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"Prove that 4 points $A, B, X, Y$, no 3 collinear, are concyclic if and only if $\measuredangle XAY = \measuredangle XBY$

(Where $\measuredangle$ stands for directed angle $\mod 180^\circ$)"

I'm quite confused with this one, how does a proof look like that this actually matches the "normal" cyclic quadrilateral theorem? This is part of the book Euclidean Geometry in Mathematical Olympiads by Evan Chen.

Thanks in advance! :)

  • $\begingroup$ There's two cases for the 'normal' theorem depending on the position of $A,B,X$ and $Y$ relative to each other. Your question can be answered by by looking at the different position cases, and seeing that in all instances, $\measuredangle XAY=\measuredangle XBY$. $\endgroup$
    – user574848
    Jul 16, 2019 at 10:56
  • $\begingroup$ Yeah, my confusion is more about how to the directed angle cyclic quadrilateral theorem implies the 'normal' one. $\endgroup$
    – smtf11
    Jul 16, 2019 at 15:25

1 Answer 1


Let $\Phi$ be a circumcircle of $\Delta AXY.$

  1. Let $B$ be placed inside $\Phi$ and $BX\cap\Phi=\{X,B'\}$.

Thus, $$\measuredangle XB'Y=\measuredangle XAY=\measuredangle XBY,$$ which is a contradiction because $\measuredangle XBY>\measuredangle XB'Y.$

  1. Let $B$ be placed outside $\Phi.$

We can get a contradiction by the same way.

Id est, $B\in\Phi$ and we are done!


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