Consider a convex quadrilateral with vertices at $๐,~๐,~๐$ and $๐$ and on each side draw a square lying outside the given quadrilateral, as in the picture below. Let $๐,~๐,~๐$ and $๐ $ be the centers of those squares:
a) Find expressions for $๐,~๐,~๐$ and $๐ $ in terms of $๐,~๐,~๐$ and $๐$.
b) Prove that the line segment between $๐$ and $๐$ is perpendicular and equal in length to the line segment between $๐$ and $๐ $.
I think this problem has been asked before, but they don't give any good hints. I don't really know where to start. I tried finding $p$ first by finding $(p-a)$ and $(p-b)$. I tried another way by translating $a$ to the origin. I haven't been able to go farther than this.
I think I have an idea for part $b$ using similar triangles and things, but part a is really confusing.Thank you!
I translated the square with $p$ as its center so that a would be at the origin. So $b$ would then be $bโa$ and $p$ would be $pโa$, right? $pโa$ is half of the diagonal. So then $(pโa)=(bโa)\cdot\frac{\sqrt2}{2}$.
Rotating by $-\frac{\pi}{4}$ would give us $$\frac{(bโa)2}{โ2}\cdot e^{โi\frac{\pi}{4}}=\frac{(bโa)\sqrt2}{2}\cdot\left(\frac{\sqrt2}{2}โi\frac{\sqrt2}{2}\right)=\left(\frac{(bโa)}{2}โi\frac{(bโa)}{2}\right)=\frac{bโaโbi+ai}{2}$$ Therefore, $pโa=bโaโbi+ai2$ and when we translate everything back we get $$p=bโaโbi+ai2+aโนp=bโaโbi+ai+(2a)2โนp=b+aโbi+ai2.$$ I can do a similar process for the rest of the points, right?
Does it matter which point I translate to the origin?