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fquares of each right line FG, GH, HI, IK, KF; that is, thrice the fum of the fquares of the fides A B, B C, C D, DE, E A of the five-fided right-lined figure will be equal to the fum of the fquares of the diagonals, together with .four times the fum of the fquares of the right lines orderly joining the middle points of the diagonals.

Therefore, &c. Which was to be demonftrated.

SCHOLIU M.

There are other the like properties of right-angled figures of fix or more fides. In all which fome number of times the fum of the fquares of the fides will be equal to fome number of times the fum of the fquares of the diagonals, together with fome number of times the fquares of the proper right lines joining the middle points of the diagonals of the figures.

EUCLID's

EUCLI D's

ELEMENTS.

BOOK III.

I.

"E

DEFINITION S.,

QUAL circles are those whofe diameters, or the right lines that be drawn from their centres, are equal.

2. A right line is faid to touch a circle, when meeting it and being produced does not cut it.

3. Circles are faid to touch one another, which meeting one another do not cut one another.

4. Right lines in a circle are faid to be equally diftant from the centre, when perpendiculars drawn to them from the centre are equal.

5. And that line is fald to be farther from the centre, upon which the greater perpendicular falls.

6. A fegment of a circle is a figure contained under a right line and a part of the circumference of a circle".

7. An angle of a fegment is that which is contained under a right line and the circumference of a circle.

8. An angle in a fegment is the angle contained under two right lines drawn from any affumed point in the cir

a This is rather an axiom than a definition; for what else is it but taking for granted this propofition, viz. thofe circles are equal whofe diameters or femidiameters are equal.

From hence it follows that the fegment of a circle is threefold, viz. a femicircle when the right line paffes through the centre, a greater fegment, and a leffer.

This definition feems to be of no great ufe, at least in thefe elements.

cumference

cumference of the circle, to the ends of the line, being the base of that segment.

9. But when the right lines containing fuch an angle do cut off part of the circumference of the circle, the angle is faid to ftand upon that part cut off.

10. A fector of a circle is the figure comprehended between two right lines drawn from the centre of the circle and that part of the circumference of the circle which is between them.

II. Similar fegments of circles are those which receive equal angles, or whereof the angles in them are equal.

PROPOSITION I. PROBLEM.To find the centre of a given circle.

Let ABC be a given circle: It is required to find the centre of the circle A B C.

Let any right line A B be drawn in it; and [by 10. 1.] divide the fame in two equal parts at D; and draw DC. [by II. 1.] at right angles to AB, which produce to E. Then divide c E into two equal parts at F: I fay, the point F is the centre of the circle AB C.

A

FG

E

D B

For if it be not, let, if poffible, G be the centre, and draw GA, GD, GB. Then because A D. is equal to DB, and DG is common, the two fides A D, DG will be equal to the two fides GD, D B each to each. And [by def. 15. 1.] the base GA is equal to the base GB, for they are both drawn from the centre G. Therefore the angle A D G [by 8. 1.] is equal to the angle G D B. But fince a right line ftanding upon a right line makes the adjoining angles equal to one another; each of these angles. [by 10. def. 1.] are right angles: Therefore the angle G D B is a right angle. But F DB is a right angle too: Wherefore the angle F D B is equal to the angle GDB, the greater equal to the leffer, which is impoffible. Therefore G is not the centre of the circle A B C. After the fame manner we demonftrate that no other point befides F can be the centre. Therefore the point F is the centre of the circle A B C. Which was to be done.

Ceral.

Corol. From hence it is evident, If any right line in a circle cuts another into two equal parts, at right angles, the centre of the circle will be in the cutting line.

D

H

G

The praxis of this problem may be fomewhat fhorter, thus: Open the compaffes to any convenient diftance that will cut the given circle, and fetting one foot in the circumference at the point A describe an arch B C with the other foot, cutting the circumference of the given B circle in the points B, C, and about the centres B, c, with the diftance AB or A c defcribe two arches intersecting one another in the point D. Draw the right line AD cutting the circumference of the circle in H. Bifect A H in G, and then will G be the centre of the given circle. The demonftration cannot be here given, because it depends upon the xxivth propofition of this book.

PROP. II. THEOR.

A

If any two points be taken in the circumference of a circle, the right line which joins them will fall within the circle.

Let ABC be a circle, in whose circumference let any two points A, B be taken: I fay, the right line A B joining thofe two points falls within the circle.

For if not, let it fall without, if poffible, as A E B ; and [by 1. 3.] find the centre D, of the circle A B C, join a D, DB, and continue out DF to E.

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D

Then because DA is equal to D B, the angle D A E [by 5. 1.] will be equal to the angle DBE; and because one fide A E B of the triangle DA E is produced, the angle DEB [by 16. 1.] will be greater than the angle DA E. But the angle D A E is equal to the angle DBE: Therefore the angle DEB will be greater than the angle D BE. But [by 18. 1.] the greater fide is oppofite to the greater angle. Therefore DB is greater than DE. But

A

B

F

E

DF is equal to the right line DB: Wherefore D F is greater

than

han DE, the leffer than the greater, which is impoffible.. Therefore the right line drawn from the point A to B does not fall without the circle, nor does it fall in the circumference. Confequently it must neceffarily fall within the circle.

If therefore any two points be taken in the circumference of a circle, the right line which joins them will fall within. the circle. Which was to be demonstrated.

This theorem may be demonftrated affirmatively, thus Let the right line A B touch two points A, B in the circumference of the circle A B, whofe centre is c: I fay, the right line AB falls within the circle, fo that all its intermediate points will be within the circle. For take any one of its intermedi

A

ate points D, and from the centre draw the B right lines CA, CD, C B. Then because the two fides CA, CB of the triangle CA B are equal; the angles CAB, CBA [by 5. 1.] will be equal to one another. But [by 16. 1.] the angle CDA is greater than the angle C BA, the external than the internal. Therefore the fame angle CDA will be greater than the angle CAD, and fo the fide CA will be [by 19. 1.] greater than the fide CD. Wherefore fince CA extends from the centre to the circumference, the right line C D will not extend to the circumference. Confequently the point D will fall within the circle. The fame may be demonftrated of any other intermediate point of the right line AB. Wherefore the whole right line A B falls within the circle.

Some have thought this propofition needs no demonstration, it being fo very felf-evident. But it must either be a demonftrable propofition, or an axiom, or an indemonftrable one. Euclid did not care to make it an axiom because of increafing their number too much; and therefore did rightly in demonstrating it.

PROP. III. THEOR.

If a right line drawn thro' the centre of a circle divides a right line not drawn thro' the centre into two equal parts, it will cut the fame at right angles; and if it cuts the fame at right angles, it will cut it into two equal parts.

Let there be a circle A B C, and let the right line cD drawn in it thro' the centre divide the right line A B not

drawn

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