Recently, I study geometry on the sphere in Patrick Ryan's Geometry book.
A line on the unit sphere $S^2$ is defined as \begin{equation} l=\{x\in S^2: <x,z>=0\} \end{equation} for some point $z\in S^2$. A point $z$ is called the pole of $l$. A reflection to line $l$ which is defined as \begin{equation} \Omega_{l}x=x-2<x,z>z \end{equation} where $z$ is the pole of $l$ and $x\in S^2$.
My problem is proving the following theorem: Let $\alpha$, $\beta$, and $\gamma$ be three lines through a point $P$. Then there is a unique line $\delta$ through $P$ such that \begin{equation} \Omega_{\alpha}\Omega_{\beta}\Omega_{\gamma}=\Omega_{\delta}. \end{equation}
My partial answer: Let the pole of $\alpha$, $\beta$, and $\gamma$ be $N_{\alpha},N_{\beta}$ and $N_{\gamma}$ respectively. Then after some calculation I get
\begin{equation} \Omega_{\alpha}\Omega_{\beta}\Omega_{\gamma}=x-2<x,N_{\gamma}>N_{\gamma}-2<x,N_{\beta}>N_{beta}-2<x,N_{\alpha}>N_{\alpha} +4<x,N_{\gamma}> <N_{\gamma},N_{\beta}>N_{\beta}+4<x,N_{\gamma}><N_{\alpha},N_{\beta}>N_{\alpha}+ +4<x,N_{\beta}><N_{\alpha},N_{\beta}>N_{\alpha} -8<x,N_{\gamma}><N_{\gamma},N_{beta}><N_{\gamma},N_{\alpha}>N_{\alpha} \end{equation}