Let $A,B,C$ be the vertices of a triangle. Find the center of the rotations. $R_{C,\pi} \circ R_{A,\frac{\pi}{2}}$ 
Let  $A,B,C$ be the vertices of a triangle. Find the center of the
  rotations. $R_{C,\pi} \circ R_{A,\frac{\pi}{2}}$

Hey everyone, I am learning about euclidean rigid motions and my book only explains theory but does not focus on computations therefore, half the problems are computations and the other half are proofs. I am having a hard time with this problem, so if anyone can give me any steps or hints that would be great and help me start of this problem. 
The only thing the book tells me that I can use for this problem is the composition of these two rotations is also another rotation. 
 A: If $\Delta$ is a straight line, I denote by $S_\Delta$ the reflection about $\Delta$, and if $P_1$ and $P_2$ are two distinct points, I denote by $(P_1P_2)$ the straight line through the points $P_1$ and $P_2$. I can write:
$$
R_{C,\pi}=S_{(A_1C)}\circ S_{(AC)},\quad R_{A,\frac\pi2}=S_{(AC)}\circ S_{(AC_1)},
$$
with 
$$
A_1=R_{C,\frac\pi2}(A),\,  C_1=R_{A,-\frac\pi4}(C).
$$
It follows that
$$
R_{C,\pi}\circ R_{A,\frac\pi2}=S_{(A_1C)}\circ(S_{(AC)}\circ S_{(AC)})\circ S_{(AC_1)}=S_{(A_1C)}\circ S_{(AC_1)}.
$$
Notice that
\begin{eqnarray}
\overline{(\overrightarrow{AC_1},\overrightarrow{A_1C})}&=&\overline{(\overrightarrow{AC_1},\overrightarrow{AC})}+\overline{(\overrightarrow{AC},\overrightarrow{A_1C})}\\
&=&-\overline{(\overrightarrow{AC},\overrightarrow{AC_1})}-\overline{(\overrightarrow{CA},\overrightarrow{CA_1})}\\
&=&\frac\pi4+\frac\pi2=\frac{3\pi}{4}
\end{eqnarray}
i.e. $\overline{(\overrightarrow{AC_1},\overrightarrow{A_1C})} \ne 0 \mod \pi$. Therefore the two lines $(A_1C)$ and $(AC_1)$ intersect at exactly one point; let's call it $D$. Hence
$$
R_{C,\pi}\circ R_{A,\frac\pi2}=S_{(A_1C)}\circ S_{(AC_1)}=R_{D,2\cdot\frac{3\pi}{4}}=R_{D,\frac{3\pi}{2}}.
$$
