Prove that if $YA\cdot YB=YX\cdot YM$ then $(A,B;X,Y)=-1$ . Given 4 collinear points $A,X,B,Y$  in this order. Let $ M$ be the midpoint of $AB$.  Prove that if $$YA\cdot YB=YX\cdot YM$$  then $(A,B;X,Y)=-1$ .
I think its related with orthogonality but couldn't proceed much, so went with a different route.
I tried directly solving , so I took $YA\cdot YB=YX\cdot YM \implies (AX+XY)YB=(BX+BU)(XY+MX)\implies AX\cdot BY= BX(XY+MX)+BY\cdot MX \implies AM\cdot BY=BX\cdot MY \implies BM\cdot BY=BX\cdot MY$.
But, I am not able to proceed. Any hints ?
Thanks in advance!
 A: 
Given $YA\cdot YB=YX\cdot YM$, we are able to construct the picture as above. Let ${AH\over HB}=r$, then $${AY\over BY} = {AY\over HY}\times {HY\over BY} = r^2$$
$${AX\over BX} = {AM\times AX\over BM\times BX} ={AZ\times AH\over BW\times BH} = ({AH\over BH})^2=r^2$$
Therefore $${AX\over BX} = {AY\over BY}$$
A: By an affine transformation you may map $Y$ to $0$ and $X$ to $1$ (this does not affect the claim). Let $a$ and $b$ be the (new) coordinates of $A$ and $B$. Then
$$ a  b =  m= \frac12(a+b)  \Leftrightarrow
\frac{a-1}{b-1}\frac{b}{a}=-1.$$
This also holds for complex scalars b.t.w.
A: Here's another proof (your approach of directly solving, just breaking the factors differently):
$$YA\cdot YB = YM\cdot YX \\ \implies (AX+XY)BY=(YA-AM)(YB+BX) \\ \implies AX\cdot BY +XY\cdot BY= AY\cdot BY - AM\cdot BY - AM\cdot BX + AY\cdot BX \\ \implies  AX\cdot BY - AY\cdot BX = AY\cdot BY-XY\cdot BY+AM\cdot (-BX-BY)\\= AY\cdot BY - XY\cdot BY - AM\cdot XY\\ = AY\cdot BY - XY\cdot (AM+BY) \\=AY\cdot BY -XY\cdot (BM+BY) \\ = AY\cdot BY - XY\cdot MY =0 \ (given)\\ \implies AX\cdot BY - AY\cdot BX =0$$
A: First, we note that the angle-bisector theorem implies that (assuming that $A',X',B',Y'$ are collinear in that order)
$$X',Y' \text{ are the points where the internal and external bisectors of $\angle C'$ cut $A'B'$ in $\Delta A'B'C'$ }\iff (A',B';X',Y')=-1 \qquad \qquad \qquad (1)$$.
Now,

$A,B,X$ are given as in your problem, so $M$ is fixed, and the product relation fixes $Y$ on the line containing $A,X,B$.

So, proving your problem is equivalent to proving that in any triangle $A'B'C'$ with points named as in $(1)$, the point $Y$ lying on the ray $A'X'B'$ satisying $YA'\cdot YB'=YM'\cdot YX'$ is precisely the point $Y'$ ($M'$ midpoint of $A'B'$).
So, we can just take a triangle (and hence assume it's sidelengths $a',b',c'$ to be given) with the naming convention in $(1)$, find lengths of $Y'A',Y'B',Y'M',Y'X'$ from the angle-bisector theorem in terms of $a',b',c'$ to show $Y'A'\cdot Y'B'=Y'M'\cdot Y'X'$ as an identity in $a',b',c'$.
