I think I've seen this on AoPS before, can't find post tho...

https://artofproblemsolving.com/community/c6h1896p5928

We will prove the following : Consider two lines intersecting in one point forming an X , the bottom and top angle have measure $na $ and $n$ points inside the top angle . Then it is possible to draw an angle with measure $a$ from each of these points such that the bottom angle is entirely covered from these angles. [asy][asy] draw((0,0)--(6,6),red); draw((6,0)--(0,6),red); dot((3,4)); dot((4,5)); [/asy][/asy] We induct on $n$. Case $n=1$ is obvious. Label the two lines $l_0,l_n $ and draw consider the $n+1$ lines $l_0,l_1,l_2,...,l_n$(anticlockwise) such that the angle between $l_i,l_{i+1}$(call it $a_i$) is $a$. Also label the points $P_0,P_1,...,P_{n-1}$ . We need a translation of $a_0,...,a_{n-1}$ to $P_0,P_1,...,P_{n-1}$ that covers the bottom angle. [asy][asy] draw((0,0)--(5,5),red); draw((1,0)--(4,5),red); draw((2,0)--(3,5),red); draw((3,0)--(2,5),red); draw((4,0)--(1,5),red); draw((5,0)--(0,5),red); [/asy][/asy] If there are no points in $a_0top$ ( or $a_{n-1}top$) : Consider the point $P$ that is closer to $l_1$ draw a line $l'_1$ parallel to $l_1$ and use the rest $n-1$ points to cover the bottom angle between $l'_1,l_n$ then translate $a_0bottom$ to $P$. If there is at least one point inside both $a_0$ and $a_{n-1}$ then : Set $d_k=$(number of points between $l_0,l_k$) - $k$ $d_1 \ge 0$ , $d_{n-2}\le 0$ and we have that $d_{i+1}\ge d_i-1$ so there is $d_m=0$ so induct for angles between $l_0,l_m $ and $l_m,l_{n-1}$ as they contain $m$ and $n-m$ points respectively. We illustrated the case when angle $na$ is convex but the above works for any $na\le 2\pi$ such as the case $a=2\pi/n$ .