# Show that if two triangles built on parallel lines, with equal bases have the same perimeter only if they are congruent.

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.

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?

• But you have gone to the case AC<A'C', so from BC=B'C' you don't get congruent. – coffeemath Apr 6 at 19:10
• Yeah, sorry. Let's just say that |A'C'| + |B'C'| > |AC| + |BC| for that one too, which is false. – Ban Apr 6 at 20:21
• If $BC=B'C'$ and $AB=A'B'$ then the perimeters are not equal. I hope someone posts an elementary solution not involving conics or functions. – Paracosmiste Apr 7 at 17:51

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 $$AB$$.

• This argument is clear except between the two right-angled cases, when $\angle ABC$ or $\angle CAB$ is a right angle. Then it's not clear whether the perimeter is increasing or decreasing, as one of the sides is getting longer while the other is getting shorter. Calculus or algebra (reduction to quadratic) could be used here. As such, this is only a partial solution – man and laptop Apr 7 at 14:38
• @manonlaptop I thought about that case. It's intuitively clear that at any point the longer leg is changing faster than the shorter leg. (As you commented, you only need that observation when the triangle is acute.) You could provide a rigorous algebraic proof. I'm sure you could do it synthetically with Euclid. – Ethan Bolker Apr 7 at 14:44

As an alternative proof, because the triangles are built on parallel lines, they have the same area. Using Heron's Formula $$A = \frac{1}{4}\sqrt{(AB + AC + BC)(-AB + AC + BC)(AB - AC + BC)(AB + AC - BC)}$$ 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 $$f(x) = \sqrt{1 + x^2} + \sqrt{1 + (1-x)^2} + 1$$.

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 $$BC+AC>BC'+AC'$$. $$~~~\square$$