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?

Since my figure is very misleading. I still can't solve it. this is what I get from this:this
I know the triangles are simular but I can't figure out how to actually show that they're perpendicular.

  • $\begingroup$ your picture is misleading because all angles between the radii seem like 45 degrees. In order not to conclude false statements, better to come up with a graph where the points are not evenly spaced out. $\endgroup$ – imranfat Aug 7 '16 at 22:16
  • 1
    $\begingroup$ Your image should look like i.stack.imgur.com/NcnOm.png $\endgroup$ – Andrei Aug 7 '16 at 22:52
  • $\begingroup$ Thank you so much. My drawing was very misleading, as imranfat said. I'll try solving this out by myself now. $\endgroup$ – Supersalt Aug 7 '16 at 22:59
  • $\begingroup$ I'm sorry. I was unable to solve it, I'm having trouble to understand how to show that something is perpendicular. I assume I could show that the triangles are simular. I'll update the thread in a moment. $\endgroup$ – Supersalt Aug 8 '16 at 0:43

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: $$\measuredangle A_1MB_1+\measuredangle C_1MD_1=\frac{\measuredangle A_1MB_1+\measuredangle B_1MC_1+\measuredangle C_1MD_1++\measuredangle D_1MA_1}{2}=\frac{360^{\circ}}{2}$$


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 $A,C$.


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