Prove that this is a rhombus.

Question

Show that a convex quadrilateral with vertices ABCD is a rhombus if the following triangles have equal perimeters: ABP, BCP, CDP, DAP. P is where the diagonals cross.

$AP = a$

$BP = b$

$CP = c$

$DP = d$

$AB + a + b = BC + b + c = CD + c + d = DA + d + a$

This means there are 6 equations:

1. $AB + a + b = BC + b + c$

$AB + a = BC + c$

2.

$BC + b + c = CD + c + d$

$BC + b = CD + d$

3.

$CD + c + d = DA + d + a$

$CD + c = DA + a$

4.

$DA + d + a = AB + a + b$

$DA + d = AB + b$

5.

$AB + a + b = CD + c + d$

6.

$BC + b + c = DA + d + a$

I was hoping that massaging these equations would lead to all sides being equal, or at least that a=c or b=d. But all that I was able to get was:

1.-3. :

$AB + CD = BC + DA$

No matter how I combine any of the equations, it all leads to this one. Well, except for these two:

$AB + a + b + BC + b + c = CD + c + d + DA + d + a$

$AB + a + 2b + c + BC = CD + c + 2d + a + DA$

$AB + 2b + BC = CD + 2d + DA$

$AB + a + b + DA + d + a = BC + b + c + CD + c + d$

$AB + a + b + DA + a = BC + b + c + CD + c$

$AB + 2a + DA = BC + 2c + CD$

But if I combine these with anything else it always leads to $AB + CD = BC + DA$

Answer 1

Try to prove that by exhaustion.

It is true that rhombuses meet the requirement. Squares, which are just particular rhombuses, can also do that.

Excluding the above, we then consider ‘non-trivial’ rectangles (say ABCD with $DA \ne AB$). Obviously, they will not satisfy the given conditions.

Next, we go on to investigate parallelograms. If rectangles cannot go through, parallelograms will also not go through for the same reason.

Then, kite, trapezoids, … , non-trivial quadrilaterals.

With respect to a non-trivial quadrilateral, the diagonals cut the figure into 4 triangles. Will the smallest triangle has the same perimeter as the largest?

Answer 2

Not my proof, but I like it.

Let $|AP|<|PD|$ and $|CP|<|PB|$. Then rotate both $A, C$ through $P$.

$\triangle A'PC'$ has the same perimeter as $\triangle DPB$, so $C'=B$ and $A'=D$. $P$ is then the midpoint of both the diagonals, and it follows that $$|AB|=|BD|=|DC|=|CA|$$