Given a polyhedral angle $O.A_1A_2\dots A_n$ (where $O$ is the vertex), I need to show that the sum of dihedral angles with sides $OA_1,\dots,OA_n$ is bigger than $(n-2)\pi$.
My try is to draw a plane meeting $OA_i$ at $B_i$ and show that the sum of dihedral angles is bigger than the sum of the angles of polygon $B_1B_2\dots B_n$. But it seems that each dihedral angle is not bigger than the corresponding angle of the polygon and I do not know how to continue.
Can someone help me? Thanks a lot!
On
Thanks to the hint of @timon92, we only need to show that $\sum_{i=1}^n \angle P_iPP_{i+1}<2\pi$. Without loss of generality, one can assume that $PP_1=PP_2=\cdots=PP_n$. Let $Q$ be the perpendicular projection of $P$ on the plane $P_1P_2\dots P_n$ then $QP_1=QP_2=\cdots QP_n$. Consider two isosceles triangles $PP_1P_2$ and $QP_1P_2$: they have common side $P_1P_2$ and $PP_1>QP_1$, thus $\angle P_1PP_2<\angle P_1QP_2$. Similarly for the remaining angles, we get $$\sum_{i=1}^n \angle P_iPP_{i+1}<\sum_{i=1}^n \angle P_iQP_{i+1}=2\pi.$$
Hint: pick a point $P$ inside the polyhedral angle. Let $P_1. P_2, \ldots, P_n$ be the perpendicular projections of $P$ onto planes $OA_1A_2, OA_2A_3, \ldots, OA_nA_1$, respectively. For every $i$ the dihedral angle at $OA_i$ is equal to $\pi - \angle P_{i-1}PP_i$. The problem boils down to proving that $\sum_{i=1}^n \angle P_iPP_{i+1} > 2\pi$.