+ II. I.

4. T.

through the other Points, and will be a Circle defcribed about the Triangle ABC.

Secondly, Let DF, EF, meet each other in the Point F, in the Side B C, as in the fecond Figure; and join AF. Then we prove, as before, that the Point F is the Centre of a Circle deferibed about the Triangle ABC.

Laftly, Let the Right Lines D F, EF, meet one another again in the Point F, without the Triangle, as in the third Figure; and join AF, FB, FC. And becaufe A D is equal to D B, and D F is common, and at Right Angles, the Bafe A F fhall be equal to the Bafe B F. So likewife we prove, that CF is alio equal to A F. Wherefore B F is equal to CF. And fo again, if a Circle be defcribed on the Centre F, with either of the Distances F A, F B, FC, it will pass through the other Points, and will be described about the Triangle ABC; which was to be done.

Corall, If a Triangle be Right-angled, the Centre of the Circle falls in the Side oppofite to the Right Angle; if acute-angled, it falls within the Triangles, and if obtule-angled, it falls without the Triangle.

To infcribe a Square in a given Circle.

LET ABCD be a Circle given. It is required to inferibe a Square within the fame..

Draw A C, D B, two Diameters of the Circle, cutting one another at Right Angles + and join AB, BC, CD, DA.

Then, because BE is equal to ED (for E is the Centre), and E A is common, and at Right Angles to BD, the Bafe B A fhall be equal to the Bafe AD; and for the fame Reafon BC, CD, BA, and A D, are all equal to each other. Therefore the quadrilateral Figure A B C D is equilateral. I fay, it is alfo Rectangular. For, because the Right Line DB is a Diameter of the Circle ABCD, therefore

BAD

BAD will be a Semicircle. Wherefore the Angle
BAD is a Right Angle. And for the fame Rea- 31. 3.
fon, every one of the other Angles ABC, BCD,
CDA, is a Right Angle. Therefore, ABCD is a
rectangular quadrilateral Figure: But it has alfo been
proved to be equilateral. Wherefore, it shall neceffa-
rily be a Square, and is infcribed in the Circle ABCD;
which was to be done.

PROPOSITION VII.
PROBLEM:

To defcribe a Square about a given Circle.
ET ABCD be a Circle given. It is required
to defcribe a Square about the fame.

LE

Draw A C, B D, two, Diameters of the Circle,

cutting each other at Right Angles +; and thro' the † 11. 1. Points A, B, C, D, draw * F G, GH, HK, K F, * 17. 3· Tangents to the Circle A B C D.

128. I.

† 34. 1.

Then, because F G touches the Circle A B C D, and E A is drawn from the Centre E to the Point of Contact A, the Angles at A will be + Right Angles + 18. 3. For the fame Reason, the Angles at the Points B, C, D, are Right Angles. And fince the Angle A E B is a.Right Angle, as alfo E BG, GH fhall t be parallel to A C, and for the fame Reafon, A C to K F. In this manner we prove likewife, that GF and H K are parallel to B E D; and fo G F is parallel to H K. Therefore G K, GC, AK, F B, B K, are Parallelograms; and fo GF+ is equal to HK, and GH to FK. And fince A C is equal to B D, and AC equal to either G H or F K; and BD equal to either GF, or H K; GH, or F K is equal to G F, or H K. Therefore F G H K is an equilateral quadrilateral Figure: I fay, it is alfo equiangular. For, because G B E A is a Parallelogram, and AEB is a Right Angle; then A G B fhall be also a Right Angle. In like manner we demonftrate, that the Angles at the Points H, K, F, are Right Angles. Therefore the quadrilateral Figure F G H K is rectangular; but it has been proved to be equilateral likewife. Wherefore, it must neceffarily be a Square, and is defcribed about the Circle ABCD; which was to be done. PRO

PROPOSITION VIII.

་

PROBLEM.

To defcribe a Circle in a given Square.

LET the given Square be ABCD. It is required

to defcribe a Circle within the fame.

Bifect the Sides A B, AD, in the Points F, E; and draw + EH through E, parallel to A B, or DC; and KF through F, parallel + to BC, or AD. Then AK, K B, AH, HD, AG, GC, BG, GD, are all Parallelograms, and their oppofite Sides are equal. And because DA is equal A B, and A E is half of AD, and A F half of A B, A E fhall be equal to AF; but the oppofite fides are also equal. Therefore F G is equal to G E. In like manner we demonftrate, that GH, or GK, 'is equal to either F G, or G E. Therefore GE, GF, GH, G K, are equal to each other: And fo a Circle being defcribed about the Centre G, with either of the Dftances G E, GF, GH, GK, will alfo pass through the other Points, and shall touch the Sides D A. AB, BC, CD, because the Angles at E, F, H, K, are Right Angles. For if the Circle fhould cut the Sides of the Square, a Right Line drawn from the End of the Diameter of a Circle at Right Angles, will fall within the Circle; which is abfurd. Wherefore a Circle defcribed about the Centre G, with either of the Distances G E, GF, GH, GK, will not cut D A, a B, B C, C D, the Sides of the Square. Wherefore, it shall neceffariy touch them, and will be defcribed in the Square ABCD; which was to be done.

PROPOSITION IX.
PROBLEM.

To defcribe a Circle about a Square given.

LET ABCD be a Square given. It is required

to circumfcribe a Circle about the fame.

Join AC, BD, mutually cutting one another in the Point E.

And

་

And fince D'A is equal to A B, and A C is common, the two Sides DA, A C, are equal to the two Sides B A, A C; but the Base DC is equal to the Bafe B C. Therefore the Angle DAC will be * 8. 1. equal to the Angle BAC And confequently the Angle D A B is bifected by the Right Line A C. In the lame manner we prove, that each of the other Angles ABC, BCD, CDA, are bifected by the Right Lines A C, D B.

Then, because the Angle D A B is equal to the Angle ABC, and the Angle EAB is half of the Angle DA B, and the Angle A E B half of the Angle ABC; the Angle EA B fhall be equal to the Angle EBA: and fo the Side E A is † equal to the Side E B. † 6.1. In like manner we demonftrate, that each of the Right Lines E C, E D, is equal to each of the Right Lines E A, E B. Therefore the four Right Lines EA, EB, EC, E D, are equal between themfelves. Wherefore, a Circle being defcribed about the Centre E, with either of the Distances E A, E B, EC, ED, will also pass through the other Points, and will be defcribed about the Square ABCD; which was to be done.

PROPOSITION X.
PROBLEM.

To make an Ifoceles Triangle, having each of the
Angles at the Bafe double to the other Angle.

CUT

*

UT* any given Right Line A B in the Point C, 11. 2. fo that the Rectangle contained under A B and BC be equal to the Square of A C; then about the Centre A, with the Distance A B, let the Circle BDE be described; and + in the Circle B DE apply the Right + 1 of this. Line B D equal to AC; which is not greater than the Diameter. This being done, join D A, D C, and defcribe a Circle A CD about the Triangle A'D C. Then, because the Triangle under A B and B C is equal to the Square of A C, and A C is equal to B D, the Rectangle under A B and BC fhall be equal to the Square of B D. And because fome Point B, is taken without the Circle A C D, and from that Point there fall two Right Lines, B C A, BD, to the Cir

I 5 of this.

37.3.

† 32. 3.

5. I.

cle, one of which cuts the Circle, and the other falls on it; and fince the Rectangle under A B and B C is equal to the Square of B D, the Right Line B D fhall *touch the Circle AC D. And fince D.B touches it, and C D is drawn from the Point of Contact D, the Angle BDC is equal to the Angle in the alternate Segment of the Circle, viz. equal to the Angle DAC. And fince the Angle BDC is equal to the Angle DAC; if CDA, which is common, be added, the whole Angle B D A is equal to the two Angles ‡32. 1. CDA, DAC. But the outward Angle B C D is t equal to CDA and DAC. Therefore BDA is equal to BCD. But the Angle BDA* is equal to the Angle CBD, because the Side A D is equal to the Side AB. Wherefore D BA fhall be equal to BCD: And fo the three Angles B D A, D B ́A, BCD, are equal to each other. And fince the Angle DBC is equal to the Angle BCD, the Side B D ist equal to the Side D C. But BD is put equal to CA. Therefore C A is equal to C D. And fo the Angle CDA is equal to the Angle D A C. Therefore the Angles CDA, DAC, taken together, are double to the Angle D A C. But the Angle BCD is equal to the Angles CDA, and D A C. Therefore the Angle BCD is double to the Angle D A C. But BCD is equal to B·D A, or D B A. Wherefore BDA, or DB A, is double to DA B. Therefore, the focetes Triangle ABD is made, having each of the Angles at the Baje double to the other Angle; which was to be done..

+6.1.

15. I.

10 of this.

PROPOSITION XI.
PROBLEM.

To defcribe an equilateral and equiangular Penta-
gon in a given Circle.

LETABCDE be a Circle given. It is required to describe an equilateral and equiangular Pentagon in the fame.

Make an Ifoceles Triangle F GH, having * each of the Angles at the Bafe H double to the other Angle F; and defcribe the Triangle ADC in the * 2 of this. Circle ABCDE, equiangular † to the Triangle

FGH;