Identifying compositions of reflections, and rotations in a hexagon Let $ABCDEF$ be a regular hexagon that is oriented clockwise (so that a rotation from $A$ to $B$ to
$C$ to $D$ to $E$ to $F$ is clockwise).
i) Identify $R_{D,120} \circ R_{A,60}$
which are two rotations meaning the end result will also be a rotation of $180$ but I can't seem to find the center.
ii) Identify $R_{F,180} \circ \rho_{ED} \circ R_{D,120}$
This one is confusing since it is a rotation composed with a reflection composed with a different rotation. I overheard two people mentioning how the composition is a glide reflection but I thought glide reflections where compositions of reflection and translation.
If anyone can help me with these two problems
UPDATE: Here is my work for ii)

Identify $R_{F,180} \circ \rho_{ED} \circ R_{D,120}$
I started out with:
$$\rho_{ED} \circ R_{D,120}$$
where I let:
$$\rho_{ED}=\rho_{m}$$
$$ R_{D,120}=\rho_{m} \circ \rho_{n}$$
$$ \text{ m: be represented by the line ED }$$ 
$$ \text{ n: be represented by the line that goes through D and A}$$ 
Then we have:
$$\rho_{ED} \circ R_{D,120}=\rho_{m} \circ (\rho_{m} \circ \rho_{n})=(\rho_{m} \circ \rho_{m} ) \circ \rho_{n}=\rho_{n}   $$
So now we have:
$$R_{F,180} \circ \rho_{n}$$
I let $$R_{F,180}=\rho_{k} \circ \rho_{r}$$
where:
$$ \text{ k: be represented by the line BF}$$ 
$$ \text{ r: be represented by the line EF }$$ 
$$R_{F,180} \circ \rho_{n}=(\rho_{k} \circ \rho_{r}) \circ \rho_{n}=\rho_{k} \circ (\rho_{r} \circ \rho_{n})=\rho_{BF} \circ \tau_{BF}=\gamma_{BF}$$
which is a glide reflection
 A: Denote by $O$ the middle of the segment $[A,D]$. You can check that all the triangles $OAF, OBA, OCB, ODC, OED, OFE$ are equilateral triangles. I shall denote by $(AD)$ the straight line through $A$ and $D$, etc. It is obvious that $(AD)=(DA)$. The notation $\mathrm{Id}$  stands for the identity map.
i) In order to identify $R_{D,120}\circ R_{A,60}$ we can make the following decompositions:
$$
R_{D,120}=\rho_{\Delta_1}\circ \rho_{DA},\quad R_{A,60}=\rho_{AD}\circ\rho_{\Delta_2},
$$
with
$$
\Delta_1\cap(DA)=\{D\},\, \angle(DA,\Delta_1)=60,\, \Delta_2\cap(AD)=\{A\},\, \angle(AD,\Delta_2)=30.
$$
Since $\angle(\overrightarrow{DA},\overrightarrow{DE})=60$, we deduce that $\Delta_1=(DE)$. Similarly we have $\Delta_2=(AE)$ because $\angle(\overrightarrow{AE},\overrightarrow{AD})=30$.
Hence
$$
R_{D,120}\circ R_{A,60}=\left(\rho_{DE}\circ \rho_{DA}\right)\circ\left(\rho_{AD}\circ\rho_{AE}\right)=\rho_{DE}\circ\underbrace{\left(\rho_{DA}\circ\rho_{AD}\right)}_{\mathrm{Id}}\circ\rho_{AE}=\rho_{DE}\circ\rho_{AE}=\rho_{DE}\circ\rho_{AE}.
$$
Now notice that $(DE)\cap(AE)=\{\color{red}{E}\}$ and
$$
\angle(\overrightarrow{EA},\overrightarrow{ED})=\angle(\overrightarrow{EA},\overrightarrow{EO})+\angle(\overrightarrow{EO},\overrightarrow{ED})=30+60=90=\frac{\color{red}{180}}{2},
$$
and therefore
$$
R_{D,120}\circ R_{A,60}=R_{\color{red}{E},\color{red}{180}}.
$$
ii) In order to identify $R_{F,180}\circ\rho_{\color{blue}{ED}}\circ R_{\color{blue}{D},120}$, we first make the following decomposition: 
$$
R_{D,120}=\rho_{\color{blue}{DE}}\circ\rho_{\Delta_3},
$$ 
with 
$$
\Delta_3\cap(DE)=\{\color{blue}{D}\},\, \angle(\Delta_3,DE)=60.
$$
Therefore $\Delta_3=(DA)$, because $OED$ is an equilateral triangle. It follows that
\begin{eqnarray}
R_{F,180}\circ\rho_{ED}\circ R_{D,120}&=&R_{F,180}\circ\left(\rho_{\color{blue}{ED}}\circ R_{\color{blue}{D},120}\right)=R_{F,180}\circ\left[\rho_{\color{blue}{ED}}\circ\left(\rho_{\color{blue}{DE}}\circ\rho_{\color{green}{DA}}\right) \right]\\
&=&R_{F,180}\circ\left[\underbrace{\left(\rho_{\color{blue}{ED}}\circ\rho_{\color{blue}{DE}}\right)}_{\mathrm{Id}}\circ\rho_{\color{green}{DA}}\right]=R_{F,180}\circ\rho_{\color{green}{DA}}.
\end{eqnarray}
Since $\overrightarrow{FB}\perp\overrightarrow{FE}$, i.e. $\angle(\overrightarrow{FB},\overrightarrow{FE})=90$ we can write:
$$
R_{F,180}=\rho_{FB}\circ\rho_{FE}.
$$
Thus
$$
R_{F,180}\circ\rho_{ED}\circ R_{D,120}=\left(\rho_{FB}\circ\rho_{FE}\right)\circ\rho_{DA}=\rho_{FB}\circ\underbrace{\left(\rho_{FE}\circ\rho_{DA}\right)}_{T_{\overrightarrow{BF}}}=\rho_{FB}\circ T_{\overrightarrow{BF}},
$$
i.e. 
$$
R_{F,180}\circ\rho_{ED}\circ R_{D,120}=\rho_{BF}\circ T_{\overrightarrow{BF}}=T_{\overrightarrow{BF}}\circ\rho_{BF} 
$$
is a glide reflection.
A: To do this problem geometrically, starting with the ABCDEF hexagon, reflect it repeatedly through its sides to produce a "tiling" by hexagons. Now do some calculations: what vertex of the tiling is the image of, say, $B$ under the composition $R_{D,120} \circ R_{A,60}$? What vertex is the image of $C$? of $D$? Once you have done a few of these calculations, you should be able to infer what map this composition is.
