Question 143
Let $A_{1} A_{2} \ldots A_{n}$ be a regular $n$ corner. The points $B_{1}, \ldots, B_{n-1}$ are defined as follows:
- For $i=1$ or $i=n-1$, $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$;
- For $i \neq 1, i \neq n-1$, let $S$ be the intersection of $A_{1} A_{i+1}$ and $A_{n} A_{i}$. The point $B_{i}$ is then the intersection of the bisector of $\measuredangle A_{i} S A_{i+1}$ with $A_{i} A_{i+1}$.
Prove that is true
$$
\measuredangle A_{1} B_{1} A_{n}+\measuredangle A_{1} B_{2} A_{n}+\ldots+\measuredangle A_{1} B_{n-1} A_{n}=180^{\circ}
$$
- For $i=1$ or $i=n-1$, $B_{i}$ is the midpoint of side $A_{i} A_{i+1}$;
- For $i \neq 1, i \neq n-1$, let $S$ be the intersection of $A_{1} A_{i+1}$ and $A_{n} A_{i}$. The point $B_{i}$ is then the intersection of the bisector of $\measuredangle A_{i} S A_{i+1}$ with $A_{i} A_{i+1}$.
Prove that is true
$$
\measuredangle A_{1} B_{1} A_{n}+\measuredangle A_{1} B_{2} A_{n}+\ldots+\measuredangle A_{1} B_{n-1} A_{n}=180^{\circ}
$$