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ELEMENTS

OF

GEOMETRY.

BOOK IV.

A

DEFINITIONS.

I.

RECTILINEAL figure is said to be inscribed in an- Book IV. other rectilineal figure, when all the angles of the in

scribed figure are upon the sides of the

figure in which it is inscribed, each

upon each.

II.

In like manner, a figure is said to be described about another figure when all

the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

III.

A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are upon the circumference of the circle.

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A rectilineal figure is said to be described about a circle,
when each side of the circumscrib-

ed figure touches the circumference
of the circle.

V.

In like manner, a circle is said to be
inscribed in a rectilineal figure,
when the circumference of the
circle touches each side of the fi-
gure.

VI.

A circle is said to be described about
a rectilineal figure, when the cir-
cumference of the circle passes
through all the angular points of
the figure about which it is de-
scribed.

VII.

A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

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.

Draw BC the diameter

of the circle ABC; then, if BC be equal to D, the thing required is done; for in the circle ABC a straight line BC is placed equal to D: But, if BC be a 3. 1. not equal a to D, it must be

greater; cut off from it CE
equal to D, and from the

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centre C, at the distance CE, describe the circle AEF,

and join CA: Therefore, because C is the centre of the Book IV. circle AEF, CA is equal to CE; but D is equal to CE; 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. Which was to be done.

PROP. II. PROB.

In a given circle to inscribe a triangle equiangular to a given triangle.

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.

a

Draw the straight line GAH touching the circle in a 17. 3. the point A, and at the point A, in the straight line AH,

make the angle HAC equal to the angle DEF; and at b 23. 1.

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c

the point of contact, the angle HAC is equal to the c 32. 3. angle ABC in the alternate 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 d 32. 1. EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC. Which was to be done.

Book IV.

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, a 23. 1. make 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, b 17. 3. NCL touching the circle ABC: Therefore, because

c

LM, MN, NL touch the circle ABC in the points A, B, C, to which from the centre are drawn KA, KB, KC, c 18, 3. the angles at the points A, B, C, are right angles. And because the four angles of the quadrilateral figure AMBK are equal to four right angles, for it can be divided into two triangles; and because two of them, KAM, KBM

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Б

are right angles, the other two AKB, AMB are equal to two right angles: But the angles DEG, DEF are liked 13. 1. wise equal to two right angles; therefore the angles AKB, AMB are equal to the angles DEG, 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

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