3. Circles DAC, ABE and alfo (FBG, ABE) are faid to touch each other, when they meet each other, fo as not to cut one another.

The Circle BFG cuts the Circle FGH.

A

B

B

5. ASegment ABC of a Circle is a Figure contained under a right Line (AC) and a part (ABC) of the Circumference of a Circle.

6. An Angle CAB, of a Segment is that Angle which is contained under a right Line CA, and an Arch AB of a Circle.

7. An Ang. (ABC) is faid to be in a Segment (ABC) when fome Point B is taken, in the Circumference thereof, and from it right Lines (AB, CB) are drawn to the Extremes of the right Line AC, which is the Bafe of the Segment: then the Ang. ABC contained under the Lines fuppofed to be drawn (AB, CB) is faid to be an Ang. in a Seg

ment.

8. But when the right Lines (AB, BC) comprehending the Ang. ABC, do receive any Circumference of the Circle (ADC): then the Ang. ABC is faid to Stand upon that Circumference.

A

B

9. A Sector of a Circle (ACB). is a Figure comprehended under the right Lines AC, BC drawn from the Center C, and the Circumference AB between them.

10. Similar Segments ABC, DEF of a Circle, are those which include equal Angles (ABC, DEF)

or

or whereof the Angles (ABC, DEF) are in them

equal.

F

E

B

D

To find the Centre F of a given Circle ABC.

Draw a right Line (AC) any how in the Circle, which bifect in E: through E draw a Perpendicular DB, and biC fect the fame in F. Then the Point F fhall be the Cen❤ ter.

If you deny it, let G, fome Point without the Line DB,be theCenter (fince that cannot be divided equally in any point but F)and draw the Lines GA, GC, GE. Now if G be the Centre, then · GAGC and AEEC by Conft. and the a 15 def.1. Side GE common. Therefore the Ang. GEA = GEC, and confequently a right one. Whence the Ang. GEC FEC. Which is abfurd.

COROL

с

b b

Hence if a right Line (BD) bifes any right Line AC at right Angles in a Circle, the Center fhall be in the Line BD that bifects the o

ther.

The

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10 def. 1. 12 ax.

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b

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19. I.

A

E

The Centre of a Circle is eafily found, by applying the Ang. of a Square to the Circumference thereof.

For if the right Line DE that joins the Points D,E, in which the Sides of the Square QD, QE

cut the Circumference, be bifected in A, the Point A fhall be the Center. The Demonftration of this depends upon Prop. 31. of this Book.

PRO P. II.

D

B

If any two Points (A, B) be taken in the Circumference of a Circle (CAB) the right Line AB joining those two Points, ball fall within the Circle.

In the right Line AB take any Point D, and from the Center C draw CA, CD, CB. Because

a

a 15. def. CA CB, therefore the Ang. AB. But the Ang. CDBA, therefore Ang. CDBB, whence CB CD. But CB only extends from the Center to the Circumference, therefore CD does not come fo far: Whence the Point Dis within the Circle. The fame may be proved of. any other Point in the Line AB. And therefore the whole Line AB falls within the Circle, Q. E. D.

CORO L.

Hence if a right Line touches a Circle, fo as not to cut it, it touches the fame but in one Point.

PROP.

PROP. III.

F

C

D

In a Circle (EABC) if a right Line BD drawn through the Center, bifects any other Line AC not drawn through the Center, it fhall also cut it at right Angles and if it cuts it at right Angles, it shall also bifect the fame.

a

=

Draw the Lines EA, EC from the Center E. Hyp. 1. Becaufe AF FC, and EA b EC, and the Side EF is common; the Angle EFA EFC, and confequently right Angles. Q. E. D.

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Hyp 2. Because the Ang. EFA = EFC, and the Ang, EAF = ECF, and the Side EF common; therefore AF8FC, whence AC is cut into two equal parts. Q. E. D.

COROL

Hence if a Line drawn from the Vertical Angle in any Equilateral or Ifofceles Triangle bifects the Bafe; that Line is perpendicular to it. And on the contrary, a Perpendicular drawn from the vertical Angle, bifects the Bafe.