I've been on this problem for a while now, but haven't been able to get through part b. I was able to get the first part, though.
Consider a convex quadrilateral with vertices at $a, b, c$ and $d$ and on each side draw a square lying outside the given quadrilateral, as in the picture below. Let $p, q, r$ and $s$ be the centers of those squares:
a) Find expressions for $p, q, r$ and $s$ in terms of $a, b, c$ and $d$.
b) Prove that the line segment between $p$ and $r$ is perpendicular and equal in length to the line segment between $q$ and $s$.
For part a, I had, if we were trying to find p for instance: $$p = a + \frac{\sqrt2}2 e^{-\pi i/4} (b-a)$$
I just don't know what to do with b from here.
Just to note, I'm a precalculus student
$ p = \frac{a+b}{2} + \frac{a-b}{2} i$ , $q = \frac{b+c}{2} +\frac{b-c}{2}i $ , $ r = \frac{c+d}{2} + \frac{c-d}{2}i $ , $s = \frac{d+a}{2} + \frac{d-a}{2}i$
$ 2(r-p) = (c-a)(1+i)+(d-b)(1-i)$ , $ 2(s-q) = (d-b)(1+i) + (a-c)(1-i)$
$ \therefore 2(s-q) = 2(r-p) \cdot i $
Use that $ i = e^{\frac{i\pi}{2}}$