On congruent chords Let $C_i=C(A_i,r_i)$ two secant circles intersecting each other at $R,S$, with $r_1\neq r_2$. Let $M$ be the median point of $A_1A_2$. Let $t\perp RM$ at $R$, intersecting the circles at $X,Y$. 
I'd like to show that $XR=YR$.
It's enough to prove that $MX=MY$. But I'm not able to do this.

 A: After a rotation and stretch, we may assume that $M=(0,0),A_1=(-1,0),A_2=(1,0).$ Let then $R=(a,b)$, so that the distances between $R$ and the $A_i$ determine the two radii.
The perpendicular bisector to segment $MR$ at the point $R$ is parametrically $(a-bt,b+at).$ We may then get the values of $t$ corresponding to the points $X,Y.$ For $X$ we solve
 $$(a-bt+1)^2+(b+at)^2=(a+1)^2+b^2,$$
which is $(a^2+b^2)t^2-2bt=0$ having the solutions $t=0$ (which gives $R$), and $t=2b/(a^2+b^2)$ which is then the $t$ parameter value for $X$.
In the same way for $Y$ we are to solve
 $$(a-bt-1)^2+(b+at)^2=(a-1)^2+b^2 ,$$
which is $(a^2+b^2)t^2+2bt=0$ with solutions $t=0$ giving $R$ again, and $t=-2b/(a^2+b^2)$ as the $t$ parameter for $Y$. 
Now the $t$- parameters $2b/(a^2+b^2),\ 0, -2b/(a^2+b^2)$ for $X,\ R,\ Y$ respectively are equally spaced, showing that $R$ is the midpoint of segment $XY$ as desired.
Not a geometric proof, but at least the algebra is fairly simple.
A: 
The perpendicular bisector of $\overline{XR}$ (likewise, $\overline{YR}$) meets the center $A$ (respectively, $B$) and is parallel to $\overline{RM}$. As the family of parallels cuts $\overline{AB}$ into congruent segments, it must also cut $\overline{PQ}$ into congruent segments. Consequently, $\overline{XP} \cong \overline{PR} \cong \overline{RQ} \cong \overline{QY}$, so that $\overline{XR} \cong \overline{YR}$.
