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
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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.