Forty-five points are chosen along line $AB$, all lying outside of segment $AB$. Prove that the sum of the distances from these points to point $A$ is not equal to the sum of the distances of these points to point $B$.
The answer given behind the book is not clear. Here is the answer given at the back:-
"For any point $X$ lying outside segment $AB$, the difference $AX-BX=+AB$ If we assume that the sum of the distances from $A$ and from $B$ are equal, then the expression $\pm AB \pm AB +\dots+AB$, in which there are $45$ addends, is zero. This is impossible."
Please explain me this answer. According to me, the answer can be found out by the fact that there is an extra point (say, $X$) on one side (say, $B$). And that point's distance is greater for the opposite point(A) whereas short for the nearer side($B$).
Is my reasoning correct? If not, then please correct it.
