= Solution
Regard the upper half-disc, with cyclic boundary vertices $-1,0,x,1$, as a <conformal quadrilateral>. Map it to a rectangle of width $w$ and height $h$, so the boundary arcs $[-1,0]$ and $[x,1]$ become its vertical sides. The rectangle formulas for the <conformal modulus of a quadrilateral> give
$$
d=\frac wh,\qquad d_*=\frac hw,\qquad\boxed{d\,d_*=1}.
$$
The remaining two boundary arcs, $(0,x)$ and the semicircle, become the horizontal sides. Thus these are precisely the dual joining families in the same quadrilateral.
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