Let $X$ be a Hadamard manifold of negative curvature $K_X \le -1$. With respect to the visual metric (with respect to the origin $o$) on the boundary at infinity $\partial X$, consider a visual ball $\mathscr{B}(\xi,r) $ with center $\xi$ and radius $r>0$.Let $x$ be a finite point lying on the geodesic joining $o$ to $\xi$. Let $B(x,l)$ be a ball in $X$ and I want to show that the shadow of $B(x,l)$ with respect to $o$ contains $\mathscr{B}(\xi,r)$. Then is the following lines of argument correct?
It is enough to show that given any $\eta$ belonging to the visual ball, the geodesic joining $o$ to $\eta$ intersects the ball $B(x,l)$. By definition of the visual metric and by angle comparison theorem we get an upper bound on the Riemannian angles between $\xi$ and other $\eta$ (subtended at $o$) in the given visual ball, in terms of $r$, say $\theta(r)$. Then to show the containment it is enough to show that the supremum over all Riemannian angles between $x$ and $y$ (subtended at $o$) where $y$ varies over $B(x,l)$ is $\ge \theta(r)$.