Trjtdtk

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?

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.

Imagine that $AB$ is fixed on the bottom line and $C$ varies from way off left to way off right. The perimeter of the triangle is a decreasing function until triangle $ABC$ is isosceles, then increasing. It's clear from the symmetry that it takes on every value greater than its minimum value at just two points symmetrical with respect to the perpendicular bisector of $AB$.

As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula
$$
A = frac14sqrt(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)
$$
and a bit of algebra, you can show that either $AC = A'C'$ and $BC = B'C'$ or $AC = B'C'$ and $BC = A'C'$. In both cases $ABC cong A'B'C'$.

Let's say the distance between the two lines is $1$. Put an $x$ axis on the lower line, and a $y$ axis through the first point in the triangle. This ensures that the bottom left point of the triangle has coordinate $(0,0)$. Place the $(1,0)$ coordinate on the bottom right corner of the triangle.

If the third point of the triangle is on $(x,1)$, then the diameter is $$f(x) = sqrt1 + x^2 + sqrt1 + (1-x)^2 + 1$$.

Observe that $f(x)$ has a line of symmetry at $x=0.5$. In other words, if you do the substitution $u = 1-x$ you get the same function.

Next observe by plotting or by differentiation that the function is monotonically decreasing when $x < 0.5$ and increasing when $x > 0.5$.

By the previous paragraph, if a triangle exists with a certain diameter, at most only one other triangle can exist with that diameter. Moreover, the paragraph previous to that says that this other triangle can be reflected at the line $x=0.5$ to yield the first.

Q.E.D.

This is the simplest answer; it doesn't use any calculus or conics.

Let $A^*$ be the symmetric of $A$ with respect to the line passing through $C$. $BC+AC=BC+CA^*$ which is minimum iff $CA=CB$. If $C'$ is another point on the segment between $C$ and line $BA^*$, we can show (using this answer) that $BC+AC>BC'+AC'$. $~~~square$