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It's straightforward to show that if quadrilateral $ABCD$ has its vertices lying on a common circle, then its opposite angles complement one another. enter image description here

That is $\angle DAB + \angle DCB =\angle ADC + \angle ABC=\pi$

Is this condition sufficient for $ABCD$ to be cyclic?

If we draw a circumcircle to the $\Delta ABC$ and pick a point $X$ on the arc $AC$ such that $AXCB$ is a quadrilateral whose sides do not intersect, then by necessary condition, we immediately have $\beta = \pi-\alpha$. We also know that $\gamma =\pi -\alpha$.

Is it necessary for $D$ to lie on the common circle with $X$ in order for $\beta = \gamma$?

So, assume for a contradiction $D$ does not lie on the circumcircle, it's either outside or inside.

enter image description here

How does one make progress? I read everywhere that these conditions are equivalent, but haven't found a proof to the converse

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Let $DC$ intersects our circle in the points $E$ and $C$.

Thus, $\measuredangle AEC=180^{\circ}-\alpha=\measuredangle D$, which is contradiction because $\measuredangle AEC>\measuredangle D$.

By the same way we can get a contradiction if $D$ placed inside the circle.

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  • $\begingroup$ smacks forehead , it's so obvious :( even more obvious than the forward implication $\endgroup$ – Alvin Lepik May 29 '17 at 11:47
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We know that $\hat{B}+\hat{D}=\pi$ and want to prove that $D$ is on the circle.

By contradiction, suppose $D$ is not a point of the circle, for instance it's outside the circle. Connect $D$ with the centre $O$ and call $E$ the intersection of $OD$ with the circle. Quadrilateral $ABCE$ is inscribed in the circle therefore $\hat{E}$ is supplementary of $\hat{B}$. Now consider triangles $CED$ and $AED$. The external angles are $\alpha>\beta$ and $\gamma>\delta$ thus $\alpha +\gamma>\beta+\delta$ while $\alpha +\gamma=\hat{E}\cong \hat{D}=\beta+\delta$ being $\hat{D}\cong \hat{E}$ because they are supplementary of $\hat{B}$. Contradiction.

In a similar way we get a contradiction if $D$ is internal. Must be $\beta >\alpha$ while we have shown that $\alpha=\beta$. Again contradiction.

If $D$ is on a line tangent to the circle we have just a particular case of the first case, so there is nothing more to say.

The hypothesis that $D$ is not on the circle lead to contradictions so $D$ lies on the circle.

QED

enter image description here

enter image description here

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  • $\begingroup$ @kulisty Come and see my proof :) $\endgroup$ – Raffaele Oct 9 '17 at 14:16
  • $\begingroup$ There is exactly the same problem in your proof as in the previous one. How do you know that point $E$ doesn't lie on the opposite side of $AC$ ? i.stack.imgur.com/NaSZc.png $\endgroup$ – Kulisty Oct 9 '17 at 17:10
  • $\begingroup$ If this happens then point $C$ must be elsewhere because $C$ and $A$ are supplementary and cannot be both obtuse. So we can take the intersection of $DC$ with circle as point $E$ and go on with the proof as before imgur.com/nPRfI39 $\endgroup$ – Raffaele Oct 9 '17 at 17:40
  • $\begingroup$ Sorry but I don't quite get your argument (although I think it's correct). One of the angles $A$ and $C$ (say $C$) is not obtuse. How does it imply that $CD$ intersects the circle in $E$ in such way that $E$ is between $C$ and $D$ ? $\endgroup$ – Kulisty Oct 9 '17 at 18:39
  • $\begingroup$ @Kulisty Honestly I used GeoGebra when I realized this fact. Now it's late and I am tired, but I don't think it is difficult to prove. Anyway I would have been honoured to have a student like you attending my classes: very smart observations! I wish I had more students like you! $\endgroup$ – Raffaele Oct 9 '17 at 18:45

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