20. In isosceles triangle $\mathrm{ABC}$, there is a point $\mathrm{D}$ on the base $\mathrm{AB}$. It is known that $A D=a, D B=b(a<b) . \triangle A C D$ and $\triangle B C D$ have incircles that touch line $C D$ at points $M$ and $N$, respectively. Find $M N$.
20. As shown in Figure 22, $\mathrm{MN}=\mathrm{DN}-\mathrm{DM}$.

Let $P$ be the semiperimeter of $\triangle B C D$. Since $D N=D F, C N=C_{E}$, and $B F=B E$, we have $P=D N + B C$, which means $D Y=P-B C$.
From this, we get $D Y=P-AC$ (A.1), and since $\mathbf{A C}=$
$B C$, we obtain $M N=P-p$.
Furthermore, because $\triangle B D C$ and $\triangle A D C$ have equal corresponding sides (common side $\mathrm{CD}$ and $\mathrm{AC}=\mathrm{BC}$), the difference in their perimeters equals the difference in the third sides, thus,
$$
M N=P-p=\frac{B D-A D}{2}=\frac{b-a}{2} .
$$