Given a cyclic quadrilateral $$ABCD$$ and $$A_1, B_1, C_1, D_1$$ be the centers of arcs above chords $$AB,BC,CD,DA$$ in that order,prove that $$A_1C_1 \bot B_1D_1$$
I tried observing cyclic quadrilateral $$A_1B_1C_1D_1$$ but couldn't seem to get anything,anyway here's a picture,if you want it bigger open it in new tab. Thanks in advance
Let's say $\beta = D_1\hat C_1A_1$, and $\gamma = C_1\hat D_1B_1$. You have that $D\hat CB = 2\beta$, because while $D_1\hat C_1A_1$ insists on $\widehat{D_1A_1}$, $D\hat CB$ insists on $\widehat{DB}$, which is 2 times bigger than $D_1A_1$. Since $ABCD$ is cyclic the sum of two opposite angles equals $\pi$, so $D\hat AB = \pi-D\hat CB = \pi-2\beta$. Similarly $D\hat AB = 2C_1\hat D_1B_1 = 2\gamma = \pi-2\beta$ for the previous equation. So $$2\gamma = \pi-2\beta$$ $$2\gamma + 2\beta = \pi$$ $$\gamma + \beta = \frac{\pi}{2}$$ it means that $C_1\hat ED_1 = A_1\hat EB_1 = \pi - (B_1\hat D_1C_1 + D_1\hat C_1A_1) = \pi - (\gamma + \beta ) = \pi - \frac{\pi}{2} = \frac{\pi}{2}$. It follows that both $D_1\hat EA_1$ and $C_1\hat EB_1$ are $\frac{\pi}{2}$, and $A_1C_1$ and $D_1B_1$ are perpendiculars.
• no, I'm sorry, I should have been more detailed...if 2 $arcs$ are in proportion so are the angles which insist on them...I'm correcting the answer (e.g $\widehat{ABC}$ when an arc is mentioned) – sirfoga Mar 2 '14 at 9:26
• @kingW3 Law of sines applied in a circle: in fact consider a general chord $\overline{AB}$ on which angle $\alpha$ insists. You have that $\overline{AB} = 2r\sin\alpha$ – sirfoga Mar 18 '14 at 18:57