Book III.

PROP. XXVIII. THEOR..

N equal circles, equal straight lines cut off equal circumferences, the greater equal to the greater, and the less to the lefs.

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 lefs BGC, EHF: the greater BAC is equal to the greater EDF, and the less BGC to the lefs EHF. Take a K, L the centres of the circles, and join BK, KC, EL, a 1. 3. LF: And because the circles are equal, the straight lines from

their centres are equal; therefore BK, KC are equal to EL, LF; and the base BC is equal to the base EF; therefore the

b

angle BKC is equal to the angle ELF: But equal angles stand b 8. 1. upon equal circumferences, when they are at the centres ; c 26. 3. 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.

IN equal circles equal circumferences are fubtended by equal straight lines.

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

Take

Book III.

a1.3.

Take a K, L the centres of the circles, and join BK, KC, EL, LF: And because the circumference BGC is equal to the

b27.30

€ 4. I.

a to. I.

b 4. I. c 28.3.

circumference EHF, the angle BKC is equal b to the angle ELF: And because the circles ABC, DEF are equal, the ftraight lines from their centres are equal: Therefore BK, KC are equal to EL, LF, and they contain equal angles: Therefore the bafe BC is equal to the bafe EF. Therefore, in equal circles, &c. Q. E. D.

T

PROP. XXX. PRO B.

'O bisect a given circumference, that is, to divide it into two equal parts.

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

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 bifected in the point D.

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

с

A

D

lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD, DB are each of them d Cor. 1. 3. lefs than a femicircle; becaufe DC paffes through the centre d:

Wherefore the circumference AD is equal to the circumference
DB: Therefore the given circumference is bifected in D.
Which was to be done.

PROP.

Book III.

IN

PROP. XXXI. THEOR.

N a circle, the angle in a femicircle is a right angle; but the angle in a fegment greater than a femicircle is less than a right angle; and the angle in a fegment less than a femicircle 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 femicircle BAC is a right angle; and the angle in the fegment ABC, which is greater than a femicircle, is less than a right angle; and the angle in the fegment ADC, which is less than a femicircle, 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; alfo, because AE a 5. 1. is equal to EC, the angle EAC is equal to ECA; wherefore the whole angle BAC is equal to the two angles ABC, ACB: But FAC, the exterior angle of the triangle ABC, is equal b to the two angles ABC, ACB; therefore the angle BAC is equal to the angle FAC, B

and each of them is therefore a

right angle: Wherefore the angle BAC in a femicircle is a right an

E

D

b 32. I.

C

E

c 10.Def. I.

gle.

And because the two angles ABC, BAC of the triangle ABC are together lefs d than two right angles, and that BAC d 17. 1. is a right angle, ABC must be lefs than a right angle; and therefore the angle in a fegment ABC greater than a femicircle, is less than a right angle.

And because ABCD is a quadrilateral figure in a circle, any two of its oppofite angles are equal e to two right angles; there- e 22. 3. fore the angles ABC, ADC are equal to two right angles; and ABC is less than a right angle; wherefore the other ADC is greater than a right angle.

Befides, it is manifeft, that the circumference of the greater fegment ABC falls without the right angle CAB, but the circumference of the lefs fegment ADC falls within the right angle CAF. And this is all that is meant, when in the • Greek

Book III. Greek text, and the tranflations from it, the angle of the greater fegment is faid to be greater, and the angle of the lefs 'fegment is faid to be lefs, than a right angle.'

a II. I.

b 19.3.

3.

d 32. I.

COR. From this it is manifeft, 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 fame two; and when the adjacent angles are equal, they are right angles.

To

PROP. XXXII. THEOR.

F 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, fhall be equal to the angles which are in the alternate fegments of the circle.

Let the ftraight 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 fhall be equal to the angles in the alternate fegments 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

D

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 right angles to the touching line from the point of contact B, the center of the circle is b in BA; therefore the angle ADB in a femicircle is a right angle, and confequently the other two angles 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, ABD: Take from D

€ 22. 3.

B

F

thefe equals the common angle

ABD; therefore the remaining angle DBF is equal to the angle BAD, which is in the alternate fegment of the circle; and because ABCD is a quadrilateral figure in a circle, the oppofite angles BAD, BCD are equale to two right angles; therefore

the

95

the angles DBF, DBE, being likewise equal f to two right an- Book II. gles, are equal to the angles BAD, BCD; and DBF has been proved equal to BAD: Therefore the remaining angle DBE f 13. 1. is equal to the angle BCD in the alternate fegment of the circle. Wherefore, if a straight line, &c. Q.E. D.

PROP, XXXIII. PRO B.

UPON a given straight line to describe a fegment of see N.

a circle, containing an angle equal to a given rec

tilineal angle.

Let AB be the given straight line, and the angle at C the given rectilineal angle; it is required to defcribe upon the given ftraight line AB a fegment of a circle, containing an angle. equal to the angle C.

First, let the angle at C be a right angle, and bifecta AB in F, and from the centre F, at the distance FB, describe the femicircle AHB; therefore the angle AHB in a femicircle is bequal to the right angle at C.

A

F

H

a 10. I

B

b 31.3.

But, if the angle C be not a right angle, at the point A, in the ftraight line AB, make the angle BAD equal to the angle c 23. I.

C, and from the point A draw d AE at right angles to AD; bifecta AB in F, and

from F draw d FG at right angles to AB, and join GB: And because AF is equal to FB, and FG common to the triangles AFG, BFG, the two fides AF, FG are equal to the two BF, FG; and the angle AFG is equal to the angle BFG; therefore the

bafe AG is equal e to the bafe GB; and the circle defcribed e 4. I. from the centre G, at the distance GA, fhall pass through the point B; let this be the circle AHB: And becaufe from the point A the extremity of the diameter AE, AD`is drawn at

right