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Book IV,

PROP. IV. · PROB.

To

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infcribe a circle in a given triangle.

b 12.3.

Let the given triangle be ABC; it is required to inscribe a circle in ABC.

Bisect * the angles ABC, BCA by the straight lines BD, a 9. 1. CD meeting one another in the point D, from which drawb DE, DF, DG perpendiculars

А to AB, BC, CA. Then because the angle EBD is equal to the angle FBD, the angle ABC being bifected by BD; and because the right angle E BED is equal to the right angle BFD, the two triangles EBD, FBD have two angles of the one equal to two angles of the other; and the fide BD, which B В.

F is opposite to one of the equal angles in each, is common to both ; therefore their other fides are equal c; wherefore DE is equal to DF. For the same c 26.1. reason, DG is equal to DF; therefore the three straight lines DE, DF, DG are equal to one another, and the circle described from the centre D, at the distance of any of them, will pass through the extremities of the other two, and will touch the straight lines AB, BC, CA, because the angles at the points E, F, G are right angles, and the straight line which is drawn from the extremity of a diameter at right angles to it, touches d the circle : Therefore the straight lines AB, BC, CA, do d Cor.16.3. each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done.

PROP.

Book IV.

PROP. V.

PROB.

To describe a circle about a given triangle.

Let the given triangle be ABC; it is required to describe

a circle about ABC. a 10. I.

Bisect a AB, AC in the points D, E, and from these points b11. 1. draw DF, EF at right anglesb to AB, AC; DF, EF produced

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will meet one another; for, if they do not meet, they are paral. lel, wherefore AB, AC, which are at right angles to them, are parallel, which is absurd : Let them meet in F, and join FA; also, if the point F be not in BC, join BF, CF: then, because

AD is equal to DB, and DF common, and at right angles to C 4. 1. AB, the base AF is equal c to the base FB.

In like manner, it may be shown that CF is equal to FA; and therefore BF is equal to FC, and FA, FB, FC are equal to one another; wherefore the circle described from the centre F, at the di. stance of one of them, will pass through the extremities of the other two, and be described about the triangle ABC, which was to be done.

Cor. When the centre of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle ; but, when the centre is in one of the fides of the triangle, the angle opposite to this side, being in a femicircle, is a right angle; and, if the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle:

Wherefore,

Wherefore, if the given triangle be acute angled, the centre Book IV. of the circle falls within it; if it be a right angled triangle, the centre is in the fide opposite to the right angle ; and, if it be an obtufe angled triangle, the eentre falls without the tri, angle, beyond the side opposite to the obtufe angle.

PRO P.

VI.

PRO B.

To infcribe a square in a given circle.

Let ABCD be the given circle; it is required to inscribe a square in ABCD.

Draw the diameters AC, BD at right angles to one ano. ther, and join AB, BC, CD, DA; because BE is equal to ED, E being the centre, and because EA is at right angles to

A BD, and common to the triangles ABE, ADE; the base BA is equal a to the base AD; and, for the same reason, BC, CD are

E

a 4. 1.

B В each of them. equal to BA or

D AD; therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD, being a diameter of the circle, ABCD, BAD is a semicircle ; wherefore the angle BAD is a right b angle; for b 31. 3. the same reason each of the angles ABC, BCD, CDA is a right angle ; therefore the quadrilateral figure ABCD is rectangular, and it has been thown to be equilateral ; therefore it is a square ; and it is inscribed in the circle ABCD. Which was to be done.

PRO P. VII. ,

PRO B.

Tod

O describe a square about a given circle.

Let ABCD be the given circle; it is required to describe a {quare about it.

I 2

Draw

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Book IV. Draw two diameters AC, BD of the circle ABCD, at right

angles to one another, and through the points A, B, C, D a 17. 3. draw a FG, GH, HK, KF touching the circle ; and because

FG touches the circle ABCD, and EA is drawn from the

centre E to the point of contact A, the angles at A are right b 18. 3. bangles ; for the same reason, the angles at the points, B, C,

D are right angles ; and because the angle AEB is a right angle, as likewise is EBG, GH,

F C 28. 1. is parallel c to AC; for the same

reason, AC is parallel to FK, and
in like manner GF, HK may
each of them be demonstrated to

E

BI
be parallel to BED; therefore
the figures GK, GC, AK, FB,

BK are parallelograms; and GF
& 34. 1. is therefore equal d to HK, and
GH to FK ; and because AC is H

K equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK: GH, FK are each of them equal to GF or HK; therefore the quadrilateral figure FGHK is equilateral. It is also rectangular; for GBEA be. ing a parallelogram, and AEB a right angle, AGB is likewife a right angle : In the same manner, it may be shown that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular; and it was de. monstrated to be equilateral; therefore it is a square ; and it is described about the circle ABCD. Which was to be done.

P R O P. VIII.

PRO B.

Toi

O inscribe a circle in a given square. :

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Let ABCD be the given square; it is required to inscribe a circle in ABCD.

Bisect a each of the fides AB, AD, in the points F, E, and b 31. 1. through E draw b EH parallel to AB or DC, and through F

draw FK parallel to AD or BC; therefore each of the figures

AK, KB, AH, HD, AG, GC, BG, GD is a parallelogram, 6 34. I. and their opposite sides are equalc; and because AD is equal

to AB, and that AE is the half of AD, and AF the half of Book IV.
AB, AE is equal to AF; wherefore the fides opposite to
these are equal, viz. FG to GE; in the same manner, it may
be demonstrated, that GH, GK,
are each of them equal to FG or A

E

D GE; therefore the four straight lines GE, GF, GH, GK, are equal to one another; and the circle described from the centre G,

F

K at the distance of one of them, will pass through the extremities of the other three ; and will also touch the straight lines AB, BC, CD, DA, because the angles at

B

d 29. 1. the points E, F, H, K are right d angles, and because the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle e; therefore each of the straight lines AB, BC, CD, DA e 16. 3. touches the circle, which is therefore inscribed in the square ABCD. Which was to be done.

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PRO P. IX.

PROB.

To describe a circle about a given square.

a $. I.

Let ABCD be the given square ; it is required to describe a circle about it.

Join AC, BD, cutting one another in E; and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two sides DA, AC are equal to the two BA, AC, and the-base DC is equal to the base BC; wherefore the angle DAC is equal A A a to the angle BAC, and the angle DAB is bilected by the straight line AC. In the same manner, it may be

E demonstrated, that the angles ABC, BCD, CDA are severally bifected by the straight lines BD, AC; therefore,

B because the angle DAB is equal to the angle ABC, and the angle EAB is the half of DAB,

and

I 3

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