The author posted a question in Homework
tell me how to solve the problem? and got a better answer
Response from Дима[+++]
Take any three points. They almost always form a triangle except when the three points lie on the same line. Suppose we take some three points and it turns out that they lie on the same line. Then by the axiom one can draw a plane through this line and the fourth point we did not take, and only one of them. Since the three points lie on a line and the line lies on the plane, all three points lie in this plane. Moreover, the fourth point also lies in this plane, so all four points lie in the same plane, we obtain the contradiction with the condition. So this case is impossible, so any three points are vertices of a triangle, what was required to prove?
Take any three points. They almost always form a triangle except when these three points lie on the same line. Suppose we take some three points and it turns out that they lie on the same line. Then by the axiom through this line and the fourth point not taken we can draw a plane and only one. Since three points lie on a line and the line lies on the plane then all three points lie in this plane. Moreover the fourth point belongs to this plane too, so all four points belong to the same plane, we obtain the contradiction with the condition. So this case is not possible, so any three points are vertices of the triangle.
We didn't get past that.
proof: since the points lie in the same plane you can draw three straight lines between them which lie in the same plane=> h., etc.
I don't know any geometry.
I think there was a theorem that you can draw a plane through any 3 points and only one plane and hence. you get a triangle.
If you connect any three points on the same plane you get a triangle - some kind of theorem