How to prove the directed angle cyclic quadrilateral theorem?

Question

Hello stackExchange users!

"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! :)

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 $\measuredangle XBY>\measuredangle XB'Y.$

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

We can get a contradiction by the same way.

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