IN

N equal circles, equal ftraight lines cut off equal circumferences, the greater equal to the greater,

and the lefs 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 lefs BGC to the lefs EHF.

Book III.

Take a K, L the centers of the circles, and join BK, KC, EL, a. 1. 3i LF. and because the circles are equal, the straight lines from

their centers are equal, therefore BK, KC, are equal to EL, LF; and the base BC is equal to the bafe EF; therefore the angle BKC is equal to the angle ELF. but equal angles ftand upon b. 8. 1. equal circumferences, when they are at the centers; therefore c. 26. 3: 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.

equal circles equal circumferences are fubtended
by equal straight lines.

Let ABC, DEF be equal circles, and let the circumferences BGC, EHF also be equal; and join BC, EF. the ftraight line BC is equal to the flraight line EF.

Book III.

a. I. 3.

Take a K, L the centers of the circles, and join BK, KC, EL, LF. and because the circumference BGC is equal to the circum

b

b. 27.3. ference EHF, the angle BKC is equal to, the angle ELF. and

1

C. 4. I.

2. 1O. I.

because the circles ABC, DEF are equal, the straight lines from

c

their centers are equal; therefore BK, KC are equal to EL, LF, and they contain equal angles. therefore the base BC is equal © to the base EF. Therefore, in equal circles, &c. Q. E. D.

it.

PROP. XXX. PROB.

10 bifect a given circumference, that is, to divide it into two equal parts.

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

Join AB, and bifecta 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.

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

are equal to the two BC, CD; and
the angle ACD is equal to the angle
BCD, because each of them is a right
angle; therefore the base AD is equal

b. 4. I.

b to the bafe BD. but equal straight

A

C

B

c. 28. 3. lines cut off equal circumferences, the greater equal to the greater, and the lefs to the lefs, and AD, DB are each of them less than a d. Cor. 1. 3. femicircle; because DC paffes through the center . wherefore the circumference AD is equal to the circumference DB. therefore

the given circumference is bifected in D. Which was to be done.

PROP. XXXI.

THEOR.

Book III.

IN

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 lefs than a femicircle is greater than a right angle.

Let ABCD be a circle, of which the diameter is BC, and center E; and draw CA dividing the circle into the fegments 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.

a

A

F

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; alfo, because AE 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, and each of them

B

2. 5. 3.

C

E

b. 32. T.

fore the angle BAC in a femicircle is a right angle.

And because the two angles ABC, BAC of the triangle ABC

c. 10.Def. I.

are together less than two right angles, and that BAC is a right d. 17. I. angle, ABC must be less than a right angle; and therefore the

angle in a fegment ABC greater than a femicircle, is lefs than a right angle.

And becaufe ABCD is a quadrilateral figure in a circle, any

two of its oppofite angles are equal 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 circum

Book III. ference 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 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 less than a right angle.'

2, II. 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.

IF

Fa ftraight 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 ftraight 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 fegment DAB, and the angle DBE to the angle in the segment BCD.

A

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, CB1 and because the ftraight 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 b. 19. 3. is bin BA; therefore the angle ADB in a femicircle is a right angle, and confequently the other two angles

C. 31. 3.

c

ABD; therefore the remaining angle DBF is equal to the angle

BAD, which is in the alternate fegment of the circle; and because Book I'i. ABCD is a quadrilateral figure in a circle, the oppofite angles

e

BAD, BCD are equal to two right angles; therefore the angles e. 22. 3. DBF, DBE, being likewise equal f to two right angles, are equal f. 13. 1. 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 fegment of the circle. Wherefore, if a ftraight line, &c. Q. E. D.

UPON

PROP. XXXIII. PROB.

[PON a given straight line to defcribe a fegment See N. of a circle, containing an angle equal to a given rectilineal 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 straight 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 bifect AB in F, and from the center F, at the distance FB, defcribe the femicircle AHB; therefore the angle AHB in a semicircle is equal to the right angle at C.

H

2. IO. I.

A F Bb. 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, c. 23. I.

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 base AG is equal

e

to the bafe GB; and the circle defcribed from the center G, at

d. II. I.

C. 4.