Book III. 22. 3.) equal to two right angles; wherefore the angles ABC, ADC are equal to two right angles; but ABC is lefs than a right angle; therefore the remainder ADC is greater than a right angle; and it is in a fegment lefs than a femicircle.

I fay also that the angle of a greater fegment, the angle contained by the circumference ABC and the ftraight line AC, is greater than a right angle; but the angle of the segment lefs than a femicircle, the angle contained by the circumference ADC and the ftraight line AC is less than a right angle: and it is manifeft from hence. For because the angle contained by the straight lines BA, AC is a right angle; therefore the angle contained by the circumference ABC and the straight line AC is greater than a right angle (by com. not. 9.): Again because the angle contained by the straight lines CA, AF is a right angle; therefore the angle contained by the straight line AC and the circumference ADC is less than a right angle.

OTHERWISE. A demonstration that BAC is a right angle. Because the angle AEC (by 32. 1.) is double of the angle BAE; for it is equal to the two inward and oppofite angles; and AEB is alfo double of the angle EAC; wherefore the angles AEB, AEC are double of the angle BAC; but the angles AEB, AEC are (by 13. 1.) equal to two right angles; wherefore the angle BAC is a right angle. Which was to be demonstrated.

Cor From this it is manifeft that if one angle of a triangle be equal to the other two it is a right angle: for this reafon because the adjacent angle is equal to the fame two: but (by def. 10. 1.) when the adjacent angles are equal, they are right angles.

If any straight line touch a circle; and if from the contact any. straight line be drawn to the circle, cutting the circle; the angles which it makes with the touching line will be equal to the angles in the alternate fegments of the circle.

For let any straight line EF touch the circle ABCD in the point B; and from the point B let BD any ftraight line be drawn to the circle ABCD cutting it; I fay that the angles which BD makes

with the touching line EF will be equal to the angles in the alter- Book III. nate fegments of the circle, that is, that the angle FBD is equal

to the angle conftituted in the fegment DAB; and the angle EBD

is equal to the angle in the fegment DCB.

For let BA be drawn from the point B at right angles to EF; and let C any accidental point be taken in the circumference BD; and let AD, DC, CB be joined.

A

D

C

B

F

And because a certain straight line EF touches the circle ABCD in the point B ; and from the contact at B the straight line AB has been drawn at right angles to the touching line ; the center of the circle ABCD is (by 19. 3.) in the straight line AB: therefore the angle ADB, being in a E femicircle, is (by 31. 3.) a right angle; therefore BAD, ABD, the remaining angles, are equal to one right angle but ABF is also a right angle (by conft.); wherefore the angle ABF is equal to the angles BAD, ABD; let the common angle ABD be taken away; therefore the remaining angle DBF is equal to the angle in the alternate fegment of the circle, viz. the angle BAD: And because ABCD is a quadrilateral figure infcribed in a circle, its oppofite angles are (by 22. 3.) equal to two right angles; therefore the angles DBF, DBE are equal to BAD, BCD ; of which BAD has been demonftrated to be equal to DBF; therefore the remaining angle DBE is equal to the angle in the alternate fegment DCB of the circle, viz. the angle DCB.

Wherefore if any straight line touch a circle; and if from the contact any straight line be drawn to the circle, cutting the circle; the angles which it makes with the touching line will be equal to the angles in the alternate fegments of the circle. Which was to be demonftrated.

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

Let the given straight line be AB; and the angle at C, the given rectilineal angle: it is required, upon the given straight line AB,

to

Book III. to defcribe a fegment of a circle containing an angle equal to the angle at C but the angle at C is either an acute angle or a right angle or an obtuse angle.

First let it be an acute angle, as in the first figure; and let the angle BAD be made, with the straight line AB and at the point A in it, equal to the angle at C; wherefore alfo the angle BAD is acute and let AE be drawn, from the point A, at right angles to AD; and let AB be cut in halves at F; and from the point F let FG be drawn at right angles to AB; and let GB be joined.

bed; and let it be ABE; and let BE be joined. Wherefore because from the extremity of the diameter AE; from the point A ; AD is at right angles to AE; therefore AD touches the circle (by cor. to 16. 3.). And fince a certain ftraight line AD touches the circle ABE; and from the contact at A a certain straight line AB hath been drawn to the circle ABE; therefore the angle DAB is equal to the angle AEB in the alternate segment of the circle; but the angle DAB is equal to the angle at C; therefore the angle at C is equal to the angle AEB: wherefore upon the given straight line AB, a fegment of a circle hath been defcribed viz. AEB, containing the angle AEB equal to the given angle at C.

But let the angle at C be a right angle; and again let it be required to defcribe upon AB a segment of a circle containing an angle equal to the right angle at C: Again let the angle BAD be made equal to the right angle at C, as it is in the second figure; and let AB be cut in halves in F; and with the center F, and at

the

the distance of either of the lines AF, FB let the circle AEB be de- Book III. scribed. Therefore the straight line AD touches the circle ABE; because the angle at A is a right angle; and the angle BAD is equal to the angle in the segment AEB; for it is also a right angle (by 313.) being in a femicircle: but the angle BAD is equal to the angle at C: wherefore, again a fegment of a circle viz. AEB has been described upon AB, containing an angle equal to the angle at C.

But let the angle at C be an obtufe angle: and let the angle BAD be made equal to it, with the straight line AB and at the point A, as it is in the third figure; and let AE be drawn at right angles to AD; and again let AB be cut in halves at F; and let FG be drawn at right angles to AB; and let GB be joined.

And again, because AF is equal to FB and FG common; certainly the two 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 to the base GB; wherefore a circle described with the center G; and at the distance AG, will alfo país through B; let it pafs as AEB: And because AD has been drawn at right angles to the diameter AE from its extremity: therefore AD (by cor. 16.3.) touches the circle AEB: and from the contact at A, the straight line AB hath been drawn; wherefore the angle BAD is equal to AHB the angle contained in the alternate fegment of the circle : but the angle BAD is equal to the angle at C: therefore also the angle in the fegment AHB is equal to the angle at C. Wherefore upon the given straight line AB, a fegment of a circle; viz. AHB, has been described, containing an angle equal to the angle at C. Which was to be done.

To cut off a fegment from a given circle, containing an angle equal to a given rectilineal angle.

Let ABC be the given circle; and the given rectilineal angle, the angle at D: it is required to cut off a segment from the circle ABC, containing an angle equal to the angle at D.

VOL. I.

N

Let

Book III.

Let EF be drawn touching the circle ABC in the point B; and let the angle FBC be made with the ftraight line EF, and at the point B in it, equal to the angle at D.

A

Wherefore because a certain straight line
EF touches the circle ABC, and from the
contact at B, BC hath been drawn cutting
it; therefore (by 32. 3.) FBC is equal to
the angle contained in the alternate segment
BAC but the angle FBC is equal to the
angle at D; wherefore also the angle in the E B F
segment BAC is equal to the angle at D.

Wherefore from a given circle ABC a fegment BAC has been cut off, containing an angle equal to the given rectilineal angle at D. Which was to be done.

If in a circle two ftraight lines cut one another, the rectangle contained by the fegments of the one is equal to the rectangle contained by the fegments of the other.

For in the circle ABCD let the two straight lines AC, BD cut one another in the point E; I say that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB.

If AC, BD pafs through the center fo that E be the center of the circle ABCD: it is manifeft, AE, EC, DE, EB being equal, that the rectangle contained by AE, EC is equal to the rectangle contained by DE, EB.

But let AC, DB not pass through the center; and let the center of the circle ABCD be taken (by 1. 3.); and let it be F; and from the point F, let FG, FH be drawn perpendiculars to AC, DB; and let FD, FA, FE be joined.

And becaufe GF a certain straight line through the center cuts a certain ftraight line AC not through the center at right angles; it will alfo cut it in halves (by 3. 3.); wherefore AG is equal to GC: wherefore because the straight line AC has been cut into equal fegments at the point G; and into unequal fegments at the point E; therefore (by 5. 2.) the rectangle contained by AE and

EC