Using lemma, This can be proved easily $lemma$. Let $XAB$ be a triangle such that $XA=XB$. $D, C$ be points on $XA, XB$ respectly such that $DC \parallel AB$. and $S= AC \cap BD$ . and $N$ be the intersection of the bisector of $\angle CSB$ and $BC$. Finally, $M$ be the mid point of $BC$. Then $D, N, M, A$ are on one circle. $pf)$ Let $AN \cap CD = E$, $DM\cap AB = F$. And the length of $DC=p$. It can be easily shown that $BF=CE=p$. so $EF\parallel XB$. Therefore, $\angle EAF=\angle FDE$. and from $\angle CAB=\angle BDC$, We get $\angle CAN = \angle MDB$. cuz $\angle CAD= \angle DBC$, $\angle NAD = \angle DMN$. so lemma is proved. Now, Let $C_i$ be the mid points of $A_i A_{i+1}$. Using lemma, We get $\angle A_1 C_i A_n = \angle A_1 B_i A_n$. With using similarity, We can concentrate angles into $B_1$. So, the sum of angles is $180$. $Q.E.D.$

Pascual2005 wrote: Let $B_1$ and $B_n$ be the midpoints of its sides $A_1A_2$ and $A_{n-1}A_n$. $B_n$ should be $B_{n-1}$. Pascual2005 wrote: . Also, for every $i\in\left\{2,3,4,\ldots ,n-1\right\}$. $i\in\left\{2,3,4,\ldots ,n-1\right\}$ should be $i\in\left\{2,3,4,\ldots ,n-2\right\}$.

Let $C_i$ be the midpoint of $A_iA_{i+1}$. The central claim is that $\angle A_1B_iA_n=\angle A_1C_iA_n$. We only need to consider when $B_i\ne C_i$, so when $A_1A_n$ and $A_iA_{i+1}$ intersect at $X$ outside of the polygon. Note that $\angle XSB_i=90^\circ$, so $(XB_i;A_iA_{i+1})=-1$. Hence, $XA_i\cdot XA_{i+1}=XB_i\cdot XC_i$, from which we obtain $A_1A_nB_iC_i$ is cyclic by Power of a Point. The claim now follows. To finish, note that $\angle A_1C_iA_n=\angle A_iMA_{i+1}$ by symmetry, where $M$ is the midpoint of $A_1A_n$, so the angles sum to $\angle A_1MA_n=180^\circ$, as desired.

The pith of this problem is this lemma: Lemma. Let $ABCD$ be an isosceles trapezoid, so that $\overline{AB}\parallel\overline{CD}$ and $BC=DA$. Let $S=\overline{AC}\cap\overline{BD}$, and suppose the angle bisector of $\angle BSC$ intersects $\overline{BC}$ at $N$. If $P$ is the midpoint of $\overline{BC}$, then $ADPN$ is cyclic. Proof. Let $T=\overline{AD}\cap\overline{BC}$ (if they are parallel the proof is trivial). Since $\overline{ST}$ bisects $\angle ASB$, by a well-known lemma, $-1=(B,C;T,N)$. By the Midpoints of Harmonic Bundles Lemma, $TN\cdot TP=TB\cdot TC=TA\cdot TD$, and the desired result follows. $\blacksquare$ [asy][asy] size(4.5cm); defaultpen(fontsize(10pt)); pair A, B, C, D, SS, NN, P, T; A=(-3, 4); B=(3, 4); C=(6, 0); D=(-6, 0); SS=extension(A, C, B, D); NN=extension(SS, SS+(1, 0), B, C); P=(B+C)/2; T=extension(A, D, B, C); draw(B -- T -- A -- B -- C -- D -- A -- C, red); draw(B -- D, red); draw(T -- SS -- NN, orange); draw(circumcircle(A, D, P), orange); clip((-100, -1) -- (100, -1) -- (100, 100) -- (-100, 100) -- cycle); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, E); dot("$D$", D, W); dot("$S$", SS, S); dot("$N$", NN, NE); dot("$P$", P, E); dot("$T$", T, N); [/asy][/asy] Let $M$ be the midpoint of $\overline{A_1A_n}$. Since $A_1A_n=A_iA_{i+1}$ and $A_1A_iA_{i+1}A_n$ is cyclic, $A_1A_iA_{i+1}A_n$ must be an isosceles trapezoid, whence by our lemma, $$\sum_{i=1}^{n-1}\angle A_1B_iA_n=\sum_{i=1}^{n-1}\angle A_1M_iA_n=\sum_{i=1}^{n-1}\angle A_iMA_{i+1}=\angle A_1MA_n=180^\circ,$$and we are done. $\square$ [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pair A1, A2, A3, A4, A5, A6, S2, S3, S4, B1, B2, B3, B4, B5, B6, M, M2, M4; A1=dir(90); A2=dir(30); A3=dir(330); A4=dir(270); A5=dir(210); A6=dir(150); S2=extension(A1, A3, A6, A2); S3=extension(A1, A4, A6, A3); S4=extension(A1, A5, A6, A4); B1=(A1+A2)/2; B2=extension(A1+S4-S3, S2, A2, A3); B3=(A3+A4)/2; B4=extension(A6+S2-S3, S4, A4, A5); B5=(A5+A6)/2; M=(A1+A6)/2; M2=(A2+A3)/2; M4=(A4+A5)/2; pair X=A1+S4-S3; pair O=(X+S3)/2; draw(circle((0, 0), 1), red); draw(circle(O, length(O)), orange); draw(A1 -- A2 -- A3 -- A4 -- A5 -- A6 -- A1, red); draw(A3 -- A1 -- A5 -- A1 -- A4 -- A6 -- A2 -- A6 -- A3, red); draw(B2 -- X -- B3 -- X -- B4, orange+dashed); //draw(B1 -- A6 -- B2 -- A1 -- B3 -- A6 -- B4 -- A1 -- B5, orange+dotted); //draw(A6 -- M2 -- A1 -- M4 -- A6, fuchsia+dotted); draw(B1 -- M -- B2 -- M -- B3 -- M -- B4 -- M -- B5, pink); label("$A_1$", A1, A1); label("$A_2$", A2, A2); label("$A_3$", A3, A3); label("$A_4$", A4, A4); label("$A_5$", A5, A5); label("$A_6$", A6, A6); label("$S_2$", S2, NE); label("$S_3$", S3, E); label("$S_4$", S4, W); label("$B_1$", B1, unit(B1)); label("$B_2$", B2, E); label("$B_3$", B3, unit(B3)); label("$B_4$", B4, unit(M4)); label("$B_5$", B5, W); label("$M$", M, dir(75)); label("$M_2$", M2, E); label("$M_4$", M4, unit(M4)); label("$X$", X, unit(X)); dot(A1); dot(A2); dot(A3); dot(A4); dot(A5); dot(A6); dot(S2); dot(S3); dot(S4); dot(B1); dot(B2); dot(B3); dot(B4); dot(B5); dot(B6); dot(M); dot(M2); dot(M4); dot(X); dot(O); [/asy][/asy]

We use complex numbers with $\omega$ a primitive $n$th root of unity, and $A_k = \omega^k$. [asy][asy] size(8cm); pair A(int k) { return dir(90 + 180/11 - 360*k/11); } pair S = extension(A(1), A(5), A(4), A(0)); pair B = extension(S, incenter(S, A(4), A(5)), A(4), A(5)); dot("$A_1 = \omega^1$", A(1), A(1), blue); dot("$A_2 = \omega^2$", A(2), A(2), blue); dot(A(3), blue); dot("$A_k = \omega^k$", A(4), A(4), blue); dot("$A_{k+1} = \omega^{k+1}$", A(5), A(5), blue); dot(A(6), blue); dot(A(7), blue); dot(A(8), blue); dot(A(9), blue); dot(A(10), blue); dot("$A_n = \omega^n = 1$", A(11), A(11), blue); draw(S--B, heavygreen); dot("$B_1$", midpoint(A(1)--A(2)), dir(A(1)+A(2))); dot("$B_{n-1}$", midpoint(A(10)--A(11)), dir(A(10)+A(11))); dot("$B_k$", B, dir(B), deepgreen); dot("$S_k$", S, dir(S), deepgreen); draw(A(1)--A(5), deepgreen); draw(A(4)--A(0), deepgreen); draw(A(1)--A(4), lightred + dashed); draw(A(0)--A(5), lightred + dashed); filldraw(A(1)--A(2)--A(3)--A(4)--A(5)--A(6)--A(7)--A(8)--A(9)--A(10)--A(0)--cycle, invisible, blue); [/asy][/asy] Claim: For each $2 \le k \le n-2$ we have \[ b_k = \frac{2\omega^{2k+1} - \omega^{k+2} - \omega^k} {\omega^k - \omega + \omega^{k+1} - 1}. \] Proof. Note that $A_1 A_k A_{k+1} A_n$ is an isosceles trapezoid. Since the triangles $\triangle A_1 A_k S_k$ and $\triangle A_n A_{k+1} S_k$ are similar and isosceles, we have \[ \frac{|A_{k+1} B_k|}{|A_k B_k|} = \frac{|A_{k+1} S_k|}{|A_k S_k|} = \frac{|A_{k+1} A_n|}{|A_k S_1|} = \frac{\omega^{k+1}-1}{\omega^k-\omega} \]with the right hand side being a positive real number, owing to the fact that $\overline{A_1 A_k} \parallel \overline{A_n A_{k+1}}$ and point in the same direction. Thus, \[ b_k = \frac {(\omega^k-\omega) \cdot A_{k+1} + (\omega^{k+1}-1) \cdot A_k} {(\omega^k-\omega) + (\omega^{k+1}-1) } \]which is what we wanted to prove. $\blacksquare$ We can now compute the desired angle: \begin{align*} \frac{b_k-1}{b_k-\omega} &= \frac{ 2\omega^{2k+1}-\omega^{k+2}-\omega^k - \omega^{k+1} - \omega^k + \omega + 1} {2\omega^{2k+1}-\omega^{k+2}-\omega^k - \omega^{k+2} - \omega^{k+1} + \omega^2 + \omega} \\ &= \frac {2\omega (\omega^k)^2 - (\omega^2+\omega+2) \omega^k + (\omega+1)} {2\omega (\omega^k)^2 - (2\omega^2+\omega+1) \omega^k + \omega(\omega+1)} \\ &= \frac {(\omega \cdot \omega^k - 1)(2 \omega^k - (\omega+1))} {(\omega^k - \omega)(2\omega \cdot \omega^k - (\omega+1))} \\ &= \frac{1}{\omega} \cdot \frac{\omega^{k+1}-1}{\omega^{k-1}-1} \cdot \frac{2\omega^k - (\omega+1)}{2\omega^{k+1} - (\omega+1)}. \end{align*}Unsurprisingly, this telescopes when multiplied: \begin{align*} \prod_{k=2}^{n-2} \frac{b_k-1}{b_k-\omega} &= \frac{1}{\omega^{n-3}} \cdot \frac{(\omega^{n-1}-1)(\omega^{n-2}-1)} {(\omega-1)(\omega^2-1)} \cdot \frac{2\omega^2-(\omega+1)}{2\omega^{n-1}-(\omega+1)} \\ &= \frac{1}{\omega^{-3}} \cdot \frac{\left( \frac{1}{\omega}-1 \right) \left( \frac{1}{\omega^2}-1 \right)} {(\omega-1)(\omega^2-1)} \cdot \frac{2\omega^2-(\omega+1)}{2\omega^{n-1}-(\omega+1)} \\ &= \frac{2\omega^2-(\omega+1)}{\frac{2}{\omega}-(\omega+1)} = -\frac{\omega(2\omega+1)(\omega-1)}{(\omega+2)(\omega-1)} = \frac{\omega(2\omega+1)}{\omega+2} \end{align*}On the other hand, the complex number with angle $\angle A_1 B_1 A_n = \angle A_1 B_{n-1} A_n$ is \[ \frac{1 - \frac{\omega + \omega^2}{2}}{\omega - \frac{\omega + \omega^2}{2}} = \frac{2-(\omega+\omega^2)}{\omega-\omega^2} = \frac{(2+\omega)(1-\omega)}{\omega(1-\omega)} = \frac{2+\omega}{\omega}. \]So we finally arrive at \[ \frac{\omega(2\omega+1)}{\omega+2} \left( \frac{2+\omega}{\omega}\right)^2 = \frac{(2+\omega)(2\omega+1)}{\omega} = 2(\omega+\omega^{-1}) + 4 \]which is real. This proves that the requested sum $\sum_{i=1}^{n-1} \angle A_1B_iA_n$ is an integer multiple of $\pi$. It remains to do the annoying check that this is less than $2\pi$. One way is to note that for $k = 2, \dots, n-2$ we have \begin{align*} \angle A_1 B_k A_n &= \pi - (\angle A_n A_1 B_k + \angle A_1 A_n B_k) \\ &< \pi - (\angle A_n A_1 A_{k+1} + \angle A_1 A_n A_k) \\ &= \frac{2\pi}{n} \end{align*}and $\angle A_1 B_1 A_n < \pi - \angle A_2 A_1 A_n = \frac{2\pi}{n}$ as well. Thus the sum in question is less than $\frac{n-1}{n} \cdot 2\pi$ as required.

For each $i$, let $M_i$ be the midpoint of $\overline{A_iA_{i+1}}$, and let $T_i = \overline{A_1A_n} \cap \overline{A_iA_{i+1}}$. Note that $-1 = (T_iB_i;A_{i+1}A_i)$, so $T_iB_i \cdot T_iM_i = T_iA_{i+1} \cdot T_iA_i = T_iA_n \cdot A_iA_1$, so $A_nB_iM_iA_1$ is cyclic. Now $\angle A_1B_iA_n = \angle A_1M_iA_n = \angle A_iM_nA_{i+1}$, so summing cyclically we find $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n = \angle A_1M_nA_n = 180^{\circ}$.

Here is mine (may have typing issues I'll check soon) Name midpoint of $A_1 A_n$ as $N$. by symmetry $\angle A_1B_{n-1}A_n =\angle A_{n-1}NA_{n}$ and Similarly $\angle A_1B_1A_n =\angle A_2NA_1$ $\textbf{Claim:- }$ ($\forall i\neq 1,n,n-1$) $\angle A_1B_iA_n =\angle A_iNA_{i+1}$ Proof: Consider $(A_1A_iA_{i+1}A_n)$ that is a isosceles trapizode. Call midpoint of $A_iA_{i+1} = M_i$ by symmetry $\angle A_iNA_{i+1}=\angle A_1MA_n $ So it's equivalent to $\angle A_1B_iA_n = \angle A_1M_iA_n$ Enough to prove $A_1A_nB_iM_i$ is cyclic Call $K_i=\overline{A_1A_n} \cap \overline{A_iA_{i+1}}$ Now notice $(K_i,B_i,A_i,A_{i+1})=-1 \implies K_iB_i \cdot K_iM_i = K_iA_i \cdot K_iA_{i+1}=K_iA_1 \cdot K_iA_n$ So PoP gives the cyclic result. Now notice summing all these angle gives us $A_1NA_n = 180$ $\boxed{\mathcal{Q.E.D}}$

Attachments:

This diagram gives a nudge for how to synthetically prove that $A_1A_nB_iM_i$ is cyclic. Using the point names in the diagram, we see that $OS_2B_2M_2$ is cyclic by easy angle chasing. We also know that $X_2$ is on $\overline{OS_2}$ by symmetry. Thus, we have \[X_2A_1 \cdot X_2A_7=X_2O \cdot X_2S_2=X_2B_2 \cdot X_2M_2\]by Power of a Point. This can be generalized.

Let $M_i$ denotes midpoint of $A_iA_{i+1}$ with cyclic numeration. Claim. $A_1,A_n,B_i,M_i$ are concyclic for $i\notin \left\{ 1,n-1,n\right\} .$ Proof. Let $C_i=A_1A_n\cap A_iA_{i+1}.$ We easily conclude $$(A_iA_{i+1}B_iC_i)=-1\implies |C_iB_i|\cdot |C_iM_i|=|C_iA_i|\cdot |C_iA_{i+1}|=|C_iA_1|\cdot |C_iA_n|\text{ } \Box$$ Now we obtain $\sum_{i=1}^{n-1} \measuredangle A_1B_iA_n=\sum_{i=1}^{n-1} \measuredangle A_1M_iA_n=\sum_{i=1}^{n-1} \measuredangle A_{i+1}M_nA_i=\measuredangle A_nM_nA_1=\pi \text{ } \blacksquare$

Let $M_i$ be the midpoint of $A_iA_{i+1}$ for $i\in \{1,2,...,n-1\}$, while $M$ is the midpoint of $A_1A_n$. We claim that the points $A_1,A_n,B_i$ and $M_i$ are concyclic: Proof: for simplicity, take an isosceles trapezoid $BDEC$ such that $BC\parallel DE$. Let $H$ be the midpoint of $EC$, $A=BD\cap EC$, $F=BE\cap EC$ and $G\in \overline{EC}$ such that $\angle GFE= \angle GFC$. By interior-exterior angle bisectors perpendicularity , we have $\angle AFG=90$, Applying Right Angles and Bisectors theorem we get that $(A,G;E,C)=-1$, so by Harmonic Bundles and Midpoint theorem we have $$AG\cdot AH=AE\cdot AC=AD\cdot AB$$Hence by power of the point $A$ we derive the needed result.$\square$ $$\sum_{i=1}^{n-1} \angle A_1B_iA_n = \sum_{i=1}^{n-1} \angle A_1M_iA_n = \sum_{i=1}^{n-1} \angle A_iMA_{i+1} = \angle A_1MA_n = 180$$where the second equality holds by symmetry. $\blacksquare$

Attachments:

Let $M_i$ be the midpoint of $A_iA_{i+1}$ for $1\le i\le n-1.$ The key claim is that for each $i,$ the quadrilateral $A_1A_nM_iB_i$ is cyclic. To show this, first it is obvious for $i=1,n-1$ and $B_i=M_i.$ Otherwise extend $A_1A_n$ and $A_iA_{i+1}$ to meet at $X$ (if they don't meet it's easy to see $M_i=B_i.$) Now notice $XS_i$ is the external bisector of $\angle A_iS_iA_{i+1},$ so by harmonic bundles or whatever $B_i$ and $X$ are inverses in the circle with diameter $A_iA_{i+1},$ so $M_iB_i\cdot M_iX=M_iA_i^2.$ Subtracting this from $XM_i^2$ we get $XB_i\cdot XM_i=XA_i\cdot XA_{i+1}=XA_1\cdot XA_n,$ proven. Now let $M$ be the midpoint of $A_1A_n.$ We have $\sum_{i=1}^{n-1}\angle A_1B_iA_n=\sum_{i=1}^{n-1}\angle A_1M_iA_n=\sum_{i=1}^{n-1}\angle A_iMA_{i+1}=180^\circ.$