deletions | additions       diff --git a/Furthermore_we_claim_that_i__.tex b/Furthermore_we_claim_that_i__.tex  index 390397d..8e456d9 100644  --- a/Furthermore_we_claim_that_i__.tex  +++ b/Furthermore_we_claim_that_i__.tex 
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Furthermore we claim that $|i,j| < |i,k|$. This is true because we always get incrementally closer, that is \(|v_{x-1}, t| < |v_x, t||\). This follows from the fact that our path essentially defines a |i,k|$ (2).  This is true because all edges get us incrementally closer, that is \(|v_{x-1}, t| < |v_x, t||\). To see this, observe the included figure.  Essentially the points \(v_{x},v_{x+1},t\) define a triangle. From the image it is clear that the interior angle of \(v_{x+1}\) is obtuse. If it were less, that would imply that `t` was inside our circle-segment -- but it can't be, because if it was closer to our \(v_x\) we would have traveled to it. As the angle of \(v_{x+1}\) is obtuse, the side opposite it \(v_x \rightarrow t\) must be the unique hypotenuse \(\square\).  We briefly prove this inductively on the length of the path to `k`. Our base case is a path of length one, that is Given that,  we are RTP. $|i, v_1| < |i, k|$. This follows from can apply  the way lemma  we form our paths. Clearly $v_1$ and $k$ both have to be part of the same wedge. However, there is an edge from $i$ to exactly the closest point in each edge and our path takes us from $i$ to $v_1$.  We assume that this applied for the path up until the second-last edge. That is, $|i, v_{-2}| < |i, k|$.  We now prove that given this assumption, $|i,j| < |i,k|$ holds. proved above.