mon, be taken from both, the remaining Rectangle contained under BE and EF, is equal to the Square of EH. But the Rectangle under BE and EF is the Parallelogram BD, because EF is equal to ED. Therefore the Parallelogram BD is equal to the Square of EH; but the Parallelogram BD is equal to the Right-lined Figure A. Wherefore the Rightlined Figure A is equal to the Square of EH. And fo there is a Square made equal to the given Rightlined Figure A, viz. the Square of EH; which was to be done.

The END of the SECOND Book.

EUCLID's

EUCLID's

LEMENTS.

BOOK III.

DEFINITION S.

E

QUAL Circles are fuch whofe Diameters are equal; or from whofe Centers the Right Lines that are drawn are equal. II. A Right Line is faid to touch a Circle when touching the fame, and being produced, does

not cut it.

II. Circles are faid to touch each other, which
Touching do not cut one another.

V. Right Lines in a Circle are faid to be equally
diftant from the Center, when Perpendiculars
drawn from the Center to them be equal.
V. And that Line is faid to be farther from the
Center, on which the greater Perpendicular falls.
VI. A Segment of a Circle is a Figure contained
under a Right Line, and a Part of the Cir-
cumference of a Circle.

VII. An Angle of a Segment is that which is con-
tained by a Right Line, and the Circumference
of a Circle.

VIII. An

63

# 10. I.

† 11. 1.

VIII. An Angle is faid to be in a Segment, when fome Point is taken in the Circumference thereof, and from it Right Lines are drawn to the Ends of that Right Line, which is the Bafe of the Segment; then the Angle contained under the Lines drawn, is faid to be an Angle in a Segment. IX. But when the Right Lines containing the Angle do receive any Circumference of the Circle, then the Angle is faid to stand upon that Cir cumference.

X. A Sector of a Circle, is that Figure comprehended between the Right Lines drawn from the Center, and the Circumference contained between them. XI. Similar Segments of Circles are thofe which include equal Angles, or whereof the Angles in them are equal.

PROPOSITION I.

PROBLEM.

To find the Center of a Circle given.

L

ET ABC be the Circle given. It is required to find the Center thereof.

D; and let DC be

Let the Right Line AB be any how drawn in it, which * bifect in the Point drawn from the Point D at which let be produced to E. Then if EC be bifected in F, I fay, the Point F is the Center of the Circle ABC.

Right Angles to AB,

*

For if it be not, let G be the Center, and let GA, GD, GB, be drawn. Now becaufe DA is equal to DB, and DG is common, the two Sides AD, DG, are equal to the two Sides GD, DB, each to each; Def. 15. 1. and the Bafe GA is equal to the Bafe GB; for they are drawn from the Center G. Therefore the Angle ADG is equal to the Angle GD B. But when a Right Line ftanding upon a Right Line, makes the adjacent Angles equal to one another, each of the +Def. 10. 1. equal Angles will be a Right Angle. Wherefore the Angle GDB is a Right Angle. But FDB is alfo a

# 8. I.

*

2

Right