6. THEOREMS ON THE AREAS OF SIMILAR POLYGONS.

(1) The areas of similar triangles are in the same ratio as the squares of any two homologous sides. § 314.

(2) The areas of similar polygons are in the same ratio as the squares of any two homologous sides. § 315.

(3) The areas of similar polygons are in the same ratio as the squares of any two homologous diagonals. § 316.

(4) The area of the square described on the hypotenuse of a right triangle is equal to the sum of the areas of the squares described on the other two sides. § 317.

(5) The area of any polygon described on the hypotenuse of a right triangle is equal to the sum of the areas of the similar polygons similarly described on the other two sides. § 318.

7. MISCELLANEOUS THEOREMS.

(1) If a given line-segment is divided internally into any two parts, the square on the whole segment is equal in area to the sum of the squares on the two parts together with twice the rectangle contained by the two parts. § 311.

(2) If a given line-segment is divided externally into any two parts, the square on the given segment is equal in area to the sum of the squares on the two parts less twice the rectangle contained by the two parts. § 312.

(3) In any triangle the square on the side opposite an acute angle is
less than the sum of the squares on the other two sides by
twice the product of either of these sides and the projection of
the other upon it. § 321.

(4) In an obtuse-angled triangle the square on the side opposite the
obtuse angle is greater than the sum of the squares on the
other two sides by twice the product of either of these sides
and the projection of the other upon it.
§ 322.

(5) If two sides of a triangle are of given lengths while the third side
varies, the triangle will have the maximum area when the first
two sides are at right angles. § 334. 1.

(6) Given two intersecting straight lines AB, AC, and a point P between them; of all line-segments which pass through P and are terminated by AB and AC, that which is bisected at P makes with AB and AC the triangle of minimum area. § 334.3.

CHAPTER V

MEASUREMENT OF THE CIRCLE

SECTION I

REGULAR POLYGONS

DEFINITIONS

335. A regular polygon is a polygon which is both equilateral and equiangular.

For example, an equilateral triangle is a regular polygon of three sides; a square is a regular polygon of four sides.

A polygon of more than three sides may be equilateral without being equiangular, or it may be equiangular without being equilateral; but in order to be classed as regular it must be both equilateral and equiangular.

That there can be regular polygons of any given number of sides will be seen from the first proposition of this chapter.

Make a diagram of a quadrilateral that is (1) equilateral, but not equiangular; (2) equiangular, but not equilateral; (3) both equilateral and equiangular; (4) neither equilateral nor equiangular.

336. A regular polygon of more than four sides may or may not be convex, but unless the contrary is stated, it will be understood that any figure under discussion is convex.

A polygon of five sides is called a pentagon; one of six sides, a hexagon; one of seven sides, a heptagon; one of eight sides, an octagon; one of ten sides, a decagon; one of twelve sides, a dodecagon.

337. POSTULATE 8. A circle may be divided into any given number of equal arcs.

The problem "to divide a circle, or any arc of a circle, into a given number of equal parts" is not always solvable by the methods of elementary geometry. The method of solution, if there is one, must of course depend on the number of such parts required. All that the above postulate affirms is that the circle may be thought of as made up of any specified number of equal parts, and the points of division may be assumed. It says nothing at all about a method of finding these points of division.

338. In the preceding chapters we have shown how to divide line-segments, angles, and arcs of circles, into certain numbers of equal parts. The following exercises will recall the methods employed.

EXERCISES

1. Divide a given line-segment into two equal parts; into four equal parts; into eight equal parts (Art. 56).

2. Divide a given line-segment into three equal parts; into six equal parts; into nine equal parts (Art. 142).

3. Divide a given line-segment into any required number of equal parts (Art. 269).

4. Divide a given angle into two, four, eight, etc. equal parts (Art. 54). 5. Divide a right angle into three equal parts (Art. 142).

6. Divide a given arc of a circle into two, four, eight, etc., equal parts (Art. 165).

The problem to divide any given angle or arc of a circle into three equal parts has been found to be impossible of solution by the methods of elementary geometry, in which we make use of the ruler and compasses only.

7. Divide a circle into two, four, eight, etc., equal arcs.

8. Divide a circle into three, six, or twelve equal arcs. Ex. 5.)

(Make use of

PROPOSITION I

339. If a circle is divided into any number of equal

arcs:

I. The chords joining the points of division, taken in order, form a regular inscribed polygon.

II. The tangents to the circle at the points of division, taken in order, form a regular circumscribed polygon.

REMARK. The number of sides of the polygon is equal in each case to the number of points of division in the circle.

Let a given circle be divided into any number of equal arcs AB, BC, CD, etc., and let the points of division be joined in order, thus forming the polygon ABCDE; also at the points of division, let the tangents to the circle be drawn, thus forming the polygon FGHJK.

It is required to prove that the polygon ABCDE is a regular polygon; and also, that the polygon FGHJK is a regular poly

gon.

Proof. First, since by hypothesis the arcs AB, BC, CD, etc., are equal, the chords AB, BC, CD, etc., are equal. (Art. 161.) Therefore the polygon ABCDE is equilateral.

=

Also, the arc EAB
Therefore, EAB =

the arc ABC. Why?

ABC.

(Art. 181.)

Similarly BCD = ABC, and so on.

Therefore the polygon ABCDE is equiangular.

That is, the polygon ABCDE being both equilateral and equiangular is regular.

Next, let the points A, B, C, etc., be joined to the centre O. Then Ls AOB, BOC, COD, etc., are all equal. Why? Therefore, s AFB, BGC, CHD, being supplements of equal angles at the centre [why ?] are also equal.

Hence the polygon FGHJK is equiangular.

Also, the line-segment AF

=

the line-segment FB. (Art. 194.) Also, the line-segment FB the line-segment BG.

=

Similarly BG GC, GC CH, and so on.

=

Therefore the side FG =

=

=

Prove.

the side GH, the side HJ, and so on. Hence the polygon FGHJK is equilateral. That is, the polygon FGHJK is also regular.

340. COROLLARY I. If the vertices of a regular inscribed polygon are joined to the mid-points of the arcs subtended by the sides of the polygon, the joining lines will form another regular polygon of twice the number of sides.

341. COROLLARY II. If tangents are drawn at the mid-points of the arcs between the points of contact of the sides of a regular circumscribed polygon, these together with the sides of the original polygon form a regular polygon of twice the number of sides.