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V. So likewise a Circle is said to be inscribed in a
Right-lined Figure, when the Circumference of
the Circle touches all the sides of the Figure in
which it is inscribed.
VI. A Circle is said to be described about a Figure,

when the Circumference of the Circle touches all the Angles of the Figure which it circumscribes. VII. A Right Line is said to be applied in a Circle,

when its Extremes are in the Circumference of the Circle.

PROPOSITION I.

PROBLEM.
To apply a Right Line in a given Circle, equal to

a given Right Line, whose Length does not exa
ceed the Diameter of the Circle.

ET the Circle given be A BC, and the given It is required to apply a Right Line in the Circle ABC, equal to the Right Line D.

Draw B C the Diameter of the Circle; then if BC be equal to D, what was required, is done: for in the Circle AB C there is applied the Right Line B C, equal to the Right Line D: But if not, the Diameter BC is greater than D, and put * CE equal to D;* 3. 1o and about the Center C, with the Distance CE, let the Circle A EF be described ; and join CA.

Then because the Point C is the Center of the Circle A EF, CA will be equal to CE; but D is equal to CE. Wherefore AC is equal to D. And so in the Circle ABC, there is applied a Right Line AC, equal to the given Right Line D, not greater than the Diameter; which was to be done.

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PROPOSITION II.

PROBLEM.
In a given Circle, to describe a Triangle equiangus

lar to a given Triangle.
ET ABC be à Circle givefi, and D E F a given

* 17.30

in the Circle ABC, equiangular to the Triangle DEF. Draw the Right Line GAH touching * the Circle

ABC in the Point A, and with the Right Line AH 1 23.1.

at the Point A, make + an Angle HAC, equal to the Angle DEF. Likewise at the same Point A, with the Line A G, make the Angle G A B equal to the Angle D F E, and join B C.

Then because the Right Line HAG touches the Circle ABC, and AC is drawn from the Point of Contact in the Circle ; the Angle HAC shall be $equal to ABC, the Angle in the alternate Segment of the Circle. But the Angle HAC is equal to the Angle DEF; therefore also the Angle ABC, is equal to the Angle DEF: For the same Reason, the Angle

ACB is likewise equal to the Angle DFE. Where+ Cor. 2. fore the other Angle BAC, shall be equal to the

other Angle EDF. And consequently, the Triangle ABC is equiangular to the Triangle DEF, and is described in the Circle ABC; which was to be done.

1 32. 3.

32. I.

PROPOSITION III.

PROBLEM.
About a given Circle to describe a Triangle, equi-

angular to a Triangle given.
ET ABC be the given Circle, and DÉF the

given Triangle. It is required to describe a Triangle about the Circle AB C equiangular to the Triangle DEF.

Produce the Side E F both Ways to the Points G and Hy and find the Center of the Circle K, and any how draw the Line KB. Then at the Point K, with

KB

KB make * the Angle BK A equal to the Angle * 23. 1.
DEG; and the Angle BKC at the fame Point ķ
on the other Side the Line KB, equal to the Angle
DFH; and thro' the Points A, B, C, let the Right
Lines LAM, MBN, NCL, be drawn touching the
Circle ABC.

Then because the Lines LM, MN, NL, touch the Circle ABC in the Points A, B, C, and the Lines KA, KB, KC, are drawn from the Center K to the Points A, B, C; the Angles at the Points A, B, C, will be + Right Angles. And because the four An- † 18. 3 gles of the quadrilateral Figure AMBK are equal to four Right Angles, (for it may be divided into two Triangles,) and the Angles KAM, KBM, are each Right Angles; therefore the other Angles A KB, AMB are equal to two Right Angles. But DEG, DEF, are equal to two Right Angles; thereforç the Angles AKB, AMB, are equal to the Angles DEG, DEF, whereof AKB is equal to DEG, Wherefore the other Angle A MB is equal to the other Angle DEF. In like Manner we demonstrate, that the Angle LNB is equal to the Angle DFE. Therefore the other Angle MLN is f equal to the other Angle 1 Cor.

2 EDF. Wherefore the Triangle LNM is equian- 32. I? gular to the Triangle DEF, and is described about the Circle ABC; which was to be done.

PROPOSITION IV.

PROBLEM.

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To inscribe a Circle in a given Triangle.
LE
ET ABC be a Triangle given. It is required

.
to inscribe a Circle in the same,
Cut * the Angles ABC, BCA, into two equal* 9.
Parts by the Right Lines BD, DC, meeting each
other in the Point D. And from this point draw
DE, DF, DG, + perpendicular to the Sides A B, † 12. the

BC, A O. G

Now because the Angle E B D is equal to the Angle FBD, and the Right Angle BED is equal to the

Right Angle BFD; then the two Triangles EBD, }, DBF, have two Angles of the one, equal to two

H 3

Angles

I 26. I.

Angles of the other, and one Side DB common to both, viz. that which fubtends the equal Angles; therefore the other sides of the one Triangle shall be equal to the other Sides of the other; and fo DE shall be equal to DF. And for the same Reason, D G is equal to DF: Therefore DE is also equal to DG And fo the three Right Lines DE, DF, DG, are equal between themselves. Wherefore a Circle described about the Center D, with either of the Distances DE, DF, DG, will also pass thro' the other Points. And the sides AB, BC, AC, will touch it; because the Angles at E, F, and Gare Right Angles. For if it fhould cut them, a Right Line drawn on the Extremity of the Diameter of a Circle at Right Angles, will fall within the Circle; which is * abfurd. Therefore a Circle described about the Center D, with either of the Diftances DE, DF, DG, will not cut the Sides A B, BC, CA; wherefore it will touch them, and will be a Circle described in the Triangle A BC. Therefore the Circle EFG is described in the given Triangle ABC; which was to be done,

16. 3.

PROPOSITION V.

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10. I.

†11. I,

PROBLEM.
To describe a Circle about a given Triangle.
ET ABC be a given Triangle. It is required to

describe a Circle about the fame.
Bisect * the Sides AB, AC, in the Points D, E;
from which Points let DF, EF, be drawn fat Right
Angles to AB, AC, which will meet either within
the Triangle ABC, or in the Side B C, or without
the Triangle.

First let them meet in the Point F within the Triangle, and join BF, FC, FA. Then because AD is equal to DB, and D F is common, and at Right Angles to AB; the Base A F will be f equal to the Base FB. And after the same Manner we prove, that the Bafe CF is equal to the Base FA. Therefore also is BF equal to CF: And so the three Right Lines FA, FB, FC, are equal to each other. Wherefore a Circle described about the Center F, with either of the

Distances

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Distances FA, FB, FC, will pass also thro' the other Points, and will be a Circle described about the Triangle ABC. Therefore describe the Circle ABC.;

Secondly, let DF, EF, meet each other in the Point F, in the Side BC, as in the second Figure, and join AF. Then we prove, as before, that the Point F is the Center of a Circle described about the Triangle ABC.

Lastly, let the Right Lines DF, EF, meet one another again in the Point F, without the Triangle, as in the third Figure; and join AF, FB, FC. And because AD is equal to D B, and DF is common, and at Right Angles, the Base A F shall be equal to the Base BF. Sọ likewise we prove, that CF is also equal to AF. Wherefore BF is equal to CF. And so again, if a Circle be described on the Center F, with either of the Distances FA, FB, FC, it will pass through the other Points, and will be described about the Triangle A B C; which was to be done.

Coroll

. If a Triangle be Right-angled, the Center of the Circle falls in the side opposite to the Right Angle; if acute-angled, it falls within the Triangle; and if obtuse-angled, it falls without the Triangle.

PROPOSITION VI.

PROBLEM.

To inscribe a Square in a given Circle.

ET ABCD be a Circle given. It is required

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Draw AC, BD, two Diameters of the Circle cutting one another at Right Angles, and join A B, BC, CD, DA.

Then because BE is equal to ED, (for E is the Center) and EA is common, and at Right Angles to BD, the Base B A shall be * equal to the Base AD; * 4. Sy and for the same Reason BC, CD, as also B A, AD, are all equal to each other. Therefore the quadrilateral Figure ABCD, is equilateral. I say it is also rectangular. For because the Right Line D B is a Diameter of the Circle ABCD, BAD, will be a SeH4

micircle.

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