Geometric realisations of trees are Cat(0) I am trying to prove the title of this post. I have already looked into this post: $\mathbb{R}$ -trees are CAT(0) space 
But this does not cover the possibility where the geodesic triangle in the tree has a vertix of degree 3 i believe. 
I have been focussing on the case where V={1,2,3,4,5} and E= {12,23,24,45} and we have the geodesic triangle connecting the vertices 1,3 and 5. 
In this case the above post does not seem to work since we require a comparison triangle with sides of length 2,3 and 3.
I have tried playing around with cosine rule, triangle inequalities etc. but nothing seems to give me some strong generalisations or insight on why the cat(0) inequality should hold. Intuitively I get that due to there being a vertix of degree 3 there is overlap between the geodesics (for example points on the edge 12 of the geodesics connecting 1,3 and 1,5), making the points on distance 1 near the corner representing vertix 1 of the comparison triangle immediately satisfying the inequality and allowing us to build some distance for points that are no longer contained in overlapping geodesics of the tree, but I can't rigorously prove why this distance whill remain large enough when moving further away from the vertix with degree 3.
Any advice would be greatly appreciated.
 A: I think you need to draw a picture. In a tree $T$ (this applies to all $\mathbb R$-trees), all triangles are "tripods"—there is some point $c$ such that the geodesic $[x,y]=[x,c]\cup[c,y]$ for all choices of $x,y$ among vertices of the triangle. Thus geodesics between points $p$ and $q$ on the boundary of the triangle travel along an edge. What's more, any point of the triangle (except for $c$, which lies in three) lies in two "edges" of the triangle.
So, fix a triangle $\triangle xyz$ in the tree, and two points, $p$ and $q$ on the edges of the triangle. In what follows, I will assume that both $p$ and $q$ are not $c$, but it should be clear how my argument works when either of them is. Draw a comparison triangle $\bar\triangle \bar x \bar y \bar z$ in $\mathbb E^2$. $p$ and $q$ each have two comparison points, call them $p_1$ and $p_2$, $q_1$ and $q_2$ respectively. Without loss of generality, $p_1$ and $q_1$ lie in the same edge of $\bar\triangle \bar x \bar y \bar z$. This means that $d_T(p,q) = d_{\mathbb E^2}(p_1,q_1)$.
Notice that the triangle $\bar\triangle vp_1p_2$ (where $v$ is the common vertex of the edges containing $p_1$ and $p_2$) is isosceles, so the angle $\angle p_1 = \angle p_2$ and thus both are acute. This, together with the law of cosines, gives you a way to calculate $d_{\mathbb E^2}(p_2,q_1)$ in terms of $\overline{p_1p_2}$, $\overline{p_1q_1}$ and $\pi - \angle p_1 > \frac\pi 2$. Because $\cos$ is negative between $\frac \pi2$ and $\pi$, we conclude, in particular, that $d_T(p,q) < d_{\mathbb E^2}(p_1,q_2)$. If $p_2$ and $q_2$ are in different edges of $\bar\triangle \bar x \bar y \bar z$, then the above discussion shows that $\overline {p_2q_2}$ and $\overline {p_1q_1}$ lie in a trapezoid, where the angles at $p_1$ and $q_1$ are obtuse, so we again conclude that $d_T(p,q) < d_{\mathbb E^2}(p_2,q_2)$.
