Recall that an $n$-simplex in $\overline{\mathbb{H}^n}$ (the closure of $n$-dim hyperbolic space) with vertices $v_0,...,v_n\in \overline{\mathbb{H}^n}$ is the closed subset of $\mathbb{H}^n$ bounded by the hyperbolic hyper-surface contain all the vertices except one. In particular, all edges are geodesic. A simplex is called ideal if all the vertices lies on the boundary.
The book Lectures on Hyperbolic Geometry by Benedetti shows that any hyperbolic $n$-simplex is contained in an ideal simplex by arguing that if we take any interior point $p$ and extend the geodesic line from $p$ to $v_i$ to the boundary with end point $v_i'$. Then obviously, the original simplex is contained in the ideal simplex with vertices $v_0',...,v_n'$.
But I cannot figure out the "obvious" part. Can anyone help me explain a little?
It becomes obvious if you think about it in the Klein model of $\mathbb{H}^n$, with the point $p$ sitting at the center of the $n$-ball. In this case, the given simplex is simply a Euclidean simplex contained in the $n$-ball, and the ideal simplex is a Euclidean simplex whose vertices lie on the boundary sphere. The containment follows from the fact that the larger simplex is convex, and contains each of the vertices of the smaller simplex.