I need some help with this exercise I found in chapter 3 of the book "The Geometry of Discrete Groups" by Beardon.
- Prove (analitically and geometrically) that for all non-zero $x,y \in \mathbb{R} ^n $, $$|x| |y-x^{*}| =|y| |x-y^{*}|, $$ where for non-zero $w\in \mathbb{R} ^n$, $w^*$ denotes $\dfrac{w}{|w|^2}$.
The analitical proof means no problem for me because:
$$|x| ^2|y-x^{*}| ^2=|x|^2 ( |y|^2 -2\dfrac{<y,x>}{|x|^2}+\dfrac{1}{|x|^2} ) =|x|^2 |y|^2 -2<y,x>+1. $$
Similarly, we get $|y| ^2|x-y^{*}| ^2 =|y|^2 |x|^2 -2<x,y>+1$. So they are the same.
Now, for the geometric proof I know I must consider the plane that contains $0, x, x^*, y, y^*$, but then I have no clue where to start, so any hint would be very appreciated. Thank you so much!
Draw a figure like the one below, and derive the equation by showing that the triangles are similar, using the given information that $\lvert x x^* \rvert = \lvert y y^* \rvert = 1$, or rather that $OX \cdot OX' = OX \cdot OY' = 1$.