Answer to Question #135108 in Geometry for J

Question

Suppose that in a certain metric geometry satisfying Axioms D − 1 to D − 4, points A, B, C, D are collinear and
AB = 4 = AC , BC = 8 , BD = 3 , CD = 5 , AD = 1

What betweenness relations follow? ( Since the ruler postulate
follows, you can use a diagram.)

Answer 1

Consider the following Metric Axioms.

D-1: Each pair of points A and B is associated with a unique real number, called the distance from A to B, denoted by AB.

D-2: For all points A and B, AB  0, with equality only when A = B.

D-3: For all points A and B, AB = BA.

D-4: For all points BD-CD=BC or BC+CD=BD.

Suppose A, B, C and D are collinear

Also, AB = 4 = AC, AD = 6, BC = 8, BD = 9, and CD = 1.

The objective is to show that there is a betweenness among the points.

AB = 4 = AC, BC = 8 confirms that B and C are on either sides of A and at equal distances.

Since BD = 9, and CD = 1, it trivially follows that D is close to C and farthest from B.

Use this information for betweenness as B – A – C – D.

Also, BC + CD = 9 = BD, BA + AC = 8 = BC confirms the details.