*

EF, together with the Square of G E, is equal to the Squares of H E and G E. And if the Square of EG, which is common, be taken from both, the remaining Rectangle, contained under B E and E F, is equal to the Square of EH. But the Rectangle under Band EF is the Parallelogram B D, becaufe E F is equal to E D. Therefore the Parallelogram B D is equal to the Square of EH; but the Parallelogram B D is equal to the Right-lined Figure A. Wherefore the Right-lin'd Figure A is equal to the Square of EH. And fo, there is a Square made equal to the given Right-lined Figure A, viz. the Square of E H; which was to be done.

The END of the SECOND BOOK,

EUCLID's

ELEMENTS.

BOOK III.

DEFINITION S.

1. EQUAL Circles are fuck whofe Diameters are equal; or from whofe Centres the Right Lines that are drawn are equal.

II. A Right Line is jaid to touch a Circle, when meeting the fame, and being produced, it does

not cut it.

III. Circles are faid to touch each other, which meeting do not cut each other.

IV. Right Lines in a Circle are faid to be equally diftant from the Centre, when Perpendiculars drawn from the Centre to them are equal. V. And that Line is faid to be farther from the Centre, 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 Circumference of a Circle.

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

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, fo 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 Circumference.

X. A Sector of a Circle is that Figure, which is comprebended between two Right Lines, drawn from the Centre, 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.

L

PROPOSITION I.

PROBLEM.

To find the Centre of a Circle given.

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

Let the Right Line A B be any how drawn

in it, which * bisect in the Point D; and let DC be † * 10. 1. drawn from the Point D, at Right Angles to A B, † 11. 1. which let be produced to E.

Then, if EC be* bifected in F, I fay, the Point F

is the Centre of the Circle ABC.

For, if it be not, let G be the Centre, and let GA, GD,G B, be drawn. Now, becaufe D A is equal to D B, and DG is common, the two Sides A D, DG, are equal to the two Sides G D, D B, each to each; alfo the Bafe G A is equal to the Bale G B, for they †Def. 15.1 are drawn from the Centre G. Therefore the Angle ADG is equal to the Angle G D B. Right Line ftanding upon a Right Line adjacent Angles equal to one another, equal Angles will be a Right Angle. t F

But when a * 8. 1.
makes the
each of the
Wherefore + Def. 10.1

the

the Angle GDB is a Right Angle. But F D B is also
a Right Angle. Therefore, the Angle FD B is equal
to the Angle G D B, a greater to a lefs, which is ab-
furd. Wherefore G is not the Centre of the Circle
ABC. After the fame manner we prove, that no
other Point, unlefs F, is the Centre. Therefore, F is
the Centre of the Circle A B C; which was to be found.

Coroll. If in a Circle any Right Line cuts another
Right Line into two equal Parts and at Right An-
gles, the Centre of the Circle will be in that cutting
Line.

PROPOSITION II.

THEORE M.

If any two Points be affumed in the Circumference of a Circle, the Right Line joining those two Points fball fall within the Circle.

LET ABC be a Circle; in the Circumference of which let any two Points A, B, be affumed. I fay, a Right Line drawn from the Point A, to the Point B, falls within the Circle.

Find D* the Centre of the given Circle, and let any Point E be taken in the Right Line A B, and let D'A, DE, D B, be joined.

Then, becaufe DA is equal to DB, the Angle
DAB will be + equal to the Angle DB A; and fince
the Side A E of the Triangle D A E is produced, the
Angle DEB will be greater than the Angle DA E.
But the Angle D A E is equal to the Angle DBE;
therefore the Angle DE B is greater than the Angle
DBE. But the greater Angle fubtends the greater
But
Side. Wherefore D B is greater than D E.
DB only comes to the Circumference of the Circle ;
therefore DE does not reach fo far. And fo the
Point E falls within the Circle. Therefore, if two
Points are affumed in the Circumference of a Circle, the
Right Line joining thofe two Points shall fall within the
Circle; which was to be demonstrated.

Coroll. Hence if a Right Line touches a Circle, it
will touch it in one Point only.

PRO