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I fhall here only just mention two theorems; one of Mr. Huygens's, whofe demonfiration is too long to be fubjoin'd. If a circle be defcribed about the centre of gravity of any right lined figure, and right lines be drawn from all the angles of the figure to any point of the circumference of the circle, the fum of the fquares of all thofe lines will always be of the fame magnitude.

And the following theorem, which was found out by me Some years ago, and communicated by me to feveral people fome years ago, viz. The fquare of the number expreffing the area of any trapezium infcribed in a circle, will be equal to the product of the four numerical differences between the number expreffing half the fum of the four fides, and each of the numbers expreffing the fides. The demonftration, which I bave by me, cannot here be conveniently annex'd.

That point upon which if the figure any how refts, fuppofing it to be heavy, it would continue without altering its fituation.

EUCLID's

EUCLI D's

ELEMENTS,

BOOK IV.

DEFINITIONS.

I, A Right lined figure is faid to be infcribed in a right lined figure, when every angle of the figure inscribed touches every fide of the figure in which it is inscribed.

2. A right lined figure is faid to circumfcribe a right lined figure, when every fide of the circumfcribed figure touches every angle of the figure about which it is circumfcribed.

3. A right lined figure is faid to be infcribed in a circle, when every angle of the infcribed figure touches the circumference of the circle.

4. A right lined figure is faid to be circumfcribed about a circle, when every fide of the circumfcribed figure touches the circumference of the circle.

5. A circle is faid to be infcribed in a right lined figure, when the circumference of the circle touches every fide of the figure in which it is infcribed.

6. A circle is faid to be circumfcribed about a right lined figure, when the circumference of the circle touches every angle of the figure about which it is circumfcribed.

7. A right line is faid to be applied in a circle, when the extremes of that line are in the circumference of the circle.

PROPO

PROPOSITION I. PROBLEM.

To apply a right line in a given circle equal to a given right line which is not greater than the diameter of the circle a.

Let the given circle be A B C, and D a given line not greater than the diameter of the circle: it is required to apply a right line in the circle A B C equal to the given right line D.

A

C

E

Draw the diameter B C of the circle A B C ; then if в C be equal to D, the thing required will be done already; B for the right line BC is applied in the circle ABC equal to the given right line D. But if not BC is greater than D, and [by 3. 1.] make C E equal to D, and from the centre c with the diftance c E [by 3. poft.] describe the circle A E F, and [by 1. poft.] join c A.

D

F

Then because the point c is the centre of the circle AEF, CA will be equal to CE: But D is equal to the right line CE; and therefore D will be equal to CA.

Wherefore the right line CA is applied in the circle ABC, equal to the given right line D, which is not greater than the diameter of the circle. Which was to be done.

a If it be required to accommodate or apply a right line lefs than the diameter of a given circle within the circle equal to a given right line which shall be parallel to another given right line, the thing may be done thus:

L BN E

D

E

Let ABC be the given circle, D the centre, and EF the right line, to which the given right line less than the diameter of the circle is to be accommodated in the circle equal; and let G be the right line to which the applied right line is to be parallel. Thro' the centre HD draw [by 31. 1.] a diameter AC of the circle parallel to the right line G: Then if the right line EF be equal to the diameter, the thing required is done. But if EF be lefs than the diameter,

A

I

M

G

K

F

diameter, having bifected it in н, cut off DI equal to E н, and DK equal to HF: So that the whole 1K be equal to the whole EF; and thro' I, K [by 11. 1.] draw L M, N o perpendicular to AC; and join LN: I fay LN is applied equal to E F, and parallel to G: For fince LM, NO are equally diftant from the centre; they will [by 14. 3.] be equal to one another and because they are divided in half at 1, K [by 3. 3.] for they are cut at right angles by the right line A c paffing thro' the centre D, their halves L 1, KN will be equal. But becaufe [by 28. 1.] LI, NK are also parallel; alfo LN, IK [by 31. 1.] will be equal and parallel: Wherefore fince IK is equal to E F and parallel to G; LN will be equal to E F, and [by 30. 1.] parallel to G : By the fame reason if м o be drawn, it will alfo be equal to E F, and parallel to G. Which was to be demonstrated.

PROP. II. PROBL.

To infcribe a triangle in a given circle, equiangular to a given triangle.

Let the given circle be ABC, and the given triangle be DEF; it is required to infcribe a triangle in the circle A B C equiangular to the triangle DE F.

Draw the right line G A H touching the circle ABC in the point A, and at the right line AH, and at the point A in it [by 23. 1.] make the angle HA C equal to the angle DEF: Alfo at the right line G A, and at the point A in it make the angle GAB equal the angle FD E, and join

B C.

Then because the right line HAG touches the circle ABC, and AC is drawn from the point of contact; the angle HAC [by 32. 3.] will be equal to the angle E A B C, which itands in the alternate segment of the cir

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cle: But the angle H A C is equal to the angle DEF; and therefore the angle A B C is equal to the angle DE F. By the fame reason, the angle A CB is equal to the angle FDE: Therefore the remaining angle B A C will be [by 32. 1.] equal to the remaining angle EFD; and fo the triangle ABC is equiangular to the triangle D E F, and [by 3. def. 4.] is infcribed in the circle A B C

Wherefore

1

Wherefore a triangle is infcribed in a circle, equiangular to a given triangle. Which was to be done.

This problem may be otherwife conftructed thus: Let ABC be the given circle, whofe centre is н, and D E F a given triangle to which a triangle is to be infcribed equiangular to the given triangle DEF: Draw any femidiameter H A, and at the centre н with the femidiameter н A make [by 23. 1.] the angle A H B double to one of the angles E DF of the given triangle. Apply the right line AB. At the point A with the right line AB make the angle BAC equal to the angle DFE of the given triangle. Apply the right line BC. Then will the triangle A B C be infcribed in the given circle A B C equiangular to the given triangle D E F.

E

C

H

D.

B

F

For fince [by conftruction] the angle B HA at the centre is double to the angle EDF; and the angle BCA at the circumference [by 20. 3.] is one half that angle BHA; therefore the angle BCA will be equal to the angle ED F. But the angle BAC [by conftruction] is equal to the angle DFE. Therefore [by 32. 1.] the remaining angle ABC will be equal to the remaining angle D E F of the given triangle. Wherefore the triangle ABC is equiangular to the triangle D E F, and is inscribed in the circle A B C.

B

D

A

H

Hence an equilateral triangle may be eafily infcribed in a given circle; for it C is but applying A D equal to the femidiameter A H of the circle; and then again, applying BD equal to AH, and joining the right line AB, which will be one fide of the equilateral triangle ABC infcribed in the circle. The demonftration is easy, from what has been faid above, and from prop. xxvii, xxviii. lib. iii.

PROP. III. PROBL.

To circumfcribe a triangle about a given circle equiangular to a given triangle.

Let ABC be the given circle, and DEF the given tri

angle;

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