Book IV. ~ See N.

a Io. 1.

PROP. V. PROB.

To defcribe a circle about a given triangle.

Let the given triangle be ABC; it is required to describe a circle about ABC.

Bifect AB, AC in the points D, E, and from these points b 11. 1. draw DF, EF at right angles to AB, AC, DF, EF produced

I.

d

ci2. Ax. 1. meet one another : For, if DE be joined, the angles DEF, EDF are less than the two right angles ADF, AEF: 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 d 4. 1. common, and at right angles to AB, the bafe AF is equal to the bafe 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 distance of one of them, fhall pass through the extremities of the other two; and be described about the triangle ABC. Which was to be done.

COR. And it is manifeft, that, 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 fegment greater than a femicircle; but, when the centre is in one of the fides of the triangle, the angle oppofite to this fide, being in a femicircle, is a right angle; and, if the centre falls without the triangle, the angle oppofite to the fide beyond which it is, being in a fegment lefs than a femicircle, is greater than a right angle: Wherefore, if the given triangle be acute angled, the centre of the circle falls within it; if it be a right angled triangle, the centre is in the fide oppofite to the right angle; and, if it be an obtuse angled triangle, the centre falls without the triangle, beyond the fide oppofite to the obtufe angle.

PROP.

PROP. VI. PROB.

To infcribe a fquare in a given circle.

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

E

BOOK IV.

a 4. I.

D

Draw the diameters AC, BD at right angles to one another ; and join AB, BC, CD, DA; because BE is equal to ED, for E is the centre, and that EA is common, and at right angles to BD; the bafe BA is equal a to the base AD; and, for the fame reafon, BC, CD are each of them. equal to BA or AD; therefore the B quadrilateral figure ABCD is equilateral. It is alfo rectangular; for the ftraight line BD, being the diameter of the circle ABCD, BAD is a femicircle; wherefore the angle BAD is a right angle; for the b 31. 8. fame reason, each of the angles ABC, BCD, CDA is a right angle; therefore the quadrilateral figure ABCD is rectangular, and it has been shown to be equilateral; therefore it is a square; and it is infcribed in the circle ABCD. Which was to be done.

PROP. VII. PROB.

b

To defcribe a fquare about a given circle.

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

Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw

B

E

A

b 18. 3.

FG, GH, HK, KF touching the circle: and because FG a 17. 3. touches the circle ABCD, and EA is G drawn from the centre E to the point of contact A, the angles at A are right angles; for the fame reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as likewife is EBG, GH is parallel to AC; for the fame reason, AC is parallel to FK; and in like manner, GF, HK may each of them be demonftrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal to HK, and GH to FK; d 34. 1.

N 2

H

C

K

C 28. I.

and

BOOK IV and because AC is 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 being a parallelogram, and AEB a right angle, d 34. 1. AGB is likewife a right angle: In the fame manner, it may be fhown, that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular, and it was demonftrated to be equilateral; therefore it is a fquare; and it is defcribed about the circle ABCD. Which was to be done.

a 10. I.

PROP. VIII. PROB.

To infcribe a circle in a given square.

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

A

E

D

G

K

Bifect each of the fides AB, AD, in the points F, E, and b 31. 1. through F. draw 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, c 34. 1. and their oppofite fides are equal; and because AD is equal to AB, and that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the fides oppofite to thefe are equal, viz. FG to GE. In the fame manner, it may be demonftrated, that GH, GK are each of them equal to FG or GE; therefore the four fraight lines GE, GF, GH, GK, are equal to one another; and the circle defcribed from the centre G, at the diftance of one of them, fhall pafs through B the extremities of the other three; and because the angles at the points E, F, H, K are right angles, and that the traight line which is drawn from the extremity e 16. 3. of a diameter, at right angles to it, touches the circle; therefore each of the ftraight lines AB, BC, CD, DA touches the circle, which therefore is infcribed in the fquare ABCD. Which was to be done.

d 29.

I.

H

d

PROP.

PROP. IX. PROB.

To defcribe a circle about a given square.

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

A

E

D

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 fides DA, AC are equal to the two BA, AC; and the bafe DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bifected by the ftraight line AC: In the fame manner, it may be demonstrated, that the angles ABC, BCD, B CDA are severally bifected by the straight lines BD, AC; therefore, because the

b

BOOK IV.

a 8. I.

angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA; wherefore the fide EA is equal to the fide EB: In the fame manner, it may be de- b 6. I. monstrated, that the ftraight lines EC, ED are each of them equal to EA or EB; therefore the four straight lines EA, EB, EC, ED are equal to one another; and the circle defcribed from the centre E, at the diftance of one of them, fhall pass through the extremities of the other three, and be defcribed about the fquare ABCD. Which was to be done.

T

PROP. X. PROB.

O defcribe an ifofceles triangle, having each of
the angles at the bafe double of the third

angle.

Take any ftraight line AB, and divide a it in the point C, fo a 11. 2. that the rectangle AB, BC be equal to the fquare of CA; and from the centre A, at the distance AB, defcribe the circle BDE,

b

in which place the straight line BD equal to AC, which is not b 1.4. greater than the diameter of the circle BDE; join DA, DC, and about the triangle ADC defcribe the circle ACD; the triangle ABD is fuch as is required; that is, each of the angles ABD, ADB is double of the angle BAD.

Because the rectangle AB, BC is equal to the fquare of AC, and that AC is equal to BD, the rectangle AB, BC is equal to

the

C 5. 4.

BOOK IV. the fquare of BD; and because from the point B without the circle ACD two ftraight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, and that the rectangle AB, BC contained by the whole of the cutting line, and the part of it without the circle is equal to d 37. 3. the fquare of BD which meets it; the straight line BD touches d the circle ACD; and because BD touches the circle, and DC is drawn from the point of contact e 32. 3. D, the angle BDC is equal

f 32. 1.

g 5. 1.

h 6. I.

to

the angle DAC in the alternate
fegment of the circle; to each of
thefe add the angle CDA; there-
fore the whole angle BDA is equal
to the two angles CDA, DAC;
but the exterior angle BCD is
equal to the angles CDA, DAC;
therefore alfo BDA is equal to
BCD; but BDA is equal to the

A

D

E

angle ABD, becaufe the fide AD is equal to the fide AB; there-
fore ABD, or DBC is equal to BCD; and confequently the fide
h
BD is equal to DC; but BD was made equal to CA; there-
fore alfo CA is equal to CD, and the angle CDA equal to
DAC; therefore the angles CDA, DAC together, are double
of the angle DAC: But BCD is equal to the angles CDA,
DAC; therefore alfo BCD is double of DAC; and BCD is
equal to each of the angles BDA, DBA; each therefore of
the angles BDA, DBA is double of the angle DAB; where-
fore an ifofceles triangle ABD is defcribed, having each of
the angles at the bafe double of the third angle. Which was
to be done.

a 10. 4.

T

PROP. XI. PROB.

O infcribe an equilateral and equiangular pentagon in a given circle.

a

Let ABCDE be the given circle; it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE. Defcribe an ifofccles triangle FGH, having each of the angles at G, H, double of the angle at F; and in the circle b 2. 4. ABCDE infcribe the triangle ACD equiangular to the triangle FGH, fo that the agle CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the angle at G or H; wherefore each of the angles ACD, CDA is double of the angle CAD: Bifect the angles ACD, CDA by the straight

€ 9. J.