Simple formulas for geodesics of the Beltrami-Klein model Exercice : Show that the geodesics of the Beltrami-Klein model are the euclidean lines ?

Let $\gamma_{p,T}(t)=\cosh(t)p+\sinh(t)T$ be a geodesic of the hyperbolic plane such that $\gamma_{p,T}(0)=p=(x,y,z),\gamma_{p,T}'(0)=T=(u,v,w)\in T_p\mathbb H$ and let $\Phi :\mathbb H\to \mathbb K :(x,y,z)\to \left(\frac{x}{z}, \frac{y}{z}\right)$ the isometry between $\mathbb H$ and $\mathbb K.$ We get
\begin{align*}
\Phi(\gamma_{p,T}(t))&= \Phi(\cosh(t)(x,y,z)+\sinh(t)(u,v,w)) \\&= \left(\dfrac{\cosh (t) x+\sinh(t)u}{\cosh(t)z+\sinh(t)w}, \frac{\cosh(t)y+\sinh(t)v}{\cosh(t) z+\sinh(t) w} \right)
\end{align*}
Which is a messy formula !
If we want to prove this by using action of a group on $\mathbb K$, what wold be this group ?
I cannot prove how this formla is an equation of an euclidean line !
If there any simple way to prove this result I will be very gratful !
 A: As you are aware, the geodesics on the hyperboloid of one sheet $H$ are sections by planes through the origin, and the Klein model on the unit disk $K$ in the plane $z = 1$ arises from the hyperboloid model by projection $\Phi:H \to K$ from the origin. Seeing that the geodesics of $K$ are Euclidean line segments from this alone is easy: Geodesics are plane sections of $H$ and of $K$.

Because the group $G$ of isometries of $H$ is the set of transformations in $O(2, 1)$ (the orthogonal group of the Minkowski inner product on Cartesian three-space) that preserve the "positive" sheet of the two-sheeted hyperboloid, and because $\Phi$ is an isometry, the isometries of $K$ are precisely the transformations $\Phi G\Phi^{-1}:K \to K$. Using this non-linear group to deduce the form of geodesics amounts to the same calculation as on the hyperboloid, but expressed in less convenient coordinates by working directly on $K$ rather than on $H$ where hyperbolic isometries are a matrix group.
In retrospect, the problem is that the geodesics in $K$ are "messy." Any group that deduces them from some simple example, such as a diameter of $K$, must itself be a "messy" group.
