Betweenness and the Crossbar Theorem. Lemma: Let A, B, and C be distinct points. If ABC, then neither ACB nor BAC.

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1 Betweenness and the Crossbar Theorem Lemma: Let A, B, and C be distinct points. If A*B*C, then neither A*C*B nor B*A*C. Suppose that both A*B*C and A*C*B. Thus AB+BC =AC, and AC +CB = AB. From this we get AB + BC +CB = AB, or in other words, AB + 2BC = AB, so BC=0, a contradiction. An exactly analogous proof works if we assume B*A*C. Theorem (Betweenness Theorem for Points): Let l be a line with distinct points A, B, and C. Let a, b, and c be the coordinates of points A, B, and C, respectively, under a coordinate function for l. Then A*B*C iff or. Suppose or. If, then,, and. Then:. Therefore, A*B*C. If, then,, and, and the same results follow. For the second half of the proof, suppose A*B*C. If either or, then part 1 of our proof would give B*A*C, a contradiction to the Lemma. Similarly, if or, we would have B*C*A, again a contradiction. The only possibilities left are or. Corollary: Let A, B, and C be three points such that B lies on. Then A*B*C if and only if AB < AC. (Mistake in Book) Let a, b, and c be the ruler coordinates for points A, B, and C, respectively. If A*B*C, then either or. If, then, so AB < AC. If, then, so and again we have AC > AB. For the second half of the proof, assume AB < AC. Since B lies on, by definition either A*B*C or A*C*B. If A*C*B, then AC +CB = AB, which is impossible since AB <AC and all distances are positive. Thus A*B*C.

2 Corollary: If A, B, and C are three distinct collinear points, then exactly one of them lies between the other two. There is a coordinate function for the line that points A, B, and C lie on. Their coordinates are distinct, and can be ordered. Thus exactly one of the three coordinates lies between the other two, and so exactly one of the points lies between the other two. Definition: Let A and B be two distinct points. The point M is a midpoint for segment if A*M*B and AM=MB. Theorem: If A and B are distinct points, there exists a unique point M such that M is the midpoint of. Let a and b be the ruler coordinates of A and B, respectively, on line. Use the ruler placement theorem to make a=0 and b>0. By the rule postulate there is a point on corresponding to coordinate. Let that point be M. Since we have A*M*B. Also,. Note that this is a typical use of the Ruler postulate to get points anywhere you want on a line, say a given distance from one end of a segment. This allows you to get a copy of one segment on another segment or on a line. You can trisect a segment, divide it into fifths, find a point away from a given point, etc.

3 Relationships Between Betweenness and PSP Theorem: Let l be a line, A be a point on l, and B be a point external to l. If A*C*B, then B and C are on the same side of l. Given a line l, and points A, B, and C with. For contradiction, suppose that B and C were on different sides of l. By PSP, l would have to intersect the segment. But since A*C*B, A is not between C and B, so A is not on the segment. Since l and intersect at only one point (namely, A), l cannot intersect on the same side of l., a contradiction. Thus B and C are Corollary: Let l be a line, A a point on l, and B a point external to l. If C is a point on and C A, then B and C are on the same side of l. Given a line l, and points A, B, and C with. Since, and C is not A, either C = B, A*C*B, or A*B*C. If C = B then the conclusion follows immediately. If A*C*B, the conclusion follows from the previous theorem. If A*B*C, the conclusion follows from the previous theorem as well, with the roles of B and C reversed. Note: These two results essentially say that if a segment or a ray has an endpoint A on a line l, then the entire segment or ray lies on one side of the line. Corollary: If a point D is in the interior of angle BAC, then the entire ray (except for point A) lies in the interior of BAC. Thus betweenness for rays is well-defined, i.e., it does not depend on the particular point D chosen to name the ray. Since D is in the interior of BAC, D is on the B-side of the C-side of. The above corollary guarantees that the entire and ray (except for A) is then on the B-side of and the C-side of, and therefore lies in the interior of of BAC.

4 Useful Little Corollary: Given angle BAC, and a point D interior to, D is interior to BAC. The first corollary above implies that D and C are on the same side of and that D and B are on the same side of, which is the definition of D being interior to the angle. Note that this implies that any crossbar of an angle lies in its interior (with the exception of its endpoints). Corollary (The Z Theorem): If points B and D lie on opposite sides of line l and distinct points A and C lie on line l, then rays have no points in common. and B and D are in separate half-planes, and by the previous corollary all of, except for A, lies in the same half-plane as B. Similarly, all of lies in the same half-plane as D. Since the two half-planes are disjoint, and since A and C are distinct, the two rays can have no points in common. Note: Because the rays have no points in common, neither do the appropriate segments that are subsets of the rays, such as and. We will use the theorem for both rays and segments as needed without further comment.

5 Useful Corollary Not in the Book (The X-Theorem): Suppose distinct points A and C are on line l and points B and D are not on l. If and intersect, then B and D are on the same side of l. Note that if and intersect, it will not be at points A or C. By the above corollary all of as B. Similarly, all of, except for A, lies in the same half-plane lies in the same half-plane as D. If point E (distinct from A or C) is in both and, then it must be in the same half-plane as D and in the same half-plane as B. Thus E, B, and D are all together in the same half-plane.

6 Theorem: Let A, B, and C be three noncollinear points and let D be a point on the line. The point D is between the points B and C if and only if the ray is between the rays and. Let A, B, and C be three noncollinear points and let D be a point on the line. Suppose that B*D*C. Then C and D are on the same side of line, and B and D are on the same side of. But this is just the definition of D being in the interior of BAC, which is the definition of being between the rays and. For the converse, suppose that ray is between the rays and. This means that D must be in the interior of BAC. So D is on the C side of and on the B side of. Since D and C are together on one side of, the line cannot intersect the segment, so no point of the line (in particular, point B) is between D and C. Similarly, since B and D are on the same side of, C cannot be on the segment, and C is not between B and D. Since B, C, and D are collinear, and one point must be between the other two, we are left with B*D*C.

7 Lemma: If A, B, C, and D are four distinct points such that C and D are on the same side of and D is not on, then either C is in the interior of BAD or D is in the interior of BAC. We will prove the or statement by assuming one case is not true and proving the other must be. So, assume D is not in the interior of BAC. We need to show C is in the interior of BAD. We already know C is on the D-side of that C is on the B-side of. Since C and D are on the same side of the B-side of and D are on opposite sides, the line, so our goal in life is now to show, it must be that D is not on (otherwise it would be interior to BAC). Since B must intersect the segment at a point we will call C. Thus the rays and intersect, and the X Theorem guarantees that C and B are on the same side of.

8 Theorem (The Betweenness Theorem for Rays): Let A, B, C, and D be four distinct points such that C and D lie on the same side of. Then µ( BAD) < µ( BAC) if and only if is between rays and. To prove one direction, assume is between rays and. Then D is in the interior of BAC. Thus, we have. Since we must have. To prove the other direction, we use the contrapositive. Assume is not between rays and. Our goal is to prove that. If it happens that D is on ray, then, and we are done. Otherwise, since D is not interior to BAC, the above lemma gives us that C is interior to BAD. In this case, we do something that feels like cheating, but isn t. We duplicate the first part of the proof with the roles of C and D exchanged, since is now between rays and. The result is that we have. This completes the proof of the contrapositive.

9 Definition: Let A, B, and C be three noncollinear points. A ray is an angle bisector of BAC if D is in the interior of BAC and µ( BAD) = µ( DAC). Theorem: If A, B and C are three noncollinear points then there exists a unique angle bisector for BAC. Exercise

10 Theorem (The Crossbar Theorem): If a point D lies in the interior of BAC, then ray meets segment at some point G interior to. Note: This is the first of our theorems that isn t pretty straightforward. We will prove it by extending the ray to a whole line and applying Pasch; but in order to do this we need to have the line go through the interior point of a segment; so we must also extend the side of the angle and create a triangle with as one side. Use the ruler postulate to find points E and F with E*A*B and F*A*D. For convenience, call the line l. Since D is interior to BAC we know that D does not lie on line or on line. Thus line l cannot intersect point C, B, or E. If it did, then it would have to be one of these two lines that cannot contain D. Moreover, since E*A*B we know A is an interior point of and l intersects at an interior point. Thus, Pasch applies to points E, C, and B and to line l. We know that l intersects either or. We must show three things to complete the proof:

11 1. is the part of l that meets either or ; that is, a) does not meet, and b) does not meet, and we show 2. does not meet. We tackle each in turn: 1a: Since F*A*D, F and D are on opposite sides of. Since D is interior to A, D is on the C-side of, so C and D are on the same side of. Thus C and F are on opposite sides of, and so by the Z Theorem, does not meet. 1b: We just showed that C and F are on opposite sides of, so another application of the Z Theorem gives us that does not meet. 2: Because E*A*B, E and B are on opposite sides of. Because D is interior to A, D is on the B-side of. Thus, D and B are on the same side, and E on the opposite side of. Since E and D are on opposite sides of, the Z theorem once again gives us that does not meet. The only remaining possibility is that ray meets segment at some point G interior to. T. I. J* * Stands for That s It, Jack! Signifies the end of a proof.

12 The book points out that the Crossbar Theorem and the theorem proved above immediately after the X Theorem can be combined into a single theorem called the Continuity Theorem. Of course, from it you can prove another theorem that was adopted as an axiom in Birkhoff s system.

Transformations Part If then, the identity transformation.

Transformations Part If then, the identity transformation. Transformations Part 2 Definition: Given rays with common endpoint O, we define the rotation with center O and angle as follows: 1. If then, the identity transformation. 2. If A, O, and B are noncollinear,