Furthermore we claim that \(|i,j| < |i,k|\) (2).

This is true because all edges get us incrementally closer, that is \(|v_{x-1}, t| < |v_x, t||\). To see this, observe Figure 4.

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 \(\square\).