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 $s$.
I managed to do part A, via finding a diagonal and then the midpoint. For part (b), I found a coord-bash algebraic solution, via assigning a lot of variables. However, is there a geometric solution?
We can start off by translating $a$ to the origin using rotations and the fact that we can scale the distance of b and a, we come up with $p=a +\frac{\sqrt{2}}{2} e^{\pi*i/4}(b-a)$ Since $\frac{\sqrt{2}}{2}e^{\pi*i/4} = \frac{1}{2} - i\frac{1}{2}$, we can rewrite the equation to get $p = \frac{1}{2}$ $(a(1+i)$ $+$ $b(1-i))$ Simplifying this to get $p = \frac{a + b + ia - ib}{2}$, we can rewrite this in a more appealing way: $p = \frac{a + b}{2} + \frac{i(a - b)}{2}$. We can use the same strategy of translating to the origin and rotating segments for all other complex points in order to define the complex numbers q, r, s. $q = \frac{b + c}{2} + \frac{i(b - c)}{2}$ $r = \frac{c + d}{2} + \frac{i(c - d)}{2}$ $s = \frac{d + a}{2} + \frac{i(d - a)}{2}$ (In order to get these, I used the fact that we had to translate to the origin and that $p$ is a 45 degree rotation from $b$, $q$ is a 45 degree rotation from $c$, r is a 45 degree rotation from $d$ and $s$ a rotation from $a$.) I also have just given a diagram for p(not sure if its correct) and I saw on the message board as long as I describe the other equation similarities(45 degree rotation) I can just leave it at that.
B: We want to prove that the line segment between $p$ and $r$ is perpendicular and equal in length to the line segment between $q$ and $s$. Translate where the two lines meet to the origin, then rotating one of the lines to get the other gives us $e^{\pi/2}(r-p)=s-q$. If $e^{\pi/2}(r-p)=s-q$, then $|r-p|=|s-q|$. The lines are perpendicular if $e^{\pi/2}(r-p)=s-q$ is true. Using values we got in a, we can see that $e^{\pi/2}(r-p)=s-q$, or $i(r-p)=s-q$ is made into $i\left(\frac{d-di+c-ci}{2}-\frac{b-bi+a-ai}{2}\right)=\frac{a-ai+d-di}{2}-\frac{c-ci+b-bi}{2}$. This gives us: $2di+2c=2a+2bi$, or $di+c=a+bi$