Geometry: Show that two lines are perpendicular I have a homework problem telling me the following;
$ A, B, C, $ and $ D $ are points on a circle. $A_1, B_1, C_1$ and $ D_1$ are midpoints to the arcs $AB, BC, CD$ and $DA$. Show that $A_1C_1$ is perpendicular to $B_1D_1$.  
Here's what I drew real quick with Geogebra:
circle
Since the angles $\angle AMB \cong \angle DMC$ and $\angle BMC \cong \angle DMA$ because vertical angles are always equal. We also know that the angles 
$$\angle AMA_1 \cong \angle BMA_1$$ 
and $$\angle BMB_1 \cong \angle CMB_1$$ because the arcs $AA_1 \cong BA_1$ and $BB_1 \cong CB_1$.  If I somehow could show that $\angle A_1MB$ and $\angle B_1MB$ are congurent I might be able to show that the sum of two angles, $|\angle A_1MB| + |\angle B_1MB| = 90^{°}$. So what I've done is drawing the line $A_1B_1$, as in the picture above, we know that the lines $A_1M \cong B_1M$ because they're the radius of the circle. Which gives us that $\angle A_1B_1M \cong \angle B_1A_1M$ because of the base angle theorem. 
But to my problem: if I somehow could show that $A_1B_1 \parallel AC$ then the we have the vertical angles
$$\angle AMA_1 \cong \angle B_1A_1M \cong A_1B_1M \cong CMB_1$$
which proves that all angles are the equal size, which gives us $|\angle A_1MB| + |\angle B_1MB| = 90^{°}$ . How do I show that they're parallel? Or is there any other way to show that $A_1C_1$ and $B_1D_1$ are perpendicular?
Update 
Since my figure is very misleading. I still can't solve it. this is what I get from this: 
I know the triangles are simular but I can't figure out how to actually show that they're perpendicular.
 A: Without loss of generality, we can take the points to be on the trigonometric circle.
Let $\alpha, \beta, \gamma, \delta$ be the trigonometric angles associated to $A,B,C,D$ respectively.
Then, clearly:
$$
\begin{gathered}
  A_{\,1}  = \left( {\cos \left( {\frac{{\alpha  + \beta }}
{2}} \right),\;\sin \left( {\frac{{\alpha  + \beta }}
{2}} \right)} \right) \hfill \\
  B_{\,1}  = \left( {\cos \left( {\frac{{\beta  + \gamma }}
{2}} \right),\;\sin \left( {\frac{{\beta  + \gamma }}
{2}} \right)} \right) \hfill \\
  C_{\,1}  = \left( {\cos \left( {\frac{{\gamma  + \delta }}
{2}} \right),\;\sin \left( {\frac{{\gamma  + \delta }}
{2}} \right)} \right) \hfill \\
  D_{\,1}  = \left( {\cos \left( {\frac{{\alpha  + \delta }}
{2}} \right),\;\sin \left( {\frac{{\alpha  + \delta }}
{2}} \right)} \right) \hfill \\ 
\end{gathered} 
$$
and the vectors associated with segments $A_1C_1$ and $B_1C_1$ will be
$$
\mathop {A_{\,1} C_{\,1} }\limits^ \to   = \left( {\begin{array}{*{20}c}
   {\cos \left( {\frac{{\gamma  + \delta }}
{2}} \right) - \cos \left( {\frac{{\alpha  + \beta }}
{2}} \right)}  \\
   {\sin \left( {\frac{{\gamma  + \delta }}
{2}} \right) - \sin \left( {\frac{{\alpha  + \beta }}
{2}} \right)}  \\
 \end{array} \;} \right)\quad \mathop {B_{\,1} D_{\,1} }\limits^ \to   = \left( {\begin{array}{*{20}c}
   {\cos \left( {\frac{{\alpha  + \delta }}
{2}} \right) - \cos \left( {\frac{{\beta  + \gamma }}
{2}} \right)}  \\
   {\sin \left( {\frac{{\alpha  + \delta }}
{2}} \right) - \sin \left( {\frac{{\beta  + \gamma }}
{2}} \right)}  \\
 \end{array} \;} \right)
$$
whose dot product gives
$$
\begin{gathered}
  \mathop {A_{\,1} C_{\,1} }\limits^ \to  \; \cdot \;\mathop {B_{\,1} D_{\,1} }\limits^ \to   =  \hfill \\
   = \left( {\cos \left( {\frac{{\gamma  + \delta }}
{2}} \right) - \cos \left( {\frac{{\alpha  + \beta }}
{2}} \right)} \right)\left( {\cos \left( {\frac{{\alpha  + \delta }}
{2}} \right) - \cos \left( {\frac{{\beta  + \gamma }}
{2}} \right)} \right) +  \hfill \\
   + \left( {\sin \left( {\frac{{\gamma  + \delta }}
{2}} \right) - \sin \left( {\frac{{\alpha  + \beta }}
{2}} \right)} \right)\left( {\sin \left( {\frac{{\alpha  + \delta }}
{2}} \right) - \sin \left( {\frac{{\beta  + \gamma }}
{2}} \right)} \right) =  \hfill \\
   = 2\left( {\cos \left( {\frac{{\alpha  - \gamma }}
{2}} \right) - \cos \left( {\frac{{\beta  - \delta }}
{2}} \right)} \right) \hfill \\ 
\end{gathered} 
$$
So for the two segments to be orthogonal we shall have:
$
\left| {\alpha  - \gamma } \right| = \left| {\beta  - \delta } \right|
$
meaning that the couple of points $B,D$ is a rotated image of the couple $A,C$.
A: Call the intersection of $A_1C_1$ and $B_1D_1$ point $P$.
$$\measuredangle A_1PB_1=\frac{\measuredangle A_1MB_1+\measuredangle C_1MD_1}{2}$$ 
We can prove that $\measuredangle A_1MB_1+\measuredangle C_1MD_1=180^{\circ}$ using the following fact: 
$$\measuredangle A_1MB_1+\measuredangle C_1MD_1=\frac{\measuredangle A_1MB_1+\measuredangle B_1MC_1+\measuredangle C_1MD_1++\measuredangle D_1MA_1}{2}=\frac{360^{\circ}}{2}$$
