Let $P=\{P\_1,P\_2\cdots P\_n\}$ be a set of $n\geq 4$ points in the plane and $P\_iP\_j$ be the line segment connecting $P\_i$ and $P\_j$ that satisfy: 


$(1)$Any three points of $P$ are not on a line;


$(2)$In the set $\{P\_1P\_2,P\_2P\_3,\cdots,P\_{n-1}P\_n,P\_nP\_1\}\setminus \{P\_iP\_{i+1}\}$,$P\_iP\_{i+1}$ only intersect with $P\_{i-1}P\_i$ and $P\_{i+1}P\_{i+2}$ ,$i=1,2,\cdots,n$(where $P\_0=P\_n,P\_{n+1}=P\_1$).


I conjecture that there must exist $1\leq i\leq n$ such that there is no point of $P$ in the interior of $\bigtriangleup P\_{i-1}P\_iP\_{i+1}$,where $P\_0=P\_n,P\_{n+1}=P\_1$.


Is that right?



a question about points and line segments in the plane
Yes, the polygon formed by the broken line $P\_1\dots P\_n$ may be triangulated by $n-3$ inner diagonals and in any triangulation there exists a triangle (at fact, at least two triangles for $n\geq 4$) of the form $P\_{i-1}P\_iP\_{i+1}$.


