# How to prove the directed angle cyclic quadrilateral theorem?

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.

• 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, $\measuredangle XAY=\measuredangle XBY$. Jul 16, 2019 at 10:56
• Yeah, my confusion is more about how to the directed angle cyclic quadrilateral theorem implies the 'normal' one. Jul 16, 2019 at 15:25

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