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THE place of all points satisfying a given condition is called the locus of points satisfying that condition.

Indeed, the word locus (Latin) means simply place (English, locality, locate, etc.); the plural is loci.

For example, The locus of points at a given distance r from a fixed point O is the circumference described about O with a radius r. This statement, although very evident, is made a theorem (28) because of the frequent reference to it.

In proving a theorem concerning the locus of points it is necessary and sufficient to prove two things:

1. That any point on the supposed locus satisfies the condition; 2. That any point not on the supposed locus does not satisfy the condition.

For if only the first were proved, there might be some other line in the locus; and if only the second were proved, the supposed locus might not be the correct one.

Three or more lines which meet in a point are said to be

concurrent.

Three or more points which lie in a line are said to be collinear.

EXERCISES. 210. Through a given point P in an angle AOB to draw a line, terminated by OA and OB, and bisected at P.

[Through P draw

a to BO cutting OA in X; on XA lay off XY = OX; draw YP.]

211. XX', YY', are two given lines through O, and P is a given point; through P to draw a line to XX', which shall be bisected by YY'. Investigate for various positions of P.

212. To construct a right-angled triangle, given the hypotenuse and one side.

213. To construct a quadrilateral, given the four sides and either interior diagonal.

Theorem 28. The locus of points at a given distance from a given point is the circumference described about that point as a center, with a radius equal to the given distance.

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2. Let OP produced, OP, meet C in B, A.

2

3. Then OP=OB=OA=r.

4. And OP2 < OB, and OP,> OA.

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Def. O, cor. 2

Ax. 8

5. .. any point on C is r distant from O, and any point not on C is not r distant from O.

Theorem 29. The locus of points equidistant from two given points is the perpendicular bisector of the line joining them.

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Theorem 30. The locus of points equidistant from two given lines consists of the bisectors of their included angles.

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Since Q may be moved, P may be considered as

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5. .'. any point P on XX' (and similarly on YY') is

equidistant from OA and OB, but any point Q on neither XX' nor YY' is unequally distant from OA and OB.

COROLLARIES. 1. If the given lines are parallel, the locus is a parallel midway between them. (Prove it.)

The student should imagine the effect of keeping points A, A' fixed, and moving O farther to the left. YY' is moved to the left, but XX' keeps its position as the lines approach the condition of being parallel.

2. The locus of points at a given distance from a given line consists of a pair of parallels at that distance, one on each side of the fixed line. (Prove it.)

Theorem 31. The perpendicular bisectors of the three sides of a triangle are concurrent.

Given a triangle of sides a, b, c, and x, y, z their respective perpendicular bisectors.

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a

Χ

Th. 17, cor. 4

Proof. 1. x and y must meet as at P.

2. Then P is equidistant from B

and C, and C and A.

3... P is on the perpendicular bisector of c.

4. I.e. z passes through P.

B

Th. 29

Why?

COROLLARIES. 1. The point equidistant from three noncollinear points is the intersection of the perpendicular bisectors of any two of the lines joining them.

Step 2.

2. There is one circle, and only one, whose circumference passes through three non-collinear points.

For, in step 1, x and y can intersect but once (post. 1), and A, B, C are equidistant from that point (step 2), and not from any other point, since x, y, z are the loci of points equidistant from B, C, and C, A, and A, B, and no other point can be on all three.

3. Circumferences having three points in common are identical.

Otherwise cor. 2 would be violated.

4. If from a point more than two lines to a circumference are equal, that point is the center of the circle.

For suppose a circumference through A, B, C, and suppose PA PB=PC. Now with center P and radius PA a circumference can be described through A, B, C.

· PA

PB = PC. (Def. O, cor. 4.)

And this is identical with the given circumference.

(Th. 31, cor. 3.)

.. its center must be identical with the given center, since a cannot

have two centers. (Def. O, cor. 7.)

Theorem 32. The bisectors of the interior and exterior angles of a triangle are concurrent four times by threes.

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Given

the ▲ ABC, and the bisectors of the interior and exterior angles, lettered as in the figure.

To prove

that these bisectors are concurrent four times by threes; that is, 3 meet at P1, 3 at P2, 3 at Ps, 3 at P4, as lettered in the figure.

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3. HBM+Z BAE <Z CBM+≤ BAC <180°.

4. .. BH and AE meet as at P1.

5. BF BH, and AG AE.

6. .. BF and AG meet as at P4.

Th. 19

Th. 17, cor.

3

Prel. th. 9

Th. 17, cor. 4

7. Also P1 is equidistant from a and c, from c and b,

1

and .. from a and b.

8... P1 lies on CT. Similarly for P4.

1

Th. 30

Th. 30

9. Similarly P2 and P3 lie on CN. .. the four points,

P1, P2, P3, P4, are points of concurrence of the bisectors.

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