Cyclic quadrilateral in metric space

Question

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?

Answer 1

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 $OA=OB=OC=OD$.