20. Let $O$ be the center of the circle touching the three sides of $B C D E$ and let $F \in(E D)$ be the point such that $E F=E B$. Then $\angle E F B=90^{\circ}-$ $\angle E / 2=\angle C / 2=\angle O C B$, which implies that $B, C, F, O$ lie on a circle. It follows that $\angle D F C=\angle O B C=\angle B / 2=90^{\circ}-\angle D / 2$ and consequently $\angle D C F=\angle D F C$. Hence $E D=E F+F D=E B+C D$. Second solution. Let $r$ be the radius of the small circle and let $M, N$ be the points of tangency of the circle with $B E$ and $C D$ respectively. Then $E M=r \cot E, D N=r \cot D, M B=r \cot (\angle B / 2)=r \tan (\angle D / 2)$, $N C=r \tan (\angle E / 2)$, and $E D=E O+O D=r / \sin D+r / \sin E$. The statement follows from the identity $\cot x+\tan (x / 2)=1 / \sin x$.