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PROP. XVII. PROBLEM

302. To construct a triangle equivalent to a given polygon.

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Construction. 1. Take three consecutive vertices, A, B, C; draw diagonal AC, and line BF AC, meeting DC prolonged at F, also line AF.

2. ▲ ACF ▲ ABC; then, polygon AFDE ≈ ABCDE (having common part ACDE), and has a number of sides less by 1.

3. Draw diagonal AD, and line EG | AD, meeting CD prolonged at G, also line AG; then, ▲ AED≈ ▲ AGD, whence ▲ AFG polygon AFDE, or to ABCDE.

Proof. 4. A ACF and ABC have same base and altitude.

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Note. By aid of §§ 302 and 298, a square may be constructed equivalent to a given polygon.

Ex. 49. The bases of a trapezoid are 8 and 10, respectively, the altitude 6. Construct an equivalent equilateral triangle.

[An equilateral triangle can be constructed when its altitude is known.]

Ex. 50. The bases of a trapezoid are 6 and 8, respectively; the area is 35. Construct the trapezoid and an equivalent equilateral triangle.

Ex. 51. Given A ABC. Draw a line from A to BC which shall divide the triangle into two triangles the ratio of whose areas is 1 to 2.

PROP. XVIII. PROBLEM

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

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Required to construct a polygon similar to AC, and having to it the ratio

n

m

Construction. 1. Let A'B' be side of square having to the square described on AB the required ratio.

(§ 301) 2. Polygon A'C', similar to AC, will have required ratio to it.

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

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Required to construct a polygon similar to M, and ≈ N.

Construction. 1. Find m and n sides of squares

M and N. (Note, § 302)

2. Polygon P, similar to M, with A'B' a fourth proportional

to m, n, and AB, will be ≈ N.

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Ex. 52. To inscribe in a given triangle a parallelogram whose area is one-half the area of the triangle. Can a parallelogram be inscribed whose area is still greater?

Ex. 53. Through a given point P, either within or without a given angle, to draw a line which shall form with the sides of the angle a triangle of given area.

Ex. 54. To construct a triangle whose angles shall be equal, respectively, to the angles of a given triangle, and whose area shall be four times the area of the given triangle.

BOOK V

REGULAR POLYGONS.-MEASUREMENT OF THE CIRCLE.-LOCI

305. Def. A regular polygon is a polygon which is both equilateral and equiangular.

PROP. I. THEOREM

306. A circle can be circumscribed about any regular polygon.

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Given regular polygon ABCDE.

To Prove that a O can be circumscribed about it.

Proof. 1. Draw circumference through A, B, C'; also, radii

OA, OB, OC, and line OD.

2. In OAB, OCD, OB = OC, AB = CD.

3. LOBA =

4. Since ABC

(?)

ABC-ZOBC, OCD =/ BCD- OCB.
≤ ▲
ZBCD (?), and ▲ OBC = ≤ OCB (?),
ZOBA ZOCD.

=

5. Then A OAB =

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6. Then circumference through A, B, C passes through D;

and similarly through E.

307. Since AB, BC, CD, etc., are equal chords of the circumscribed O, they are equally distant from 0. (§ 164) Then, a drawn with O as a centre, and a line OF to any side AB as a radius, will be inscribed in ABCDE.

Hence, a circle can be inscribed in any regular polygon.

308. Defs. 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, 04.
The apothem is the radius of the inscribed circle, OF.

309. From equal ▲ OAB, OBC, etc., we have

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Whence, the angle at the centre of a regular polygon is equal to four right angles divided by the number of sides.

PROP. II. THEOREM

310. If a circumference be divided into equal arcs, their chords form a regular inscribed polygon.

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Given circumference ACD divided into five equal arcs, AB,

BC, CD, etc., and chords AB, BC, etc.

To Prove polygon ABCDE regular.

Proof. 1. Sides of polygon are equal by § 159.

2. Each

equal arcs.

is measured by one-half the sum of three of the

(§ 192)

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