Hence AB is perpendicular to the radii at their extremities, and is tangent to the circumferences.

Upon the other side of CC', a second exterior common tangent, A'B', may be constructed, which will meet AB in CC' produced; and the problem has two solutions.

If the circles are equal, draw the radii CA and C'B perpendicular to CC'.

AB will be the common tangent required.

2d, To draw an interior common

tangent.

SOLUTION. From the centre C, with a radius equal to the sum of the radii of the given circles, describe the circumference DD'. From C' draw the tangent C'D.

Draw CD and C′B || CD. ABC'D is a rectangle, since AD || C'B, and ADC'R: hence AB is the interior common tangent required.

There will be two interior com

B

D/

mon tangents, AB and A'B', which intersect each other in a point, E, of the line CC'.

SCHOLIUM. If the two circles touch each other externally, the two interior common tangents fall together. If the circles cut each other, only the exterior tangents can be drawn. If the circles touch each other internally, the two exterior tangents fall together. If the circles are wholly within each other, there is no solution.

XXXVI.

Problem. Upon a given line to construct a segment that shall contain a given angle.

Let AB be the given line, and A the given angle.

H

SOLUTION. At the point A construct the angle BAD = A.

E

F

B

G

Draw AC AD, and at E, the middle point of AB, draw EC | AB, intersecting AC in C. From C as a centre, with a radius CA, describe the circle ABF: then AFB is the segment required.

:

PROOF. AD is a tangent, since it is perpendicular to the radius CA (11, Cor. 7) hence any angle in the segment AFB = BAD = A; for they have the same numerical measure, arc AGB (11, Cor. 10).

If the given angle be obtuse, construct BAH = A: then AGB is the required segment.

EXERCISE 1. The greatest and the least distances of any point from a circumference is measured on the diameter passing through that point.

EXERCISE 2. If from a point in a circumference chords be drawn, the locus of their middle points is a circumference.

EXERCISE 3. In any inscribed polygon, the sum of the 1st, 3d, 5th, &c., angles, is equal to the sum of the 2d, 4th, 6th, &c., angles.

EXERCISE 4. In any circumscribed polygon, the sum of the 1st, 3d, 5th, &c., sides, is equal to the sum of the remaining sides.

EXERCISE 5. If two circles cut each other, and diameters are drawn through one of the cutting points, the extremities of those diameters are in the same straight line with the other cutting point.

EXERCISE 6. The lines bisecting the angles contained by the opposite sides (produced) of an inscribed trapezium intersect at right angles.

BOOK III.

RATIO AND PROPORTION.

A

B

I.

1. Four magnitudes, A and B of one kind, and C and D of the same or another kind, are proportional, when the ratio of A to B is equal to that of C to D; that is, when

D

A

B

=

C

D

(II., 7). Thus A and B may

be two lines, C and D, two triangles. The proportion is often written A: B = C: D, and read, A is to B as C is to D.

2. The first and fourth terms, A and D, are called the extremes; the second and third, B and C, the means: the first and third, A and C, the antecedents; the second and fourth, B and D, the consequents; and the fourth term, D, a fourth proportional to the other three. Three quantities, A, B, and C, are proportional, when A: B=B: C. B is called a mean proportional between A and C.

A C

3. Since the form A: BC: D is equivalent to B D' the following transformations may be made by simple algebraic operations. Multiplying both members by BD, we have A.D B.C.

=

Hence the product of the extremes is equal to the product

of the means.

A.C, or,

If A: B = B: C, we have B2 BA.C; that is, the mean proportional between two quantities is equal to the square root of their product.

EXERCISE 1. Find the mean proportional between 6 and 24; between 5 and 10.

SCHOLIUM. It is important for the student to remember, that, in these transformations, the terms A, B, C, and D, are the numerical representatives of the four magnitudes; that is, A and B are the numbers of times a given unit is contained in the first two; C and D, the numbers of times the same or a different unit is contained in the third and fourth respectively.

4. Dividing A.DB.C successively by B.D, C.D, and A.C, we have,

Hence, if the product of two quantities is equal to the product of two other quantities, one pair of factors may be made the extremes, and the other pair the means, of a proportion. EXERCISE 2. From A.DB.C, write eight proportions. 5. Since the proportions II. and III. are true when I. is true, it follows, that, if four quantities are proportional, the first is to the third as the second is to the fourth; that is, they are proportional by alternation. Also the second is to the first as the fourth to the third; that is, they are proportional by inversion.

6. By squaring, cubing, &c., and extracting roots of the

equation

A C

=

B D'

, or, A: B= C: D, we obtain the proportions,

A2 : B2 = C2 : D2, A3 : B3 = C3 : D3, √A : √B = √C : √D, &c.; that is, if four quantities are proportional, their like powers or roots are also proportional.

7. Since

A MA
B mB

it follows that A: B = mA : m B, where

m may be a whole number or fraction; that is, equimultiples, or like parts, of two quantities, have the same ratio as the quantities themselves.

Adding unity to both members of the equation, we have,

dividing this by the given equation, we have,

, or, A+B : B = C + D : D ;

Hence, if four quantities are in proportion, they are in proportion by composition.

9. Subtracting unity from both members of

C

A

=

we

B D'

, or, A-B: B=C-D: D,

, or, A-B: A=C-D: C.

Hence, if four quantities are in proportion, they are in proportion by division.

10. Dividing the first equation in (8) by the first in (9) we obtain

or A+B: A-B=C+D: C-D.

Hence, if four quantities are in proportion, they are in proportion by composition and division.