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. slikica 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.

  • $\begingroup$ Thanks,anyway does that mean that if two chords are in proportion so are their angles?Also unrelated to that is the angle that insists on radius equal for all radius(es) in one circle?And is it equal to 45 degree? $\endgroup$ – kingW3 Mar 1 '14 at 23:35
  • $\begingroup$ 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) $\endgroup$ – sirfoga Mar 2 '14 at 9:26
  • $\begingroup$ @kingW3 Referring to the question raised in your comment, the length of a chord is proportional to the sine of that angle at the circumference. See my proof in the lemma of my answer to problem #500669. $\endgroup$ – Mick Mar 18 '14 at 17:58
  • $\begingroup$ @kingW3 Law of sines applied in a circle: in fact consider a general chord $\overline{AB}$ on which angle $\alpha$ insists. You have that $\overline{AB} = 2r\sin\alpha$ $\endgroup$ – sirfoga Mar 18 '14 at 18:57

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