(97.) THEOREM. If a straight line be divided into any number of parts, the semicircumference described on the whole line will be equal to the sum of the semicircumferences described on the parts.

ACB

AGD+DHF+FKB: AB :: AD+DF+FB.

Since the third term of this proportion is equal to the fourth term, follows that the first term is equal to the second term

hence,

it

(98.) THEOREM. If, on the sides of a triangle inscribed in a semicircle, semicircles be described, the two lunes formed thereby will together be equal to the area of the triangle.

Let ABC be a triangle inscribed in a semicircle.

On AB, BC, let semicircles ADB, BFC be described; the lunes ADBG, BFCH are together equal to the triangle ABC.

D

H

Since the areas of circles, and of course semicircles, are as the squares of their radii or diameters (B. V, Prop. xш), we have

semicircle ABC: ADB :: AC2: AB3, and

semicircle ABC: BFC :: AC: BC2; therefore,

ABC : ADB+BFC :: AC2 : AB2+BC2.

But, since the triangle ABC is right-angled, AC2=AB+BC2; therefore ABC - ADB+BFC. From these equals take away the segments AGB, BHC, and we shall have the triangle ABC = the sum of the lunes ADBG and BFCH.

1

(99.) PROBLEM. To divide a given circle into any number of parts, which shall be equal in area and equal in perimeter, and not have the portions in the form of sectors.

Let AB be the diameter of the circle, and suppose we wish to divide it into four parts. Divide the diameter into four equal parts at the points C, D, F. Then describe the semicircles upon the opposite sides of the different segments of the diameter, as exhibited in the diagram. Now if we suppose the diameter to be effaced, the four portions will fulfil the conditions required.

A

B

D

3

Semicircles are to each other as the squares of their diameters (B. V, Prop. XIII). Representing the semicircle described on BC as a diameter by 1, it follows that the one described on the diameter BD will be represented by 4, the one described on BF will be represented by 9, while the entire semicircle described on AB will be denoted by 16 hence the spaces above the diameter AB, as well as those below, will be to each other as the numbers 1, 3, 5, 7. Now when the diameter is supposed to be removed, the portions which are thus united will, in each case, be represented by 8, and therefore they are all equal.

The sum of the semicircumferences described on BC and CA is

equal to the semicircumference on AB (Art. 97), and the same for the semicircumferences described on the other segments into which the diameter AB is divided; hence each portion has for its perimeter an entire circumference.

(100.) PROBLEM. It is required to find what part of the diameter of a grindstone a given number of individuals must respectively grind off, so that they shall receive equal portions of the stone.

Let AB be the diameter of the stone, and suppose it is to be equally shared among four individuals.

Divide the radius AC into four equal parts; then upon BD, BF, BG, describe semicircumferences. Draw the radius CH at right angles to the diameter AB; and the points K, L, M, thus determined, will be the points to which they must grind respectively.

We have (B. IV, Prop. xvш, Cor. 1)

CM2

BC X CD,

CL2 = BC X CF,

CK1 = BC X CG, and

CH2= BC X CA;

therefore CM2, CL2, CK2, CH2 are to each other as CD, CF, CG, CA, or as 1, 2, 3, 4. Hence the areas of the circles whose radii are CM, CL, CK, CH are to each other as 1, 2, 3, 4 (B. V, Prop. x); therefore the stone has in this way been divided equally among four individuals. A similar method would apply for a greater number of divisions.

If we denote the radius of the stone by R, and the number of individuals by n, we shall have

BOOK SIXTH.

DEFINITIONS.

1. The common section of two planes is the line in which they meet to cut each other.

2. A line is perpendicular to a plane, when it is perpendicular to any two lines in that plane which meet it. 3. One plane is perpendicular to another, when every line in the one which is perpendicular to their common section is perpendicular to the other plane.

4. The inclination of two planes to each other, or the angle they form between them, is the angle contained by two lines drawn from any point in the common section, and at right angles to the same, one of these lines in each plane.

5. A line is parallel to a plane, when, if both are produced to any distance, they do not meet; and, conversely, the plane is then also parallel to the line.

6. Two planes are parallel to each other, when, both being produced to any distance, they do not meet.

7. A solid angle is the angular space included between three or more planes which meet at the same point.

PROPOSITION I.

THEOREM. One part of a straight line cannot be in a plane, and another part out of it,

For (B. I, Def. VII), when a straight line has two points common with a plane, it lies wholly in that plane.