Since $OA=OC$, $OB=OD$, we can say that $O$ - "two bicyclist point" for circles $(BXC)$ and $(AXD)$, then, $\angle OYX=90$. Similary we can say, that $\angle OZX=90$, and points $O, X, Y, Z$ lie on a circle. QED

Solution: Claim. $A,D,O,Z$ are concyclic. (Proof) $\angle AZX = \angle ABX = \angle ACD=\angle XZD$. Thus $\angle AZD = 2\angle ABD = \angle AOD$. Main problem: As $AO = DO$ and $XZ$ is the angle bisector of $\angle AZD$, from the claim $\angle OZX = 90^{\circ}$. Similarly $\angle OYX = 90^{\circ}$ and we are done.

actually it must be $\Delta CXD$ in place of $\Delta BXD$

Can anyone plz make figure?

leon.tyumen wrote: Since $OA=OC$, $OB=OD$, we can say that $O$ - "two bicyclist point" for circles $(BXC)$ and $(AXD)$, then, $\angle OYX=90$. Similary we can say, that $\angle OZX=90$, and points $O, X, Y, Z$ lie on a circle. QED What is byclist point and why is $\angle OYX=90$ and $\angle OZX=90$. Also TheDarkPrince wrote: Solution: Main problem: As $AO = DO$ and $XZ$ is the angle bisector of $\angle AZD$, from the claim $\angle OZX = 90^{\circ}$. Similarly $\angle OYX = 90^{\circ}$ and we are done. How did he get $\angle OZX=90^\circ$ can anyone please explain.

Trivial. Just note that $Y$ and $Z$ are the miquel points of $ABCD$.Since $ABCD$ is cyclic we get $\angle OYX = \angle OZX=90^{\circ}$.Hence the result follows.

It Kazakhstan 2001.

China Mathematical Olympiad 1992

pkrmath2004 wrote: China Mathematical Olympiad 1992 Yes indeed. https://artofproblemsolving.com/community/c6h556271p3233203 For reference.

very nice problem! Claim 1:- points $O,Y,A,B$ are concyclic Proof:- we join $XY$ to get: $\angle{AYB}=360^{\circ}-(\angle{XYA}+\angle{XYB})$ $=\angle{ADB}+\angle{ACB}$ $=2\angle{ADB}$ $=\angle{AOB}$ so we get that points $O,Y,A,B$ are concyclic $\blacksquare$ Claim 2:- Points $C,Z,O,B$ are concylic Proof:- $\angle{BZC}=\angle{BZX}+\angle{CZX}$ $=\angle{XAB}+\angle{XDC}$ $=2\angle{CAB}$ $=\angle{BOC}$ , hence we get points $C,Z,O,B$ are concyclic $\blacksquare$ Claim 3:- $\angle{XYO}=90^{\circ}$ Proof:- $\angle{XYO}=\angle{XYA}-\angle{OYA}$ $=180^{\circ}-\frac{\angle{AOB}}{2}-\left(90^{\circ}-\frac{\angle{AOB}}{2}\right)$ $=90^{\circ}$ $\blacksquare$ similarly using Claim 2 we can prove that $\angle{XZO}=90^{\circ}$ hence we get that points $O,X,Y,Z$ lie on a circle $\blacksquare$

Cute problem! Claim 1: Points $Z$,$B$,$C$,$O$ are concyclic. Proof: Notice that $\angle BZX=\angle XZC=180^{\circ}-\angle BAC$. This gives $\angle BZC=2\angle BAC=\angle BOC$, and we are done! By symmetry, we have that points $B$,$O$,$Y$, $D$ are concyclic. Claim 2: $\angle OZX=90^{\circ}$ Proof: $\angle BZO=180^{\circ}-\angle OCB= 90^{\circ}+\angle BAC$. This gives $\angle OZX=90^{\circ}$, and we are done! By symmetry, $\angle OYX=90^{\circ}$ as well, giving that $O$, $X$, $Y$, $Z$ lie on a circle.

Inner Miquel Points !?!??!!! Invert about the circumcircle. It is well known that $X$ goes to the Miquel point of $ABCD$. $Y$ and $Z$ then go to $AB\cap CD$ and $AD \cap BC$, which are collinear with the Miquel point.