# Prove that the diagonals of this shape are perpendicular and equal (Quadrangle with an isosceles right angled triangle on each side)

This is the question:

Given any negatively oriented quadrangle $ABCD$, and isosceles, right-angled, positively oriented triangles $ABX, BCY, CDZ, DAV$, with right-angles at $X, Y, Z, V$ respectively, show that the segments $XZ$ and $YV$ are perpendicular to each other and of the same length.

Now, I managed to solve an easier version where $ABCD$ is a Parallelogram, because the diagonals intersect with the two lines, so we can say that $XYZD$ is a square because its diagonals are perpendicular and the vertices form $90^{\circ}$ .

But, in this case I am not sure how to prove it, I definitely think it is proven by some rotation transformation, but I am not sure where or how exactly.My idea is that maybe if I prove that $XFZ$ can be rotated by $90^{\circ}$ into $VFY$ but how do I know they're the same length or maybe that $F=E$ somehow, that would get me somewhere close to the answer. I asked several PHD students in our college but none managed to solve it or give useful advice unfortunately as there are no numerical values.

I drew this diagram, I apologize ahead for how bad it is, but it is the first time I use that program, I hope it is clear enough, you can click on the picture to enlarge it.

We had this proof that we can use, which I drew next to the problem that if we have any given triangle with one isosceles right angled triangle on two sides then the triangle from the two $90^{\circ}$ vertices down to the mid point of the base forms also a right angled isosceles right angled triangle. And this is what I used to show that the parallelogram version forms a square.

However, here I am not sure how to implement this. Any hints or advice will be massively appreciated, am trying to solve this for a whole week now.

• The expected answer was that : we can rotate both $X$ and $Z$ points by $90^{\circ}$ about the point $F$ , so that implies that we can rotate the whole segment $XZ$ about $F$ by $90^{\circ}$. Rotation preserves length, so they're equal and $90^{\circ}$ implies that they're perpendicular. – Fred Dec 11 '17 at 10:37

You are given four different points $z_j\in{\mathbb C}$ $(j\in{\mathbb Z}_4$), making up a quadrangle of the described kind. The new point $w_1$ then is defined by $$w_1:={z_0+z_1\over2}+i{z_1-z_0\over2}={1-i\over2}z_0+{1+i\over2}z_1\ ,$$ so that we have in general $$w_j={1-i\over2}z_{j-1}+{1+i\over2}z_j\qquad(j\in{\mathbb Z}_4)\ .$$The claim now is that $$w_2-w_4=i(w_3-w_1)\ ;$$ and this should be easy to verify.