Let ABCD be a circle, of which the diameter is BC, and Book III. centre 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 segment ABC, which is greater than a femicircle, is lefs than a right angle; and the angle in the segment ADC, which is less than a femicircle, is greater than a right angle.

a

/ F

A

D

Join AE, and produce BA to F; and because BE is equal to EA, the angle EAB is equal to EBA; alfo, becaufe AE a 5. x. 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, and each of them is therefore a right angle: Wherefore the angle BAC in a femicircle is a right angle.

C

B

b 32. I.

C

E

C1o.def. I.

And because the two angles ABC, BAC of the triangle ABC are together lefs than two right angles, and that BAC is a d 17. 1. right 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 because ABCD is a quadrilateral figure in a circle, any

e

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 lefs than a right angle; wherefore the other ADC is greater than a right angle.

PROP. XXXII. THEOR.

F a ftraight line touches a circle, and from the point of contact a ftraight 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 D, and from the point D, 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 FDB is equal to the angle which is in the fegment DAB, and the angle BDE to the angle in the fegment BCD.

BOOK III.

b 19.3. c 31. 3.

d 32. 1.

From the point D draw DA at right angles to EF, and take any point C in the arch BD, and join AB, DC, CB; and

d

C

A

B

C

a 11. 1. because the ftraight line EF touches the circle ABCD in the point D, and DA is drawn at right angles to the touching line from the point of contact D, the centre of the circle is in DA; therefore the angle ABD in a femicircle is a right angle, and confequently the other two angles BAD, ADB are equal to a right angle: But ADF is likewise a right angle; therefore the angle ADF is equal to the angles BAD, ADB: Take E D F from thefe equals the common angle ADB; therefore the remaining angle BDF 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 oppofite angles BAD, BCD are e 22.3. equal to two right angles; therefore the angles BDF, BDE, f 13. 1. being likewife equal to two right angles, are equal to the angles BAD, BCD; and BDF has been proved equal to BAD : Therefore the remaining angle BDE is equal to the angle BCD in the alternate fegment of the circle. Wherefore, if a straight line, &c. Q. E. D.

U

e

f

PROP. XXXIII. PROB.

PON a given ftraight line, to describe a fegment of a circle, containing an angle equal to a given rectilineal angle.

Let AB be the given ftraight line, and the angle at C the given angle; it is required to describe 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 a 10. 1. angle, and bifect AB in F, and from the centre F, at the distance FB, defcribe the femicircle AHB; therefore

b 31. 3. the angle AHB in a femicircle is equal to the right angle at C.

C 23. I.

C

C

Λ

F

В

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 d 11. 1. 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

to the two BF, FG; and the angle AFG is equal to the angle Book III. BFG; therefore the base AG is equal to the bafe GB; and

the circle described from the centre G, at the distance GA, fhalle 4.1. pass through the point B; let this be the circle AHB: And be

cause from the point A, the extremity of the diameter AE, AD is drawn at right angles to AE, therefore AD touches the f Cor.16.3. circle; and because AB drawn from the point of contact A cuts the circle, the angle DAB is equal to the angle in the alternate fegment AHB : But the angle DAB is equal to the angle C, g 32. 3• therefore alfo the angle C is equal to the angle in the fegment AHB: Wherefore, upon the given ftraight line AB the fegment AHB of a circle is defcribed, which contains an angle equal to the given angle at C. Which was to be done.

PROP. XXXIV. PROB.

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 angle; it is required to cut off a fegment from the circle ABC that shall contain an angle equal to the angle D.

b

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

c

But

BOOK III. 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 circle ABC, containing an angle equal to the given angle D. Which was to be done.

IF

PROP. XXXV. THEOR.

F two ftraight lines within a circle cut one another, the 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.

If AC, BD pafs each of them through the centre, so that E is the centre; it is evi dent, that AE, EC, BE, ED, being all equal, the rectangle AE, EC is likewife B equal to the rectangle BE, ED.

a

E

D

But let one of them BD pafs through the centre, and cut the other AC, which does not pafs through the centre, at right angles, in the point E: Then, if BD be bifected in F, F is the centre, join AF: And because BD, which paffes through the centre, cuts AC, which does not pass through the centre, at right angles in E, a 3. 3. AE, EC are equal to one another: And because BD is cut into two equal parts in F, and into two unequal in E, the rectangle BE, ED, together with the fquare b 5.2. of EF, is equal to the fquare of FB; that is, to the fquare of FA; but the €47. 1. fquares of AE, EF are equal to the

12. I.

a 3.3.

b 5.2.

b

F

A

E

B

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

Next, let BD, which paffes through the centre, cut the other AC, which does not pafs through the centre, in E, but not at right angles: Then, as before, if BD be bifected in F, F is the centre Join AF, and from F draw FG perpendicular to AC; therefore AG is equal to GC; wherefore the rectangle AE, EC, together with the fquare of EG, is equal to the square of

AG:

AG: To each of thefe equals add the fquare of GF; therefore Book III. the rectangle AE, EC, together with the fquares of EG, GF,

D

C 47. I.

E

B

b 5.2.

is equal to the squares of AG, GF:
But the fquares of EG, GF are equal
to the fquare of EF; and the fquares of
AG, GF are equal to the fquare of
AF: Therefore the rectangle AE,
EC, together with the fquare of EF,
is equal to the fquare of AF; that is, A
to the fquare of FB: But the square
of FB is equal to the rectangle BE,
ED, together with the fquare of EF;
therefore the rectangle AE, EC, together with the square of
EF, is equal to the rectangle BE, ED, together with the fquare
of EF: Take away the common square of EF, and the remain-
ing rectangle AE, EC is therefore equal to the remaining rect-
angle BE, ED.

D

Laftly, let neither of the straight lines AC, BD pass through the centre : Take the centre F, and join FE, and produce it to the circumference in G, H: And because the rectangle AE, EC is equal, as has been fhown, to the rectangle GE, EH; and for the fame reason, the rectangle BE, A ED is equal to the fame rectangle GE, EH; therefore the rectangle AE, EC is equal to the rectangle BE, ED. Wherefore, if two ftraight lines, &c. Q. E. D.

I'

PROP. XXXVI. THEOR.

E

G

F from any point without a circle, two ftraight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, fhall be equal to the fquare of the line which touches it.

Let D be any point without the circle ABC, and DCA, DB two straight lines drawn from it, of which DCA cuts the circle, and DB touches the fame: The rectangle AD, DC is equal to the fquare of DB.

Either DCA paffes through the centre, or it does not; firft, let it pafs through the centre E, and join EB; therefore the angle EBD is a right angle: And because the ftraight line AC a 18.3.

a

is