Let $l$ be a line through $A$ different from $A B$ and join $B$ to $A, X, Y$ and $Z$ as in the above diagram. No matter how $l$ is chosen, the angles $A X B, A Y B$ and $A Z B$ always subtend the chord $A B$. For this reason the angles in the triangles $B X Y$ and $B X Z$ are the same for all such $l$. Thus the ratio $X Y: Y Z$ remains constant by similar triangles.

Note that this is true no matter how $X, Y$ and $Z$ lie in relation to $A$. Suppose $X, Y$ and $Z$ all lie on the same side of $A$ (as in the diagram) and that $\measuredangle A X B=\alpha, \measuredangle A Y B=\beta$ and $\measuredangle A Z B=\gamma$. Then $\measuredangle B X Y=180^{\circ}-\alpha, \measuredangle B Y X=\beta, \measuredangle B Y Z=180^{\circ}-\beta$ and $\measuredangle B Z Y=\gamma$. Now suppose $l$ is chosen so that $X$ is now on the opposite side of $A$ from $Y$ and $Z$. Now since $X$ is on the other side of the chord $A B, \measuredangle A X B=180^{\circ}-\alpha$, but it is still the case that $\measuredangle B X Y=180^{\circ}-\alpha$ and all other angles in the two pertinent triangles remain unchanged. If $l$ is chosen so that $X$ is identical with $A$, then $l$ is tangent to the first circle and it is still the case that $\measuredangle B X Y=180^{\circ}-\alpha$. All other cases can be checked in a similar manner.

!