Points A,B,C (fix) and X (variable) such that |AX| + |BX| = |CX| Let $A,B,C$ (fix) points of the plane. Where are points $X$ (variable) in the plane with $|AX| + |BX| = |CX|$ ?
It seems $X$ need to lie on the arc under $AB$ of the circumscribed circle of the triangle $ABC$.
How can I prove that?

Note: There is no Trapezium $AXBC$.

From the equation for $X$ we can get 
$$4 |A-X|^2 |B-X|^2 = (|C-X|^2 - |A-X|^2 - |B-X|^2)^2$$
which gives you a quartic polynomial in the coordinates of $X$. If this doesn't factor (which in general it won't), the locus of $X$ will not be an arc of a circle, nor any conic section. 
Here's a picture of it in the case $A = (0,0)$, $B=(1,0)$, $C = (0,1)$.

EDIT:  In this particular case, the curve has genus $0$ and therefore has a rational parametrization: 
$$ \eqalign{X_1 &= {\frac {64\,{t}^{3}-1024\,{t}^{2}+4096\,t}{9\,{t}^{4}-96\,{t}^
{3}+512\,{t}^{2}-4096\,t+16384}}\cr
X_2 &= {\frac {12\,{t}^{4}-160\,{t}^{3}+512\,{t}^{2}}{9\,{t}^{4}-96\,
{t}^{3}+512\,{t}^{2}-4096\,t+16384}}\cr
 0 \le &t \le 8
}$$
In most cases, it appears the curve has genus one and is an elliptic curve, with no rational parametrization.
