Question

I am submitting this for the second time, because the person that answered the first time definitely did not read my last two paragraphs. I need to DISCUSS, why the theorem works when adding a point D. I cannot draw a triangle and throw point D on it. I understand up to a certain point, but I have no idea what my professor is looking for in my answer. I have included her comments to hopefully help you help me. Thank you.

My question is based on the following:

Consider the axiomatic system and theorem below:

Axiom 1: If there is a pair of points, then they are on a line together.

Axiom 2: If there is a line, then there must be at least two points on it.

Axiom 3: There exist at least three distinct points.

Axiom 4: If there is a line, then not all of the points can be on it,

Theorem 1: Each point is on at least two distinct lines.

I have proven and understand up to 3 points, but I am struggling with explaining what happens with the 4th point.

If I use Axiom 3 to create 4th point D (the first 3 being A, B, and C), this will give me distinct lines AD, BD, CD, ADB, ADC, and BDC.

ADB, ADC, and BDC were all previously existing lines, and AD, BD, and CD are new lines, correct?

These are two separate cases because I cannot have point D on a previously existing line, and a new line, at the same time. I feel I understand up to this point. I need to discuss these two possibilities separately, but I am confused on how to go about that.

My professor states that I need to "discuss the different possibilities for how many distinct lines those are, we do not know if those are 3 distinct lines or not, this is where the different cases come in". I don't understand AT ALL what she is looking for.

Answer 1

Please think about a square or rectangle.

Explanations : According to Axiom 1 - these pair of points are on the edges of a line together and these points are connected with two another sides of the square or rectangle.

According to Axiom 2 - this is simply saying that 2 points are existing on any 1 sides(lines) of the square or rectangle 1 point at the starting edge and another point will be on the ending edge of the side(line).

According to Axiom 3 - In a square or rectangle Atleast 3 distinct points are obvious on three corners.

According to Axiom 4 - you can understand this as all the points in a rectangle or square doesn't exist on only one line, as a they have 4 sides so all the points can not exist on a single line.

According to Theorem 1 - On the four corners of the rectangle or square there is 1 obvious point and two distinct sides(lines) are drawn through this point so this point will act as edge of two distinct lines.

So in my view a square or rectangle is the solution for this question.