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Prove that if Triangles ABC = DEF in a metric geometry, then line AB contains exactly two of the points D, E, and F.

We are not allowed to use the facts: In a metric geometry, if triangles ABC=DEF, then {A,B,C} = {D,E,F}

or

In a metric Geometry, if A, B, and C are not collinear, then A is an extreme point of triangle ABC.

Naturally, I tried to do this with a proof by contradiction by assuming line AB does NOT contain exactly two of the points D,E, or F. I set up three cases for this.

1 Line AB contains none of D,E, or F; 2 Line AB contains one of D, E, or F. (Where I'd imagine the same contradiction will arise no matter which point I choose); 3 Line AB contains all three of D, E, and F.

In the first two cases, I can't find where a contradiction will arise. Please Help.

Some Definitions

The definition of a triangle that I am given is If {A,B,C} are not collinear points in a metric geometry, then the triangle ABC is the set $\bigtriangleup$ABC = $\bar{AB}$ $\cup$ $\bar{BC}$ $\cup$ $\bar{CA}$

The definition of a line segment I am given is $\bar{AB}$ = {C $\in$ $\mathscr{P}$ | A-C-B or C=A or C=B}

The definition of between I am given is A-C-B iff A, B, C are collinear and d(A,B) = d(A,C) + d(C,B); where d is the distance function for a metric geometry.

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What definition of "triangle" do you have? –  Najib Idrissi May 27 '13 at 17:15

1 Answer 1

I believe this is a valid proof, however, I used a proposition that is not obvious (I think). ** will denote when I used this assumption. I will prove it at the end.

Prove that if $\bigtriangleup$ABC=$\bigtriangleup$DEF in a metric geometry, then $\overleftrightarrow{AB}$ contains two of the points D, E, or F.

Proof I will prove this by contradiction. This can be split into three cases. $\overleftrightarrow{AB}$ contains none of D, E, and F; one of D, E, or F; all three of D, E, and F.

Suppose $\overleftrightarrow{AB}$ contains one of D, E or F. Clearly $\overleftrightarrow{AB}$ cannot intersect any other side at any other point. If $\overleftrightarrow{AB}$ intersects an adjacent side to the vertice that it contains, then the $\overleftrightarrow{AB}$ and the adjacent side are equal because they intersect at two points. If $\overleftrightarrow{AB}$ intersects the opposite side of the contained vertice, then there is a point on either side of the intersection that cannot belong to any other side or else $\overleftrightarrow{AB}$ would be equal to that side. In any case, $\overleftrightarrow{AB}$ contains more than one of D, E, or F.

Now suppose $\overleftrightarrow{AB}$ contains all three. Then D, E, and F are collinear and $\triangle$DEF is not a triangle.

Now suppose $\overleftrightarrow{AB}$ doesnt contain any of D, E or F. This means either $\overleftrightarrow{AB}$ doesnt intersect $\triangle$DEF which would mean that $\triangle$ ABC $\neq$ $\triangle$DEF since A, B are not in $\triangle$DEF. If $\overleftrightarrow{AB}$ intersects any of the sides between the vertices then we could at least 3 points on either side of the intersection that couldn't possibly be contained by the remaining sides of $\triangle$ DEF since one side would have to contain two, thus equating the lines, and thus $\overleftrightarrow{AB}$ would contain at least two of D, E or F. This is a contradiction.

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