and circumscribed circles by r, R, respectively; the radius of the escribed circle touching a, and touching b and c produced, by ra, and similarly for г, .

DEFINITION. A triangle is said to be inscribed in another when its vertices lie respectively on the sides of the other.

EXERCISES. 356. To describe a circumference with a given radius, and (1) Passing through two given points. (Combine a and c.)

(a.)

(2) Passing through one given point and touching a given line. (a, b.) (3) Passing through one given point and touching a given circle. (4) Touching two given lines. (b.)

(5) Touching a given line and a given circle. (a, b.)

357. In a given triangle to inscribe a triangle with two of its sides given, and the vertex of their included angle given. (a.)

358. To describe a circumference passing through a given point and touching a given line, or a given circle, in a given point.

(c.)

359. On a given circumference to find a point having a given distance from a given line. (b.)

360. On a given line, not necessarily straight, to find a point equidistant from two given points. (c.)

361. Describe a circumference touching two parallel lines and passing through a given point. (d, a.)

362. Find a point from which two given line-segments are seen under (or subtend) given angles. (e.) (Pothenot's problem.)

363. Construct the triangle ABC, given a, ha, ma.

364. Also, given A, a, ha.

365. Also, given ▲ A, a, ma 366. Also, given a, hb, hc.

367. Also, given A, ha, Va.

(First construct the right-angled tri

angle with side ha and hypotenuse va.)

368. Also, given ha, ma, R. (First construct the right-angled triangle with side ha and hypotenuse ma; then find the circumcenter by a, c.) 369. Also, given a, R, hb. (First construct the right-angled triangle with side h and hypotenuse a; then find the circumcenter by a.)

370. Also, given c, r, A = 90°; c, rc, ≤ A = 90°; b, r, ≤ A = 90°; or b, rc, A = 90°.

371. Describe two circles of given radii r1, r2, to touch one another, and to touch a given line on the same side of it.

1

In

III. METHOD OF CONSTRUCTION OF LOCI BY POINTS. order to apply the second method, just given, it is necessary A few of the simpler ones have already

to know the loci. been stated and used. If the locus in question is not known, and if no indication of its form and position is seen, resort may be had to the finding of certain points in the locus; when several such points have been located, the form and position of the locus may often be inferred, after which the proof frequently presents little difficulty.

The investigation is often simplified by first seeking special points that are easily found, - remarkable points, as they are called. This will be seen in ex. 372.

EXERCISES. 372. Given the base and the vertical angle of a triangle, to find the locus of its orthocenter.

D

C is evidently on an arc passing through A and B. When C is at A, so are D and E, and hence P; similarly, when C is at B, so is P; when A ABC is isosceles with AC: = BC, then both C and P are evidently on the bisector of AB. Hence three remarkable points of the locus are fixed, and one would infer that it is an arc. If it is an arc, then APB is constant; likewise its equal likewise the supplement of DPE, or C (··· ZD = ZE = rt. Z ) ; but C is constant; .. the locus of P is an arc, which is the conjugate of the locus of C.

DPE, and

373. Given the base and the vertical angle of a triangle, to find the locus of its in-center.

374. Through the extremities A, B of a given line, any parallels, AX, BY, are drawn. Find the locus of the intersection of the bisectors of the angles BAX, YBA. (Take, for the first point, that found when AX LAB.)

375. Find the locus of the points of contact of tangents drawn from a fixed point to a system of concentric circles.

376. Two opposite vertices of a given square move on two lines at right angles to each other. Find the locus of the intersection of the diagonals.

377. Find the locus of the intersection of two lines passing through two fixed points on a circumference and intercepting an arc of constant length.

IV. METHOD OF PARALLEL TRANSLATION. This is one of several methods of simplifying a construction by rearranging the figures in such a way as to suggest a solution of the problem. It will best be understood by examining exs. 378 to 380. In considering the subsequent exercises, this method may or may not be used, as seems advisable.

EXERCISES. 378. To construct a trapezoid, given the four sides.

C

Y

a

a-c

Analysis. Assume the figure drawn. Then if d is translated parallel to itself and between c and a, to the position YZ, the XYZ can easily be constructed (I, pr. 7). The process may now be reversed and the trapezoid constructed. 379. To place a line so that its extremities shall rest upon two given circumferences, the line being equal

and parallel to another line.

Analysis. If O and O' are the given circles, and AB the given line, and if OO' is translated along a line parallel and equal to AB, then either XY or X'Y' answers the conditions. Hence the process may be reversed; first describe OO", and then from Y, Y' draw YX and Y'X' and || BA.

Y

B

380. Given two parallels, XY, X'Y',

with a transversal WZ limited by XY and X'Y'; also two points A, B, not between the parallels, and on opposite sides of them. Required to join A and B by the shortest broken line which shall have MN, the intercept between XY and X'Y', parallel to WZ.

Analysis. If any MN in the figure is translated along NB parallel to its original position, until N coincides with B and M is at P, then AM1P< AM2P or AM3P (I, th. 8); hence AMINIB is the shortest broken line. Hence the process may be reversed;

M2

Mi

M1 W

X

IN

-Y、

B

first draw BP || and = ZW; then join A and P, thus fixing M1; and then draw M1N1 WZ.

381. Without stating any special theorem, the following facts concerning the quadrilateral are of value :

In the quadrilateral ABCD, suppose CD and DA translated, parallel to their original position, to C'B and BA'. Then AA'C'C contains the parts of ABCD, but arranged as follows:

1. The lines from B to the vertices of the

parallelogram equal the sides of ABCD.

2. The angles about B equal the angles of ABCD.

3. The sides of the parallelogram equal the diagonals of ABCD.

4. The angles of the parallelogram equal the angles formed by the diagonals of ABCD.

5. The angles formed by the sides of the parallelogram and the lines from B equal the angles between the sides and the diagonals of the quadrilateral.

6. The area of the parallelogram equals twice that of ABCD. 382. Through one of the two points of intersection of two circumferences to draw a line from which the two circumferences cut off chords having a given difference. (The projection of the center-segment on the required line equals half the given difference; hence translate this projection to the position OA; the right-angled ▲ OO'A can now be constructed, and the required line will be parallel to OA.)

383. In ex. 382, show that if the two chords lie on opposite sides of P, the sum replaces the difference.

384. In a given circle to draw a chord equal and parallel to a given line. 385. From a ship two known points are seen under a given angle; the ship sails a given distance in a given direction, and now the same two points are seen under another known angle. Find the positions of the ship. (On the line joining the known points, construct segments to contain the given angles; the problem then reduces to ex. 379.)

386. Construct a quadrilateral, given the diagonals, an angle at their intersection, and two opposite angles. (Ex. 381, statements 2 and 3; and III, th. 11, cor. 3.)

387. Construct a trapezoid, given the diagonals, their included angle, and the sum of two adjacent sides.

388. To construct a triangle, given A, a, and the foot of ha

389. Also, given A, a, and the orthocenter.

390. Also, given A, a, and the centroid.

BOOK IV.-RATIO AND PROPORTION.

Section 1.-Fundamental Properties.

INTRODUCTORY NOTE. The inference was drawn in Book II that a relation exists between algebra and geometry known as the Law of Homology, with the following correspondence :

And as it was assumed that a straight line may be represented by a number, so it may be assumed that any other geometric magnitude, such as an arc, a segment of a circle, an angle, the surface of a polygon, etc., may be represented by a number. With these assumptions, the fundamental properties of Ratio and Proportion may be proved either by algebra or geometry, as may be most convenient, the proof being valid for both of these subjects. The purely geometric treatment is too difficult for the beginner.

DEFINITIONS. To measure a magnitude is to find how many times it contains another magnitude of the same kind, called the unit of measure.

A ratio is the quotient of the numerical measure of one magnitude divided by the numerical measure of another magnitude of the same kind.

For example, the ratio of a line 8 ft. long to one 16 ft. long is or 1; that of one 16 ft. long to one 8 ft. long is 2.

The ratio of a to b is expressed by the symbols

a

ab, a/b, or a b. " b