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We denote $\angle BAD=\angle DBA=\alpha$ in the isosceles triangle $ABD$. Then $\angle ACB=\angle ADB=180^{\circ }-2\alpha$. Let $I$ be the incentre of the triangle $BCM$. We have $$\begin{array}{rl}\angle MIB& =180^{\circ }-\frac{1}{2}(\angle BMC+\angle CBM)=180^{\circ }-\frac{1}{2}(180^{\circ }-\angle MCB)\\ & =180^{\circ }-\frac{1}{2}(180^{\circ }-\angle ACB)=180^{\circ }-\alpha .\end{array}$$ Since the quadrilateral $BIMN$ is cyclic we can prove that $\angle BNC=\alpha$ (independently of the position of the point $N$) so the triangle $BCN$ is isosceles and $|BC|=|NC|$.

Note that the triangles $ABN$ and $DBC$ are similar because $\angle ANB=\angle DCB=180^{\circ }-\alpha$ and $\angle BAN=\angle BDC$ (suspended angles over the chord $BC$). Hence $$\frac{|NA|}{|BN|}=\frac{|CD|}{|BC|}=\frac{|CD|}{|NC|},$$ i.e. $|AN|\cdot |NC|=|CD|\cdot |BN|$.