Centroid of a Triangle in The Poincare Disk According to this note, three medians of a hyperbolic triangle in the Poincare Disk $\Bbb{D}=\{z\in\Bbb{C} : |z|\lt 1\},\,ds=\dfrac{|dz|}{1-|z|^2}$ are concurrence. But how can I prove this? 
Can we write this concurrent point in terms of vertices? 
 A: Here I follow the  @Moishe Cohen's guide lines and these conversion formulas.
Let $a, b, c$ be three points on the disk $\Bbb{D}.$  
Now the corresponding points on the upper sheet of the hyperboloid of the hyperboloid model are given by
$A=\left(\dfrac{1+|a|^2}{1-|a|^2},\dfrac{2a}{1-|a|^2}\right),$ 
$B=\left(\dfrac{1+|b|^2}{1-|b|^2},\dfrac{2b}{1-|b|^2}\right)$ and $C=\left(\dfrac{1+|c|^2}{1-|c|^2},\dfrac{2c}{1-|c|^2}\right).$  
Centroid of the triangle formed by A, B and C is given by the formula described in Moishe Cohen's answer. 
A: I suggest you learn the hyperboloid model and avoid conversion to the unit disk model until you absolutely have to. Many (but not all) things are so much more transparent when you use the hyperboloid model. My formula for the centroid in the hyperboloid model is: $$(A+B+C)/\sqrt{<A+B+C, A+B+C>},$$ where $A, B, C$ are points on the upper sheet of the hyperboloid and $<u, v>$ is the Lorentzian inner product, where the 2-sheeted hyperboloid is given by the equation $<u,u>=1$. As for the proof, you first check the midpoint formula:
$$
(P+Q)/\sqrt{<P+Q, P+Q>}
$$
and then use it to show that the centroid of a hyperbolic triangle (in the hyperboloid model) is the same as the normalization of the affine centroid of the corresponding affine triangle with the same set of  vertices.  
