20. We first establish the following lemma. Lemma. Let $A B C D$ be an isosceles trapezoid with bases $A B$ and $C D$. The diagonals $A C$ and $B D$ intersect at $S$. Let $M$ be the midpoint of $B C$, and let the bisector of the angle $B S C$ intersect $B C$ at $N$. Then $\angle A M D=\angle A N D$. Proof. It suffices to show that the points $A, D, M, N$ are concyclic. The statement is trivial for $A D \| B C$. Let us now assume that $A D$ and $B C$ meet at $X$, and let $X A=X B=a, X C=X D=b$. Since $S N$ is the bisector of $\angle C S B$, we have $$\frac{a-X N}{X N-b}=\frac{B N}{C N}=\frac{B S}{C S}=\frac{A B}{C D}=\frac{a}{b}$$ and an easy computation yields $X N=\frac{2 a b}{a+b}$. We also have $X M=\frac{a+b}{2}$; hence $X M \cdot X N=X A \cdot X D$. Therefore $A, D, M, N$ are concyclic, as needed. Denote by $C_{i}$ the midpoint of the side $A_{i} A_{i+1}, i=1, \ldots, n-1$. By definition $C_{1}=B_{1}$ and $C_{n-1}=B_{n-1}$. Since $A_{1} A_{i} A_{i+1} A_{n}$ is an isosceles trapezoid with $A_{1} A_{i} \| A_{i+1} A_{n}$ for $i=2, \ldots, n-2$, it follows from the lemma that $\angle A_{1} B_{i} A_{n}=\angle A_{1} C_{i} A_{n}$ for all $i$. The sum in consideration thus equals $\angle A_{1} C_{1} A_{n}+\angle A_{1} C_{2} A_{n}+\cdots+$ $\angle A_{1} C_{n-1} A_{n}$. Moreover, the triangles $A_{1} C_{i} A_{n}$ and $A_{n+2-i} C_{1} A_{n+1-i}$ are congruent (a rotation about the center of the $n$-gon carries the first one to the second), and consequently $$\angle A_{1} C_{i} A_{n}=\angle A_{n+2-i} C_{1} A_{n+1-i}$$ ! for $i=2, \ldots, n-1$. Hence $\Sigma=\angle A_{1} C_{1} A_{n}+\angle A_{n} C_{1} A_{n-1}+\cdots+\angle A_{3} C_{1} A_{2}=\angle A_{1} C_{1} A_{2}=180^{\circ}$.