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?
Best Answer
Fix $ A' $ and $ B' $. As $ AB = A'B' $ is fixed, the points $ C' $ for which $ ABC $ has the same perimeter as $ A'B'C' $ are the points for which $ AC + BC = A'C' + B'C' $. You recognize here the definition of an ellipse of focus $ A' $ and $ B' $. Hence the locus of $ C' $ is an ellipse. Finally, $ C' $ is in the meantime on an ellipse and on a line. These two have two intersections which give the directly and indirectly congruents triangles.
The easiest way to uncover your last case is using the ellipse argument.