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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 so 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.

EUCLI D's

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63

E U CL I D's

; LEMENTS

BOOK III.

DEFINITIONS.

E

QUAL Circles are such whose Diameters

are equal; or from whose Centers the

Right Lines that are drawn are equal.

II. A Right Line is said to touch a Circle when touching the same, and being produced, does not cut it. II. Circles are said to touch each other, which

Touching do not cut one another. V. Right Lines in a Circle are said to be equally distant from the Center, when Perpendiculars

drawn from the Center to them be equal. V. And that Line is said to be farther from the

Center, on which the greater Perpendicular falls. VI. A Segment of a Circle is a Figure contained

under à Right Line, and a part of the Circumference of a Circlc. VII. An Angle of a Segment is that which is con

tained by a Right Line, and the Circumference of a Circle.

VIII. An

Tī drawn in it, which * bisect in the Point

VIII. An Angle is said to be in a Segment, when

some Point is taken in the Circumference thereof, and from it Right Lines are drawn to the Ends of that Right Line, which is the Base of the Segment; then the Angle contained under the Lines

drawn, is said to be an Angle in a Segment. IX. But when the Right Lines containing the An

gle do receive any Circumference of the Circle, iben tbe Angle is said to stand upon that Cir

cumference. X. A Seator of a Circle, is that Figure comprebended

between the Right Lines drawn from the Center,

and the Circumference contained between them. XI. Similar Segments of Circles are those which

include equal Angles, or whereof the Angles in
them are equal.
PROPOSITION I.

PROBLEM.
To find the Center of a Circle given.
ET ABC be the Circle given. It is re-
quired to find the Center thereof.

Let the Right Line AB be any how + 11. I. D; and let DC be drawn from the Point D at

Right Angles to AB, which let be produced to E.

Then if E C be * bisected in F, I say, the Point F is the Center of the Circle ABC.

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 ; 1 Def. 15. 1. and the Base G A is I equal to the Bafe GB; for they

are drawn from the Center G. Therefore the Angle 8. 1. ADG is * equal to the Angle GD B. But when a

Right Line standing 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 also a

Right

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