92. In the inscribed quadrilateral $A B C D$, given are $|A B|=a,|A D|=b, a>b$. Find the side $|B C|$, if it is known that $B C, C D$ and $A D$ touch a certain circle, the center of which lies on $A B$.
92. From the condition of the problem, it follows that the bisectors of angles $C$ and $D$ intersect on side $A B$. Let's denote this point of intersection as $O$. We will describe a circle around $\triangle D O C$. Let $K$ be the second point of intersection of this circle with $A B$. We have

$$
\widehat{D K A}=\widehat{D C O}=\frac{1}{2} \widehat{D C B}=\frac{1}{2}\left(180^{\circ}-\widehat{D A K}\right)=\frac{1}{2}(\widehat{D K A}+\widehat{A D K})
$$

Thus, $\widehat{D K A}=\widehat{A D K}$ and $|A D|=|A K|$. Similarly, $|B C|=|B K|$; therefore, $|A D|+|C B|=|A D|+|B K|=a$.

Answer: $a-b$.