Let $\alpha=\angle D A B$. Since $|A D|=|B D|$, we also have $\angle A B D=\alpha$. By the inscribed angle theorem, we find $\angle A C D=\alpha$, while the cyclic quadrilateral theorem gives $\angle B C D=180^{\circ}-\alpha$. Therefore, $\angle B C A=180^{\circ}-2 \alpha$. The angle sum in triangle $B I M$ gives, together with the fact that $I$ is the intersection of the angle bisectors of $\triangle B C M$, that

$$\begin{gathered} \angle B I M=180^{\circ}-\angle I M B-\angle M B I=90^{\circ}+90^{\circ}-\frac{1}{2} \angle C M B-\frac{1}{2} \angle M B C \\ =90^{\circ}+\frac{1}{2} \angle B C M=90^{\circ}+\frac{1}{2} \angle B C A=90^{\circ}+90^{\circ}-\alpha=180^{\circ}-\alpha . \end{gathered}$$

Since $BIMN$ is a cyclic quadrilateral, it follows that $\angle B N M=\alpha$. The angle sum in $\triangle B N C$ now gives
$\angle N B C=180^{\circ}-\angle B C N-\angle C N B=180^{\circ}-\angle B C A-\angle M N B=180^{\circ}-\left(180^{\circ}-2 \alpha\right)-\alpha=\alpha$.
This means that

$$\angle A B N=\angle A B C-\angle N B C=\angle A B C-\alpha=\angle A B C-\angle A B D=\angle C B D .$$

Combined with $\angle N A B=\angle C A B=\angle C D B$, which holds by the inscribed angle theorem, we get $\triangle A B N \sim \triangle D B C$ (AA). Therefore,

$$\frac{|A N|}{|C D|}=\frac{|B N|}{|C B|}$$

or $|C D| \cdot|B N|=|A N| \cdot|C B|$. We know that $\angle N B C=\alpha=\angle B N M=\angle B N C$, so $\triangle B N C$ is isosceles with vertex angle $C$, thus $|C B|=|C N|$. Therefore, $|C D| \cdot|B N|=$ $|A N| \cdot|C N|$, which is what we wanted to prove.