Composition of Reflections is a Rotation 
I can prove the result is true where the lines go through the origin, but otherwise I'm finding this difficult, I also assume this isn't something that takes long to prove. Any help will be appreciated.
Thanks
 A: Let $A$ be the intersection of $L_1$ and $L_2$, and $(\widehat{\overrightarrow{AA_1},\overrightarrow{AA_2}})=\alpha$ the angle between them, where $A_i\in L_i$.
Obviously $\rho_i(A)=A$, and $\rho_i(A_i)=A_i$ for $i=1,2$.
Given a point $M \in \mathbb{R}^2$, we set  $M_1=\rho_1(M)$ and $M_2=\rho_2(M_1)$. Then
\begin{eqnarray}
\|\overrightarrow{AM_2}\|=\|\overrightarrow{\rho_2(A)\rho_2(M_1)}\|=\|\overrightarrow{AM_1}\|=\|\overrightarrow{\rho_1(A)\rho_1(M)}\|=\|\overrightarrow{AM}\|,
\end{eqnarray}
and
\begin{eqnarray}
(\widehat{\overrightarrow{AM},\overrightarrow{AM_2}})&=&(\widehat{\overrightarrow{AM},\overrightarrow{AM_1}})+(\overrightarrow{AM_1},\overrightarrow{AM_2})\mod 2\pi\\
&=&(\widehat{\overrightarrow{AM},\overrightarrow{AA_1}})+(\widehat{\overrightarrow{AA_1},\overrightarrow{AM_1}})+(\widehat{\overrightarrow{AM_1},\overrightarrow{AA_2}})+(\widehat{\overrightarrow{AA_2},\overrightarrow{AM_2}}) \mod 2\pi\\
&=&2(\widehat{\overrightarrow{AA_1},\overrightarrow{AM_1}})+2(\widehat{\overrightarrow{AM_1},\overrightarrow{AA_2}})\mod 2\pi\\
&=&2(\widehat{\overrightarrow{AA_1},\overrightarrow{AA_2}})\mod 2\pi\\
&=&2\alpha.
\end{eqnarray}
In other words $\rho_2\rho_1$ is the rotation with center $A$, and angle $2\alpha$.
Sense
$$
(\rho_i\rho_j)(\rho_j\rho_i)=\rho_i(\rho_j\rho_j)\rho_i=\rho_i\text{id}\rho_i=\rho_i\rho_i=\text{id},
$$
we have $\ro_j\rho_i=(\rho_i\rho_j)^{-1}$ therefore $\rho_1\rho_2$ is also a rotation.
A: We can write the two reflections as $\rho_1=T_ar_1T_{-a}$ and the second reflection as $\rho_2=T_br_2T_{b}$, where $r_1,r_2$ are reflections about lines through the origin and $T_a$ and $T_b$ are translations by the vectors $a$ and $b$. If we compose the two then we find: $\rho_1\rho_2=T_ar_1T_{-a}T_br_2T_{-b}=T_ar_1T_{b-a}r_2T_{-b}=T_aT_{r_1(b-a)}r_1r_2T_{-b}$. 
$r_1$ and $r_2$ are reflections about the origin, so their composition is a rotation. Now, use the rules for composing reflections to find the angle of the rotation. Then, to find the point, $c$, that the rotation is about, try to write $T_aT_{r_1(b-a)}r_1r_2T_{-b}$ as $T_c\phi T_{-c}$, where $\phi$ is the rotation $r_1r_2$.
