PROPOSITION XIX. PROBLEM.

339. To construct a polygon similar to a given polygon, and having a given ratio to it.

Let A-E be the given polygon, and let the given ratio be that of the lines m and n.

To construct a polygon similar to A-E, and having to it the ratio n: m.

Construct A'B', the side of a square which shall have to the square described upon AB the ration: m (§ 338).

Upon the side A'B', homologous to AB, construct the polygon A'-E' similar to A-E (§ 298).

Then A'-E' will have to A-E the ratio n: m.

For since A-E is similar to A'-E',

58. To construct a triangle equivalent to a given square, having given its base and the median drawn from the vertex to the base.

59. To construct a square equivalent to twice a given square.

PROPOSITION XX. PROBLEM.

340. To construct a polygon similar to one of two given polygons, and equivalent to the other.

Let M and N be the given polygons.

To construct a polygon similar to M, and equivalent to N.

Let AB be any side of M.

Construct m, the side of a square equivalent to M, and n, the side of a square equivalent to N (§ 334).

Construct A'B' a fourth proportional to m, n, and AB (§ 291).

Upon the side A'B', homologous to AB, construct the polygon P similar to M (§ 298).

Then P will be equivalent to N.

For since M is similar to P,

area M AB

area P

But by construction, we have

-2

(§ 323.)

A'B'2

EXERCISES.

60. To construct an isosceles triangle equivalent to a given triangle, having its base coincident with a side of the given triangle.

61. To construct a rhombus equivalent to a given parallelogram, having one of its diagonals coincident with a diagonal of the parallelogram.

62. To construct a triangle equivalent to a given triangle, having given two of its sides..

63. To construct a right triangle equivalent to a given square, having given its hypotenuse.

64. To construct a right triangle equivalent to a given triangle, having given its hypotenuse.

65. To construct an isosceles triangle equivalent to a given triangle, having given one of its equal sides. How many different triangles can be constructed ?

66. To draw a line parallel to the base of a triangle dividing it into two equivalent parts. (§ 320.)

67. To draw through a given point within a parallelogram a straight line dividing it into two equivalent parts.

68. To construct a parallelogram equivalent to a given trapezoid, having a side and two adjacent angles equal to one of the non-parallel sides and the adjacent angles of the trapezoid.

69. To draw through a given point in a side of a parallelogram a straight line dividing it into two equivalent parts.

70. To draw a straight line perpendicular to the bases of a trapezoid, dividing the trapezoid into two equivalent parts.

(Draw a line connecting the middle points of the bases.)

71. To draw through a given point in the smaller base of a trapezoid a straight line dividing the trapezoid into two equivalent parts. 72. To construct a triangle similar to two given similar triangles, and equivalent to their sum.

73. To construct a triangle similar to two given similar triangles, and equivalent to their difference.

BOOK V.

REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE.

341. DEF. A regular polygon is a polygon which is both equilateral and equiangular.

PROPOSITION I. THEOREM.

342. A circle can be circumscribed about, or inscribed in, any regular polygon.

Let ABCDE be a regular polygon.

I. To prove that a circle can be circumscribed about ABCDE.

Let a circumference be described through the vertices A, B, and C (§ 223).

Let O be the centre of the circumference, and draw OA, OB, OC, and OD.

Then since ABCDE is equiangular,

LABC ▲ BCD.

=

Then the circumference passing through A, B, and C also passes through D.

In like manner, it may be proved that the circumference passing through B, C, and D also passes through E.

Hence, a circle can be circumscribed about ABCDE. II. To prove that a circle can be inscribed in ABCDE. Since AB, BC, CD, etc., are equal chords of the circumscribed circle, they are equally distant from 0. (§ 164.) Hence, a circle described with O as a centre, and with the perpendicular OF from 0 to any side AB as a radius, will be inscribed in ABCDE.

343. DEF. The centre of a regular polygon is the common centre of the circumscribed and inscribed circles.

The angle at the centre is the angle between the radii drawn to the extremities of any side; as AOB.

The radius is the radius of the circumscribed circle; as OA.

The apothem is the radius of the inscribed circle; as OF. 344. COR. From the equal triangles OAB, OBC, etc., we have

ZAOB = 2 BOC = ≤ COD, etc. ($ 66.) Then each of these angles is equal to four right angles divided by the number of sides of the polygon.. (§ 37.) That is, the angle at the centre of a regular polygon is equal to four right angles, divided by the number of sides.