and at right angles to AB, the base AF is equal (1. 1.) to the base FB. In like manner, it may be shewn 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, will pass through the extremities of the other two, and be described about the triangle ABC.

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 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 obtuse angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

SCHOLIUM.

From the demonstration it is evident that the three perpendiculars bisecting the sides of a triangle, meet in the same point; that is, the centre of the circumscribed circle.

PROP. VIII. PROB.

To draw a circular segment arch of

a given span and rise.

This is, evidently, only a modification of the preceding Problem, where AB is the span and SR the rise.

Join AR, BR, and at their respective points of bisection, M, N, erect the perpendicular MO, NO to AR, BR; they will intersect at O, the centre of the circle. That OA OR=OB, is proved as before.

The joints between the arch-stones, or voussoirs, are only continuations of radii drawn from the centre O of the circle.

MN

B

ELEMENTS

OF

GEOMETRY.

BOOK IV.

DEFINITIONS.

1 A RECTILINEAL figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed

figure are upon the sides of the figure in which it is inscribed, each upon each.

2. 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.

3 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.

4. A rectilineal figure is said to be described about a circle, when each side of the circumscribed figure touches the circumference of the circle.

5. 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 figure.

6. 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.

ODO

7. Polygons of five sides are called pentagons; those of six sides, hexagons; those of seven sides, heptagons; those of eight sides, octagons ; and so on.

8. A polygon, which is at once equilateral and equiangular, is called a regular polygon.

Regular polygons may have any number of sides; the equilateral triangle is one of three sides; and the square is one of four sides.

LEMMA.

Any regular polygon may be inscribed in a circle, and circumscribed about one.

Let ABCDE, &c. be a regular polygon: describe a circle through the three points A, B, C, the centre being O, and OP the perpendicular let fall from it, to the middle point of BC: join AO and OD.

If the quadrilateral OPCD be placed upon the quadrilateral OPBA, they will coincide; for the side OP is common: the angle OPC= OPB, being right; hence the side PC will apply to its equal PB, and the point C will fall on B; besides, from the nature of the polygon, the angle PCD=PBA; hence CD will take H the direction BA, and since CD=BA, the point D will fall on A, and the two quadrilaterals will entirely coincide.

B

D

E

G

F

The distance OD is therefore equal to AO; and consequently the circle which passes through the three points A, B, C, will also pass through the point D. By the same mode of reasoning, it might be shown that the circle which passes through the points B, C, D, will also pass through the point E; and so of all the rest: hence the circle which passes through the points A, B, C, passes through the vertices of all the angles in the polygon, which is therefore inscribed in this circle. Again, in reference to this circle, all the sides AB, BC, CD, &c. are equal chords; they are therefore equally distant from the centre (Th. 14. 3.): hence, if from the point O with the distance OP, a circle be described, it will touch the side BC, and all the other sides of the polygon, each in its middle point, and the circle will be inscribed in the polygon, or the polygon circumscribed about the circle.

COR. 1. Hence it is evident that a circle may be inscribed in, or circumscribed about, any regular polygon, and the circles so described have a

common centre.

COR. 2. Hence it likewise follows, that if from a common centre, circles can be inscribed in, and circumscribed about a polygon, that polygon is regular. For, supposing those circles to be described, the inner one will touch all the sides of the polygon; these sides are therefore equally distant from its centre; and, consequently, being chords of the circumscribed circle, they are equal, and therefore include equal angles. Hence the polygon is at once equilateral and equiangular; that is (Def. 8. B. IV.), it is regular.

SCHOLIUMS.

1. The point O, the common centre of the inscribed and circumscribed circles, may also be regarded as the centre of the polygon; and upon this principle the angle AOB is called the angle at the centre, being formed by two radii drawn to the extremities of the same side AB..

Since all the chords are equal, all the angles at the centre must evidently be equal likewise; and therefore the value of each will be found by dividing four right angles by the number of the polygon's sides.

2. To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides: for the arcs being equal (see fig. Prop. XV. B. 4.), the chords AB, BC, CD, &c. will also be equal; hence, likewise, the triangles ABG, BGC, CGD, &c. must be equal, because they are equiangular; hence all the angles ABC, BCD, CDE, &c. will be equal, and consequently the figure ABCD, &c. will be a regular polygon,

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 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 (Prob. 3. 1.) 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 CF; but D is equal to CE; there

A

C

E

B

F

D

fore 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.

PROP. II. PROB.

In 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 inscribe in the circle ABC a triangle equiangular to the triangle DEF.

Draw (Prob. 2. 3.) the straight line GAH touching the circle in the point A, and at the point A, in the straight line AH, make (Prob. 9. 1.) 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, because HAG touches the circle ABC, and AC is drawn from the point of contact, the angle HAC is equal (30. 3.) to the 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 (2. Cor. 25. 1.) to the remaining angle EDF: Wherefore the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC.

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, make (Prob. 9. 1.) 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 (Prob. 2. 3.) 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 (18. 3.) 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, are right angles, the other two AKB, AMB are equal to two right angles: But the angles DEG, DEF are likewise equal (6. 1.) 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 remain