side of the line AB from C are all composed of 0 points and cannot meet the segment AC. The ray BD is composed of O points and does not meet the segment AC. The other rays whose origin is B, aside from BA and BC, meet one of the segments DC and CA. Those meeting the segment DC evidently are composed entirely of O points and those meeting the segment CA of I points. Hence the set of points [7] is composed of the points on the rays whose origin is B and which meet the segment AC.

Theorem 21. Definition. A triangle decomposes its plane into two regions one of which is convex and is called the interior. The other region is not convex

A ray

and is called the exterior.
whose origin is an interior point
meets the triangle in one and only
one point, and the interior con-
sists of all points having this
property.

Proof. Let the triangle be ABC and let [I] be the set of all points on the segments [AX] where [X] is the segment BC. By Theorem 17, any line through I except the line IX meets the triangles AXB and AXC each

A

in one point. But as B and C are on opposite sides of the line AX these points are one in each of the rays into which I decomposes the line. Hence every ray through I meets the triangle ABC once and only once.

A segment 1112 cannot contain a point of the triangle for then the ray II1⁄2 would contain at least two points of the triangle, one on the segment II2 and one on its prolongation beyond 12. Hence [1] is a convex region.

A segment one of whose end points is I must be composed entirely of I-points if it does not contain a point of the triangle; for if P is any point of the segment, the single point of the triangle on the ray IP is by hypothesis on the prolongation

of IP beyond P. From this it follows as in Theorem 19 that a broken line joining a point I to a point not in [] contains at least one point of the triangle.

A point E not in [I] or on the triangle ABC must either be exterior to BAC or on the prolongation of a segment AX beyond X. One of the latter points E is joined to a point E1 of the prolongation of AB beyond B by an interval which does not meet the triangle ABC because its ends are on the opposite side of BC from A. The prolongation of EE1 beyond E1 is composed of points in the exterior of 4BAC. Since any two points exterior to 4BAC are connected by a broken line not meeting BAC, it follows that any two points E are connected by a broken line. Hence E is a region. Since it contains points on the two prolongations of a segment AX it is not convex. Hence [I] and [E] satisfy the definitions respectively of the interior and the exterior of the triangle.

Corollary. Through each exterior point there pass lines which do not meet the triangle.

Theorem 22. Any ray BD in the interior region of 4ABC decomposes it into two convex regions, the interiors of the angles 4ABD, 4DBC. Any ray BD in the exterior region of 4ABC decomposes it into two regions at least one of which is convex. One of these is the interior or the exterior of ABD and the other the interior or the exterior of 4DBC.

B

Proof. To prove the first part of the theorem we observe that the ray BD meets the segment AC in a point P. Hence the points X on the rays joining B to the segment AP are on the opposite side of the line BD from the points [Y] on the rays joining B to the segment PC. But the sets [X] and [Y] are the angles 4ABD and DBC respectively. The proof of the second part is analogous to that just made. The details are left as an exercise for the reader.

FIG. 27.

Definition. If R is a region and there exists a set of points [B] not points of R such that every broken line joining a point of R to a point not in R contains a point B1, then [B] is called the boundary of R.

For example, a line is the boundary of each of the halfplanes it determines; an angle is the boundary of its interior and also of its exterior.

Definition. Two rays a, b, are separated by an angle 4hk if the four rays a, b, h, k have a common origin and one of a and b is interior while the other is exterior to Дhk. A set of rays having a common origin are said to be in the order {α1а 2а3α 4α5...an} if no two of the rays are separated by any of the angles а1а2, 4а2а3, . . ., 4an-1an, 4ana.

Corollary 1. A set of rays in the order (a1a2...an-1an} are also in the orders (a2a3...ana1} and {anan-1...a2a1}.

Corollary 2. Any two rays, a, b, having a common origin are in the orders {ab} and {ba}. Any three rays, a, b, c, having a common origin are in the orders {abc}, {bca}, {cab}, {acb}, {bac}, {cba}.

Theorem 23. To any finite number n>2 of rays having a common origin may be assigned notation so that they are in the order a1a2a3... an. They decompose the plane into n regions of which at most one is not convex.

Proof. The theorem is obvious for n=2.

Hence we can

prove it in general by showing that its truth for n=k implies its truth for n=k+1.

To k of the given k+1 rays let us assign notation so that they are in the order bib2... bk. They decompose the plane into k regions R1, R2, ..., Rk, whose boundaries are 4bb2, 4b2bз,..., 4bb1. The other ray, b, lies in one of the k regions determined by b1b2 . . . bk. By Theorem 22 it separates this region, R, into two regions R/R" of which one at least is convex if R; is not convex and both of which are convex if R; is convex. Hence the k+1 rays decompose the plane into k+1 regions R1, R2... R/R"... Rk of which at most one is not convex.

i

III

2

Suppose that the boundary of R is 4bbi+1. Then the boundaries of the two regions into which R; is decomposed are 4bb and bbi+1. Hence the +1 rays are in the order {b1, b2...b,bb1+1... b). By calling the first of these a1, the second a2, etc., we have assigned to them the order {αια2 . . . ακ+1}.

IV

II

VII

VI

FIG. 30.

EXERCISE 1. A set of n distinct coplanar lines meeting in a point. O decompose their plane into 2n convex regions.

EXERCISE 2. Three lines AB, BC, CA not meeting in a point decompose their plane into seven convex regions, one of which is the interior of the triangle ABC.

EXERCISE 3. A set of n lines in a plane each pair of which intersect, but no three of which pass through the same point, decompose their n(n + 1)

plane into

2

+1 convex regions.

Ba

We now introduce a new undefined term to express a relation between point pairs. The relation is called congruence. Denoting pairs of distinct points by (A, B), (C, D), etc., we write (A, B) is congruent to (C, D). Since this phrase is undefined the reader may attach to it any meaning consistent with the assumptions below. It is intended, however, to express the common notion implied by saying that the distance from A to B as measured by a tape-line is the same as the distance from C to D.

Assumption VII.* If AB then is C there exists one and only one is congruent to (C, D).

FIG. 31.

on any ray whose origin point D such that (A, B)

Assumption VIII.‡ If (A, B) is congruent to (C, D) and

B

A

B

FIG. 32.

8

(C, D) is congruent to (E, F) then (A, B) is congruent to (E, F).

Assumption IX.§ If (A, B) is congruent to (A',B') and (B, C) is congruent to (B', C') and ABC and A'B'C'} then (A, C) is congruent to (A', C′).

Assumption X. (A, B) is congruent to (B, A).

Theorem 24. If {ABC), and C' is a point on a ray A'B' such that (A, B) is congruent to (A'B') and (A, C) is congruent to (A'C') then {A'B'C'.

Proof. By Assumption VII there is a point C" on the ray B'C' such that BC is congruent to B'C". The point C" is in

Cf. Euclid, Postulate 3.

† Evidently there are two statements here, (1) the existence and (2) the uniqueness of D.

Cf. Euclid, Common Notion 1. § Cf. Euclid, Common Notion 2.