Fig. 36.

36. COR. All the angles B A C, BDC, BE C, inscribed in the same segment, are equal; because they are all measured by the half of the same arc B O C.

B

37. COR. Every angle B A D, inscribed in a semicircle, is a right angle; because it is measured by the half of the semi-circumference B C D, that is, by the fourth part of the whole circumference = 90°, that is one-half of 180°, of 360°.

B

A

38. Every angle BO C, inscribed in a segment less than a semicircle, is an obtuse angle; for it is measured by half of the arc B A C, greater than a circumference.

Fig. 37.

Fig. 38.

Fig. 39.

39. COR. The opposite angles A and C of an inscribed quadrilateral A B C D, are together equal to two right angles; for the angle B A D is measured by half the arc A BCD, the angle B C D is measured by half the arc B A D; hence the two angles B A D, BC D, taken together, are measured by the half of the circumference; hence their sum is equal to two right angles.

SECTION 5.-POLYGONS.

D

DEF. 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; the square is one of four sides.

40. To inscribe a regular polygon of a certain number of sides in a given circle.

Divide the circumference into as many equal parts as the polygon has sides; for the arcs being equal, the chords A B, B C, C D, &c., will also be equal; hence, likewise, the triangles A O B, BO C, CO D, H must be equal, because the sides are equal each to each; hence all the angles A B C, B CD, C D E, &c., will be equal; consequently the figure A B C D E, &c. will be a regular polygon.

41. To inscribe a square in a given circle.

Draw two diameters A C, B D, cutting each other at right angles; join their extremities A B C D; the figure A B C D will be a square; for the angles A O B, BO C, &c., being equal, the chords A B, B C, &c., are also equal; and the angles A B C, B CD, &c. being semicircles, are right angles.

A

Fig. 40.

Fig. 41.

42. In a given circle to inscribe a regular hexagon, and an equilateral triangle.

Suppose the problem solved, and that A B is a side of the inscribed hexagon: the radii A 0, O B, being drawn, the triangle A O B will be an equilateral. For the angle A O B is the sixth part of four right angles: therefore, taking the right angle for unity, we shall have A O B equal

equal; and the two other angles A B C, BA O, of the same triangle, are together equal 2-, or; and being mutually equal, each of them must be equal

F

Fig. 42.

to. Since the triangle A BO is equilateral, therefore the side of the inscribed hexagon is equal to the radius; hence to inscribe a regular hexagon in a given circle, the radius must be applied six times to the circumference, which will bring us round to the point of beginning. And the hexagon A B C D E F being inscribed, the equilateral triangle A ČE may be formed by joining the vertices of the alternate angles.

43. To inscribe a square or an octagon in a given circle.

For the Square.-Draw the diameters B D and A C, intersecting each other at right angles; join the points A B, B C, C D and D A, and A B CD will be the square required.

For the Octagon.-Bisect the are A B of the square in the point E, and the line A E, being carried eight times round the circumference, will form the octagon. If the arc A E be again bisected, a polygon may be

E

Fig. 43.

C

H

formed of sixteen sides, and by another bisection a polygon of thirty-two sides, &c.

44. To inscribe a pentagon or a decagon in a given circle.

For the Pentagon.-Draw the diameters A p, n m, at right angles to each other, and bisect the radius on in r; from the point s, with the distancer A, describe the arc A s, and from the point A, with the distance As, describe the arc s B; join the points A B, and the line A B being carried five times round of the circle, will form the pentagon required.

For the Decagon.-Bisect the arc

Fig. 44.

A

P

E

A E of the pentagon, in c, and the line A c, being carried ten times round the circumference, will form the decagon required.

45. About a given triangle A B C, to circumscribe a circle.

Bisect the two sides A B, B C, with the perpendiculars mo and n o; from the point of intersection o, with the distance o A, o B, or o C, describe the circle A C B, and it will a be the circle required.

D

Fig. 46.

Fig. 45.

B

777.

Fig. 47.

47. To circumscribe a square about a given circle. Draw any two diameters n o and r m at right angles to each other; through the points mo, rn, draw the lines A B, B C, C D, and D A, perpendicular to r m and and A B C D will be the square reՊՆ 0, quired.

NOTE.-If each of the quadrants be bisected and tangents drawn, the circle will be an octagon.

48. About a given circle to circumscribe a pentagon.

Find the points m n vr s, (Prob. 43) then from the centre o, to each of these points, draw the radii o n, o m, o v, o r, o s. Through the points n m draw the lines A B, BC, perpendicular to on, o m, producing them till they meet each other at B; in the same manner draw the lines CD, DE, E A, and A B C D E will be the pentagon required.

E

Fig. 48.

B

m

B

49. On a given line A B to form a regular octagon.

On the extremes of the given line A B, erect the indefinite perpendiculars A F and B E; produce A B both ways to m and n, and bisect the angles m A F and n B E, with the lines A H and B C; make A H and B C equal to A B, and draw H G, CD, parallel to A F on B E, H and also each equal to A B. From G D, as a centre, with a radius equal to A B, de

A

Fig. 49.

B

scribe arcs crossing A F, B E, in F and E, and if G F, F E, and ED be drawn, ABCDEFGH will be the octagon required.

50. On a given line A B, to form a regular polygon of any proposed number of sides.

Divide 360° by the number of sides, and subtract the quotient from 180°; make the angles A B o and B A o each equal to half the difference last found; from the point of intersection o, with the distance o A or o B, describe the circle; apply the chord A B to the cir

Fig. 50.

B

cumference the proposed number of times, and it will form the polygon required.