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angle; it is required to circumfcribe a triangle about the given circle AB C, equiangular to the given triangle

DEF.

Produce EF [by 2. poft.] both ways to the points H and G, and [by 1. 3.] find the centre K of the circle A B C, and any how drawing the right line K B, make [by 23. 1.] at the line K E, and at the point K in it, the an

gle B K A equal to the an- A

D

gle DEG, and the angle

K

CG E

TH

BK C equal to the angle

D FH; and draw [by 17. M 3.] the right lines LA M,

B N

MBN, NCL touching the circle ABC in the points A, B, C.

Then because the right lines L A M, MBN, NCL touch the circle A B C in the points A, B, C, and the right lines KA, KB, KC are drawn from the centre K to the points A, B, C, the angles at the points A, B, C, will [by 18. 3.] be right angles; and because the four angles of the quadrilateral figure A M B K, are [by 32. 1.] equal to four right angles (for the quadrilateral figure A M K в may be divided into two triangles) whereof the angles K A M, K B M are each one right angle; the remaining angles A K B, A M B will be equal to two right angles. But [by 13. 1.] the angles DEG, DEF are also equal to two right angles: Wherefore the angles AKB, AMB, are equal to the angles DEG, DEF. But A K B is equal to DEG: Therefore the remaining angle A M B, will be equal to the remaining angle DEF. After the fame manner we demonftrate, that the angle L N M is equal to the angle DFE; and fo the remaining angle M L N [by 32. 1.] is equal to the remaining angle EDF: Therefore the triangle L M N is equiangular to the triangle DEF, and [by 4. def. 4.] it is circumfcribed about the circle A B C.

Therefore a triangle is circumfcribed about a given circle equiangular to a given triangle. Which was to be done.

PROP.

PROP. IV. PROBL.

To infcribe a circle in a given triangle.

Let the given triangle be ABC: it is required to infcribe a circle in the triangle ABC.

Divide [by 9. 1.] the angles ABC, BCA each into two equal parts, by the right lines B D, C D, meeting one another in the point D, and from D draw [by 12. 1.] the perpendiculars DE, D F, D G to the fides A B, B C, C A, of the triangle.

A

E

D

G

Then because the angle ABD is equal to the angle CBD, for the angle A B C is bifected, and the right angle BED is alfo equal to the right angle B F D, there are two triangles EBD, DBF having two angles of the one equal to two angles of the other, and one fide of the one equal to one fide of the other, viz. the fide B D oppofite to one of the equal angles common to them both : Therefore [by 26. 1.] the remaining fides of the one will be equal to the remaining fides of the other; that is, DE will be equal to DF; alfo by the fame B reafon the right line D G will be equal to DF: wherefore if a circle be described with the centre D, and either of the distances D E, DF, DG, it will pass thro' the remaining points, and touch the right lines A B, B C, C A, because the angles at E, F, G, are right angles: for if it cuts them the right line drawn from the end of a diameter at right angles to it, will fall within the circle; which [by 16. 3.] is abfurd: Therefore the circle defcribed with the centre D and either of the diftances DE, DF, DC will not cut the right lines A B, BC, CA; wherefore it will touch them, and [by 5. def. 4.] be inscribed in the triangle

AB C.

F

Therefore the circle EFG is infcribed in the given triangle A B C. Which was to be done.

PROP.

PROP. V. PROBL:

To circumfcribe a circle about a given triangle.

Let A B C be a given triangle: it is required to circumscribe a circle about the given triangle A B C.

Divide [by 10. 1.] AB, A C each into two equal parts in the points D, E, and [by 11. 1.] draw D F, E F from the points D, E at right angles to A B, A C; thefe will either

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meet within the triangle ABC, in the fide B C, or without the triangle ABC.

First let them meet within the triangle in the point F, and join B F, F C, FA: Then because A D is equal to DB, and DF is both common and at right angles, the base A F [by 4. 1.] will be equal to the bafe F B. In like manner we demonftrate that CF is equal to FA, and fo BF is equal to FC: wherefore the three right lines F A, FB, FC are equal to one another: Confequently a circle described from the centre F with either of the distances F A, FB, F C will pass thro' the remaining points; and [by 6. def. 4.] the circle will be circumfcribed about the triangle A B C. Let it be circumfcribed as A B C.

Secondly, Let D F, E F meet at F in the right line B C, as in the fecond figure; and join AF. We demonftrate after the fame manner that the point F is the centre of a circle circumfcribed about the triangle A B C.

Laftly, Let D F, E F meet in the point F without the triangle ABC, as in the third figure; and join A F, FB, F C Then becaufe A D is equal to D B, and D F is common and at right angles, the bafe A F [by 4. 1.] will be equal to the base FB. Alfo after the fame manner we demonftrate that C F is equal to FA; and fo B F is equal to FC; therefore again a circle described with the centre F, and either of the distances F A, FB, FC will pass thro' the

re

remaining points, and will be circumfcribed about the triangle A B C. Let it be described as A B C.

Therefore a circle is circumfcribed about a given triangle. Which was to be done.

Corollary. Hence it is manifeft, that if the centre of the circle falls within the triangle ABC, the angle BAC being in a fegment greater than a femicircle, is less than a right angle; if it falls in the right line B C, the angle in the femicircle will be a right angle; and if it falls without the triangle A B C, the angle in the fegment less than a femicircle will be greater than a right angle. Wherefore if the given triangle ABC be an acute angled one, DE, EF will meet within the triangle; but if it has a right angle BA C, they will meet in the fide BC: If B A C be an obtufe angle, they will meet without the triangle A B C.

PROP. VI. PROBL.

To inscribe a square in a given circle.

Let the given circle be A B C D : it is required to infcribe a fquare in the circle A B C D.

Draw [by 11. 1.] two diameters A C, BD at right angles to one another; and join A B2 B C, CD, DA.

B

A

E

Then because BE is equal to ED, for E is the centre, and E A is common and at right angles, the base AB D [by 4. 1.] will be equal to the base AD; and by the fame reafon B C, CD are each equal to BA, BD; therefore the quadrilateral figure A B C D is equilateral. I fay, it is also right angled. For because the right line BD is a diameter of the circle A B C D ; BA D will be a femicircle: and fo [by 31. 3.] the angle B A D is a right angle: by the fame reafon, each of the angles ABC, BCD, CDA will be a right angle. Therefore the quadrilateral figure A B C D is right angled. And it has been proved to be equilateral too: Therefore it is a fquare, and infcribed in the circle A B C D.

Therefore the fquare A B C D is infcribed in the given circle A B C D. Which was to be done.

PROP.

PROP. VII. THEOR.

To circumfcribe a fquare about a given circle.

Let the given circle be A B CD. It is required to circumfcribe a square about the given circle A B C D.

Draw two diameters A C, B D of the circle ABCD at right angles to one another, and [by 17. 1.] thro' the points A, B, C, D draw the right lines FG, GH, HK, KF, to touch the circle ABCD in the points

A, B, C, D.

Then because F G touches the circle A B C D, and the right line EA is drawn from the centre

B

H

E

K

E to the point of contact a ; the angles G A F at A [by 18. 3.] will be right angles. By the fame reason, the angles at the points B, C, D are right angles; and because the angle A E B is a right angle, and E B G alfo a right angle; the line G H [by 28. 1.] will be parallel to AC: By the fame reason AC is parallel to F K. In like manner we demonftrate that GF, HK are each parallel to BED. Therefore G K, GC, AK, FB, BK are parallelograms; and fo GF is [by 34. 1.] equal to H K, as also G H to F K ; and because A c is equal to BD; but A c is equal to each of the lines GH, FK, and B D is equal to each of the lines GF, HK: Therefore each of the right lines HG, FK will be equal to each of the right lines GF, HK; and fo the quadrilateral figure F G H K is equilateral. I fay it is also rectangular. For because GBEA is a parallelogram, and A E B is a right angle, [by 34. 1.] AGB will be a right angle. After the fame manner we demonftrate that the angles at the points H, K, F are right angles; and fo the quadrilateral figure F G H K is right angled. But it has been proved to be equilateral; Therefore a fquare, and is circumfcribed about the circle

it is

ABCD.

Therefore a fquare is circumfcribed about a given circle. Which was to be done.

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