VI.

A circle is said to be described about a rectilineal figure, when the circumference of the circle passes through all the angular points of the figure about which it is described.

VII.

A straight line is said to be placed in a cir

cle, when the extremities of it are in the circumference of the circle.

Book IV.

PROP. I. PROB.

In a given circle to place a straight line, equal to a given straight line, not greater than the diameter of the circle.

Let ABC be the given circle, and D the given straight line, not greater than the diameter of the circle.

a

Draw BC the diameter of the circle ABC; then, if BC is equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D: But, if it is not, BC is greater than D; make CE equal to D, and from the centre C, at the distance CE, describe the circle AEF, and join CA: Therefore, because C is the centre of the circle AEF, CA is equal to CE: but D is equal to CE;

A

a 3. 1.

E

B

F

D

therefore D is equal to CA: Wherefore, in the circle ABC, a straight line is placed equal to the given straight line D: which is not greater than the diameter of the circle.

be done.

Which was to

PROP. II. PROB.

In a given circle to inscribe a triangle equiangular

to a given triangle.

Book IV.

a 17. 3.

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

Draw the straight line GAH touching the circle in the b 23. 1. point A, and at the point A, in the straight line AH, make the angle HAC equal to the angle DEF; and at the point A, in the straight line

AG make the angle

GAB equal to the
angle DFE, and join
BC: Therefore, be-
cause HAG touches
the circle ABC, and

AC is drawn from E
the point of contact,
the angle HAC is

C

c 32. 3. equal to the angle ABC in the alter

d 32. 1.

a 23. 1.

b 17. 3.

c 18. 3.

G

nate segment of the circle: But HAC is equal to the angle DEF: therefore also the angle ABC is equal to DEF: For the same reason, the angle ACB is equal to the angle DFE; therefore the remaining angle BAC is equal to the remaining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done.

PROP. III. PROB.

About a given circle to describe a triangle equiangular to a given triangle.

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

Produce EF both ways to the points G, H, and find the centre K of the circle ABC, and from it draw any straight line KB; at the point K, in the straight line KB, makea the angle BKA equal to the angle DEG, and the angle BKC equal to the angle DFH; and through the points A, B, C, draw the straight lines LAM, MBN, NCL, touching the circle ABC: Therefore, because LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, the angles at the points A, B, C, are right angles: And because the four angles of the qua

b

drilateral figure AMBK are equal to four right angles, for it Book IV. can be divided into two triangles; and because two of them

KAM, KBM are

right angles, the

other two, AKB,

AMB are equal to two right angles: But the an

L

A K

gles DEG, DEF are likewise e

the angles DEG,

B

DEF, of which AKB is equal to DEG; wherefore the remaining angle AMB is equal to the remaining angle DEF: In like manner, the angle LNM may be demonstrated to be equal to DFE; and therefore the remaining angle MLN is equal to the remaining angle EDF: Wherefore, the triangle e 32. 1. LMN is equiangular to the triangle DEF: And it is described about the circle ABC. Which was to be done.

PROP. IV. PROB.

To inscribe a circle in a given triangle.

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

A

b

See N.

Bisect the angles ABC, BCA by the straight lines BD, a 9. 1. CD meeting one another in the point D, from which draw b 12. 1. DE, DF, DG perpendiculars to AB, BC, CA: And because the angle EBD is equal to the angle FBD, for the angle ABC is bisected by BD, and because the right angle BED is equal to the right angle BFD, the two triangles EBD, FBD have E two angles of the one equal to

G

D

two angles of the other, and the

side BD, which is opposite to

one of the equal angles in each, B F is common to both; therefore

C

96

c 26. 1.

Book IV. their other sides are equal; wherefore DE is equal to DF: For the same 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 them, will pass through the extremities of the other two, and of any will also 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 d Cor. 16.3. angles to it, touches the circle: Therefore the straight lines AB, BC, CA do each of them touch the circle, and the circle EFG is inscribed in the triangle ABC. Which was to be done.

See N.

a 10. 1.

b 11. 1.

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

Bisect AB, AC in the points D, E, and from these points draw DF, EF at right angles to AB, AC; DF, EF produced

c 4. 1.

will meet one another: For, if they do not meet, they are parallel; 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 AB, the base AF is equal 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 Book IV. F, at the distance 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. And it is manifest, 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 segment greater than a semicircle; but, when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, 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, 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 side opposite to the right angle; and if it be an obtuseangled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

PROP. VI. PROB.

To inscribe 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 another; and join AB, BC, CD, DA; because BE is equal to ED, for E is the centre; and because EA is

a

B

F

D

a 4. 1.

at right angles to BD, and common to the triangles ABE, ADE; the base BA is equal to the base AD; and, for the same reason, BC, CD are each of them equal to BA, or AD; therefore the quadrilateral figure ABCD is equilateral. It is also rectangular; for the straight line BD, being the diameter of the circle ABCD, BAD is a semicircle; wherefore the angle BAD is a right angle; for the same reason b 31. 3. 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 inscribed in the circle ABCD. Which was to be done.

C

G