TITLE: Cyclic quadrilateral in metric space
QUESTION [5 upvotes]: Consider a metric space $(\Bbb M,d).$
If $X,Y,Z\in \Bbb M.$ We define cosin of angle by
$$\cos(\angle YXZ)=\frac{d(X,Y)^2+d(X,Z)^2-d(Y,Z)^2}{2d(X,Y)\cdot d(X,Z)}.$$
If we have four points $A,$ $B,$ $C$ and $D$ in $\Bbb M$ satify the indentity
$$d(A,B)\cdot d(C,D)+d(A,D)\cdot d(B,C)=d(A,C)\cdot d(B,D).$$
Then we say that $A,$ $B,$ $C,$ $D$ are concyclic.

My Question 1. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show that $$\cos(\angle BAC)=\cos(\angle BDC).$$
My Question 2. If $A,$ $B,$ $C$ and $D$ are concyclic then can we show there is a point $O$ such that $$d(O,A)=d(O,B)=d(O,C)=d(O,D).$$
My Question 3. If two questions 1 and 2 are not true then what is condition of $\Bbb M$ such that these are true?

REPLY [4 votes]: For a counterexample to 2, we can take the points $A,B,C,D$ at $0,30,60,90$ degrees on the unit circle, and then add a bump to perturb the $x$-axis.  In more detail:
Consider the surface $z=f(x,y)^2$, where
$$f(x,y)=\max\left(0,\ 0.1 - \left(x-0.5\right)^2 - y^2 \right),$$
and distances from shortest paths on the surface.
Take points
$$A=\big(1,0,0\big),\
B=\big(\frac{\sqrt{3}}{2},\frac{1}{2},0\big),\
C=\big(\frac{1}{2},\frac{\sqrt{3}}{2},0\big),\
D=\big(0,1,0\big).$$
The distances between $A,B,C$ and $D$ are all the same as in the $xy$-plane, since the surface is flat away from $(0.5,0,0)$.  So $A,B,C,D$ are concyclic in this surface.
Taking $O=(0,0,0)$, the distances $OB,OC,OD$ are also as in the $xy$-plane, and $O$ is the unique point on the surface with $OB=OC=OD$.  But the shortest path from $O$ to $A$ has to pass by the bump, so $OA>OB$, and there is no point with $OA=OB=OC=OD$.