (I. 17.) than two right angles, and that BAC is a right angle, ABC must be less than a right angle;

1.

and therefore

2.

The angle ABC 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 (III. 22.) to two right angles; therefore

1. The angles ABC, ADC, are equal to two right angles; and ABC is less than a right angle; wherefore the other

2. 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.”

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 (I. 32.) to the same two; and when the adjacent angles are equal, they are (I. Def. 10.) right angles.

PROP. XXXII.- -THEOREM.

If a straight line touch 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.

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.

From the point B draw (I. 11.) BA at right angles to EF, and take any point 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 right angles to the touching line from the point of contact B, (III. 19.)

1. The centre of the circle ABCD is in BA;

therefore (III. 31.)

2. The angle ADB in a semicircle is a right angle, and consequently (I. 32.)

3. The other two angles BAD, ABD, are equal to a right angle:

but ABF is likewise a right angle; therefore (Ax. 1.)

4. The angle ABF is equal to the angles BAD, ABD:

take from these equals the common angle ABD; therefore (Ax. 3.) 5. 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, (III. 22.)

6. The opposite angles BAD, BCD, are equal to two right angles:

but the angles DBF, DBE, are likewise equal (I. 13.) to two right angles; therefore

7. The angles DBF, DBE, are equal to the angles BAD, BCD;

and DBF has been proved equal to BAD; therefore

8. The remaining angle DBE is equal to the angle BCD in the alternate segment of the circle. Wherefore, if a straight line, &c.

Q.E.D.

PROP. XXXIII.-PROBLEM.

Upon a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle.