The idea is similar to the proof that the angle bisectors concur or that the perpendicular bisectors concur. Assume WLOG that $B C>A B>C A$. Note that $h_{a}$ and $h_{b}$ are both hyperbolas. Therefore, $h_{a}$ and $h_{b}$ intersect in four points (each branch of $h_{a}$ intersects exactly once with each branch of $h_{b}$ ). Note that the branches of $h_{a}$ correspond to the cases when $P B>P C$ and when $P B<P C$. Similarly, the branches of $h_{b}$ correspond to the cases when $P C>P A$ and $P C<P A$. If either $P A<P B<P C$ or $P C<P B<P A$ (which each happens for exactly one point of intersection of $h_{a}$ and $h_{b}$ ), then $|P C-P A|=|P C-P B|+|P B-P A|=|A B-A C|+|B C-B A|=|B C-A C|$, and so $P$ also lies on $h_{c}$. So, exactly two of the four points of intersection of $h_{a}$ and $h_{b}$ lie on $h_{c}$, meaning that $h_{a}, h_{b}$, and $h_{c}$ concur in four points.