distance between their centres 20. To draw a third circle whose radius shall be 8, and which shall be tangent to the two given circles? Can a third circle whose radius is 2 be drawn tangent to the two given circles? How will it be situated? Can one be drawn tangent to the given circles, whose radius shall be 1? Why?

SYNOPSIS.

PROP. I. Through three points.

Cor. 1. A circf. can be passed.
Cor. 2. A circf. determined by.
Cor. 3. Intersections of two circf's.

PROP. II. Two circumferences which intersect in one point.

PROP. III. Points in one circumference nearest to and farthest from the centre of another.

RELATIVE POSITIONS OF CIRCUMFERENCES.

DISTANCE BETWEEN CENTRES
OF TWO CIRCLES.

PROP. IV. Greater than sum of radii. { Cor. Converse.

PROP.

PROP.

Cor. 1. Converse.

V. Equal to sum of radii. Cor. 3. Point of tangency.

VI. Less than sum and
difference of radii.

PROP. VII. Equal to diff. of radii.

PROP. VIII. Less than diff. of radii.

PROP. IX. Perpendicular to common chord.

PROP. X. Common tangent to two circles tangent to each Cor. To all

other.

EXERCISES. Prob. To circumscribe a triangle with a circumference.
To pass circumference through three points.

SECTION VI.

OF THE MEASUREMENT OF ANGLES.

200. Angles are said to be measured by arcs, according to the principles developed in the three following propositions.

PROPOSITION I.

201. Theorem.-In the same or in equal circles, equal arcs subtend equal angles at the centre.

DEM.-In the equal circles M and N, let arc AB = arc DC; then will the

202. COR. 1.-Conversely, In the same or in equal circles, equal angles at the centre intercept equal arcs.

DEM.-If, by hypothesis, angle O' = angle O, in the equal circles M and N, arc DC arc AB. For, placing circle N upon M, so that O' shall fall on O, and O'D on its equal OA, D falls on A, and, since angle O' = angle O, O'C takes the direction OB, and, being equal to it, C falls on B. Hence, DC and AB coincide and are equal.

203. COR. 2.-A right angle at the centre intercepts a quarter of a circumference, and is said to be measured by it. Hence, a semicircumference is the measure of two right angles, and a whole circumference of four.

PROPOSITION II.

204. Theorem.—In the same or in equal circles, arcs which are in the ratio of two whole numbers subtend angles at the centre which have the same ratio, whence the angles are to each other as the arcs which subtend them.

DEM. In the equal circles M and N, let the arcs EF and IH, which subtend the angles O and O' at the centre, be in the ratio of 5 to 8; then are the angles O and O' in the ratio of 5 to 8, and we have

angle: angle O' :: arc EF: arcIH.

For, divide EF into 5 equal parts, as Ea, ab, etc., then IH can be divided into 8 such parts, le, ef, etc. Draw the radii Oa, Ob, Oc, etc., and O'e, O'f, O'g, etc.; and, since these partial arcs

are equal, the partial angles

M

E a b

FIG. 153.

N

H

which they subtend are equal,

by the preceding proposition. Now, O is composed of 5 of these angles, and O' of 8; whence

angle angle O' :: 5 : 8. But, arc EF : arc IH :: 5 : 8.

Hence, the two ratios being equal, we have

angle O' angle O arc IH: arc EF.

As the same method could be pursued in case the arcs were to each other as any other two whole numbers, the argument is general.

205. COR.-Conversely, In the same or in equal circles, angles at the centre which are in the ratio of two whole numbers are to each other as their intercepted arcs.

DEM. Thus, let angle O' be to angle O in the ratio of 8 to 5. Conceive O' divided into 8 equal partial angles, then will O be divisible into 5 such partial angles. Now, the partial angles being equal, their intercepted arcs are equal, by the preceding proposition, Cor. 1. Whence,

Hence, arc IH: arc EF :: angle O': angle O.

And the same method could be pursued with angles having the ratio of any other whole numbers.

PROPOSITION III.

206. Theorem.-In the same circle or in equal circles, two in

If that arc is not IH let it be IL, an arc less than IH, so that

angle angle O' :: arc EF : arc IL.*

:

Conceive EF divided into equal parts, each of which is less than LH, the as sumed difference between IH and IL. Then conceive one of these equal parts to be applied to IH as a measure, beginning at I. Since the measure is less than LH, at least one point of division must fall between L and H. Suppose K to be such a point. Draw O'K. Now, the arcs EF and IK are commensurable, and by the last proposition

angle : angie IO'K :: arc EF : arc IK. But we assumed that

angle angle IO'H :: arc EF : arc IL.

In these proportions the antecedents being alike, the consequents should be proportional, so that

angle IO'K should be to angle IO'H : : arc IK : arc IL.

But this proportion is false, since

angle IO'K < angle IO'H, whereas arc IK > arc IL.

In a manner altogether similar (the student should supply it) we can show that

angle O is not to angle O' :: arc EF : any arc greater than IH.

Hence, as the fourth term of the proportion cannot be less or greater than IH, it must be IH itself; and

angle Ọ: angle O' :: arc EF : arc IH. Q. E. D.

207. COR.-Conversely, In the same or in equal circles, two incommensurable angles at the centre are to each other as the arcs which they intercept.

*This is a false hypothesis, and the object of the argument following is to show its falsity.

+ This can be done by supposing EF bisected, then the halves bisected, then the fourth bisected, and this process of bisection continued until the parts are each less than LH.

DEM. In the equal circles M and N, O and O' being incommensurable angles at the centre, are to each other as the arcs EF and IH. If not, let us suppose

arc EF arc IH :: angle ✪ : angle IO ́L, an angle less than O'.

Divide O into equal partial angles, each less than LO'H, the assumed difference between 10'H and IO'L. Also conceive this angle to be applied as a measure to IO ́H, beginning at O'l. At least one line of division will fall between O'L and O'H. Let O'K be such a line. Now, as O and IO'K are commensurable, we have by (205),

But by supposition

arc EF : arc IK : : angle ✪ : angle IO’K.

arc EF arc IH: : angle ○ : angle IO'L.

Therefore, since the antecedents are the same,

arc IK should be to arc IH:: angle IO'K : angle 10'L.

But this is false, since

arc IK < arc IH, whereas angle IO'K > angle IO'L.

Whence we learn that the fourth term of the proportion cannot be less than angle IO'H. In a similar manner it can be shown (let the student do it) that it cannot be greater. Hence it must be O'H itself; and

arc EF arc IH: angle O: angle 10'H.

208. SCH.-Out of the truths developed in the three preceding propositions grows the method of representing angles by degrees, minutes, and seconds, as given in Trigonometry (PART IV., 3-6). It will be observed, that in all cases, if arcs be struck with the same radius, from the vertices of angles as centres, the angles bear the same ratio to each other as the arcs intercepted by their sides. Hence the arc is said to measure the angle. Though this language is convenient, it is not quite natural; for we naturally measure a quantity by another of like kind. Thus, distance (length) we measure by distance, as when we say a line is 10 inches long. The line is length; and its measure, an inch, is length also. So, likewise, we say the area of a field is 4 acres: the quantity measured is a surface; and the measure, an acre, is a surface also. Yet, notwithstanding the artificiality of the method of measuring angles by arcs, instead of directly by angles, it is not only convenient but universally used; and the student must know just what is meant by it. For example, a circumference is conceived as divided into 360 equal arcs, called degrees. Hence, as a right angle at the centre is subtended by one-fourth of the circumference, it is called an angle of 90 degrees. 180 degrees is the measure of two right angles, 45 degrees, of half a right angle, etc. Thus we get a perfectly definite idea of the