Book III.

PROP. XXXI.

THEOR.

IN a circle, the angle in a femicircle is a right angle; but the angle in a fegment greater than a femicircle is lefs than a right angle; and the angle in a segment less 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 lefs than a right angle; and the angle in the fegment ADC, which is less than a femicircle, is greater than a right angle.

E

A

D

b 32. I.

C

E

c Io. def. *.

Join AE, and produce BA to F; and becaufe BE is equal to EA, the angle EAB is equal to EBA; alfo, because AE s. 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 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 angle.

c

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 there, fore the angle in a fegment ABC greater than a femicircle, is lefs than a right angle,

And because ABCD is a quadrilateral figure in a circle, any two of its oppofite angles are equal to two right angles; there- 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.'

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.

2 11. I

b 19.3

C 31. 3.

d 32. I.

e 22.3.

IF a 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 fegment BCD.

A

D

C

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 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 is 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 likewife a right angle; therefore the angle ABF is equal to the angles BAD, ABD: Take from 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 becaufe ABCD is a quadrilateral figure in a circle, the opposite angles BAD, BCD are equal to

E

B

F

two

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

UPO

PROP. XXXIII. PR O B.

PON a given straight line to describe a fegment of See N. 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 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 center F, at the diftance FB, defcribe the femicircle AHB; therefore the angle AHB in a femicircle is equal to the right angle at C.

C c.

A

F

H

a 10. 1

6 31.34

B

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 e 23. 1.

C

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

سم

H

c

d II. I

E

bafe AG is equal to the bafe GB; and the circle defcribed from the center G, at the distance GA, fhall pafs through the point B.; let this be the circle AHB: And because from the A the extremity of the diameter AE, AD is drawn at

point

€ 4. I.

Book III. right angles to AE, therefore AD f touches the circle; and be

caufe AB drawn from the point

f Cor. 16.3. of contact A cuts the circle,

H

the given ftraight line AB the

a 17.3.

b 23. I.

C 32. 3.

fegment AHB of a circle is defcribed which contains an angle equal to the given angle at C. Which was to be done.

T

O cut off a fegment from a given circle which fhall contain an angle equal to a given rectilineal angle.

Let ABC be the given circle, and D the given rectilineal angle; it is required to cut off a fegment from the circle ABC that fhall contain an angle equal to the angle D.

a

A

Draw the straight line EF touching the circle ABC in the point B, and at the point B, in the ftraight line BF, make the angle FBC equal to the angle D: Therefore, because the ftraight line EF touches the circle ABC, and BC is D drawn from the point of contact B, the angle FBC is equal to the angle in the alternate fegment BAC

c

E

B F

of the circle: But the angle FBC is equal to the angle D; therefore the angle in the fegment BAC is equal to the angle D: Wherefore the fegment BAC is cut off from the given cir cle ABC containing an angle equal to the given angle D; Which was to be done.

PR P.

Book III.

PROP. XXXV. THEOR.

IF two ftraight lines within a circle cut one another, the See N. rectangle contained by the fegments of one of them

is equal to the rectangle contained by the fegments of

the other.

Let the two ftraight lines AC, BD, within the circle ABCD, cut one another in the point E; the rectangle contained by AE, EC is equal to the rectangle contained by

BE, ED.

A

If AC, BD pass each of them through the center, fo that E is the center; it is evident, that AE, EC, BE, ED, being all equal, the rectangle AE, EC is likewife B equal to the rectangle BE, ED.

Ꭰ

D

E

But let one of them BD pafs through the center, and cut the other AC, which does not pafs through the center, at right angles, in the point E: Then, if BD be bifected in F, F is the center of the circle ABCD; and join AF: And because BD, which paffes through the center, cuts the ftraight line AC, which does not pafs through the center, at right angles in E, AE, EC are equal a to one another: And because the ftraight line BD is cut into two equal parts in the point F, and into two unequal in the point E, the rectangle BE, ED together with the fquare of EF, is equal to the fquare of FB: that is, A to the fquare of FA; but the squares

of AE, EF are equal to the fquare

of FA; therefore the rectangle BE, ED, together with the fquare of EF,

F

a 3. 3.

b 5. 20

E

B

is equal to the fquares of AE, EF: Take away the common quare of EF, and the remaining rectangle BE, ED is equal to the remaining fquare of AE; that is, to the rectangle AE, EC.

C 47. I.

Next, Let BD which paffes through the center, cut the other AC, which does not pafs through the center, in E, but not at right angles: Then, as before, if BD be bifected in F, F is the center of the circle. Join AF, and from F draw FG perpen-d 12. 1,

AF,

G

dicular