Show that if two triangles built on parallel lines, as shown above, with |AB|=|A'B'| have the same perimeter only if they are congruent.
I've tried proving by contradiction:
Suppose they are not congruent but have the same perimeter, then either |AC|$\neq$|A'C| or |BC|$\neq$ |B'C'|. Let's say |AC|$\neq$|A'C'|, and suppose that |AC| $\lt$ |A'C'|.
If |BC|=|B'C'| then the triangles would be congruent which is false from my assumption.
If |BC| $\gt$ |B'C'| then |A'C'| + |B'C'| $\gt$ |AC| + |BC| which is false because their perimeters should be equal.
On the last possible case, |BC|$\gt$|B'C'| I got stuck. I can't find a way to show that it is false.
How can I show that the last case is false?