20. Prove, geometrically, that the rectangle of the sum and the difference of two straight lines is equivalent to the difference of the squares of those lines.

PROBLEMS.

21. To construct a triangle, given its angles and its area (equal to that of a given square).

Suggestion. Construct any triangle having the given angles. The problem then reduces to (29).

22. Given any triangle, to construct an isosceles triangle of the same area, whose vertical angle is an angle of the given triangle. (v. 19, Exercise.)

23. Given any triangle, to construct an equilateral triangle of the same area. (v. Exercise 21.)

24. Bisect a given triangle by a parallel to one of its sides. (v. Proposition VIII. and 28.)

25. Bisect a triangle by a straight line drawn through a given point in one of its sides. (v. 19, Exercise.)

26. Inscribe a rectangle of a given area in a given circle.

Suggestion. Draw a diagonal of the rectangle. The problem can then be reduced to inscribing in the given circle a right triangle of given area.

27. Given three points, A, B, and C, to find a fourth point P, such that the areas of the triangles APB, APC, BPC, shall be equal. (Four solutions.) (v. III., Exercise 19.)

BOOK V.

REGULAR POLYGONS. MEASUREMENT OF THE

CIRCLE.

1. DEFINITION. A regular polygon is a polygon which is at once equilateral and equiangular.

The equilateral triangle and the square are simple examples of regular polygons. The following theorem establishes the possibility of regular polygons of any number of sides.

PROPOSITION I.-THEOREM.

2. If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle; and the tangents drawn at the points of division form a regular polygon circum scribed about the circle.

Let the circumference be divided into the equal arcs AB, BC, CD, etc.; then, 1st, drawing the chords AB, BC, CD, etc., ABCD, etc., is a regular inscribed polygon. For its sides are equal, being chords of equal arcs; and its angles are equal, being inscribed in equal segments.

H

C

K

2d. Drawing tangents at A, B, C, etc., the polygon GHK, etc., is a regular circumscribed polygon.

AGB, BHC, CKD, etc., we have AB

For, in the triangles

BC= CD, etc., and

the angles GAB, GBA, HBC, HCB, etc., are equal, since each

G

H

is formed by a tangent and chord and is measured by half of one of the equal parts of the circumference (II., Proposition XV.); therefore these triangles are all isosceles and equal to each other. Hence we have the angles GHK, etc., and AG = GB = BH = HC = CK, etc., from which, by the addition of equals, it follows that HK, etc.

GH=

K

3. COROLLARY I. If the vertices of a regular inscribed polygon are joined with the middle points of the arcs subtended by the sides of the polygon, the joining lines will form a regular inscribed polygon of double the number of sides.

4. COROLLARY II. If at the middle points of the arcs joining adjacent points of contact of the sides of a regular circumscribed polygon tangents are drawn, a regular circumscribed polygon of double the number of sides will be formed.

5. Scholium. It is evident that the area of an inscribed polygon is less than that of the inscribed polygon of double the number of sides; and the area of a circumscribed polygon is greater than that of the circumscribed polygon of double the number of sides.

EXERCISE.

E

B' F

B

G

Theorem.-If a regular polygon is inscribed in a circle, the tangents drawn at the middle points of the arcs subtended by the sides of the inscribed polygon form a circumscribed regular polygon, whose sides are parallel to the sides of the inscribed polygon, and whose vertices lie on the radii drawn to the vertices of the inscribed polygon.

A

D

PROPOSITION II.-THEOREM.

6. A circle may be circumscribed about any regular polygon; and a circle may also be inscribed in it.

Let ABCD... be a regular polygon. Through A' and B', the middle points of AB and BC, draw perpendiculars, and connect O, their point of intersection, with all the vertices of the polygon and with the middle points of all the sides.

B B с

F E' E

The triangles OA'B and OB'B are equal, by I., Proposition X. OB'B and OB'C are equal, by I., Proposition VI. The angle OBB' is one-half of ABC; ... OCB' is one-half of the equal angle BCD. Hence the triangles OB'C and OCC' are equal, by I., Proposition VI. By continuing this process we may prove all the small triangles equal. O, then, is equidistant from all the vertices, and therefore with O as a centre a circle may be circumscribed about the polygon. O is also equidistant from all the sides, and therefore with O as a centre a circle may be inscribed in the polygon.

7. Definitions. The centre of a regular polygon is the common centre, O, of the circumscribed and in

scribed circles.

The radius of a regular polygon is the radius, OA, of the circumscribed circle.

The apothem is the radius, OH, of the inscribed circle.

H

The angle at the centre is the angle, AOB, formed by radii drawn to the extremities of any side.

8. The angle at the centre is equal to four right angles divided by the number of sides of the

10. Regular polygons of the same number of sides are similar.

Let ABCDE, A'B'C'DE',

be regular polygons of the same number of sides; then they are similar.

For, 1st, they are mutually equiangular, since the magnitude of an angle of

A'

either polygon depends only on the number of the sides (8 and 9), which is the same in both.

2d. The homologous sides are proportional, since the ratio AB: A'B' is the same as the ratio BC: B'C', or CD : C'D', etc.

Therefore the polygons fulfil the two conditions of similarity.

11. COROLLARY. The perimeters of regular polygons of the same number of sides are to each other as the radii of the circumscribed circles, or as the radii of the inscribed circles; and their areas are to each other as the squares of these radii. (v. III., Proposition VIII., and IV., Proposition IX.)