6 point lying on a common circle $Z$ is an interior point of segment $XY$. Three semicircles are drawn over segments $XY$, $XZ$ and $ZY$ on the same side. The midpoints of the arcs are $M1$, $M2$ and $M3$ respectively. A circle tangent to these semicircles is internally tangent to the semicircle centered at $M1$. Let their point of tangency be $T$. Show that the midpoint of segment $XY$ and the points $Z$, $M1$, $M2$, $M3$ and $T$ all lie on a common circle.

So far I've tried using coordinates and equations such as if $X=(0,0)$, $XZ=a$ and $YZ=b$ then $Z=(a,0)$, $Y=(a+b,0)$, $M=(\frac{a+b}{2},0)$, $M_1=(\frac{a+b}{2},\frac{a+b}{2})$, $M_2=(a,a)$, $M_3=(a+\frac{b}{2},\frac{b}{2})$ and after that applying the Equation of the Circle, but it seems pretty hard going in parts. Any other ideas? Or should I continue this?
 A: The result will follow if we take a particular scaling in which the length $XY$ is 2. We approach it using complex numbers for the points, and the fact that four noncollinear complex numbers $a,b,c,d$ are on a circle if and only if the cross-ratio
$$R(a,b,c,d)=\frac{(a-c)(b-d)}{(a-d)(b-c)}$$ is a real number (imaginary part zero).
Set things up so that $X=-1,\ Y=1,\ M_1=i,\ Z=t$ where $-1<t<1$ so that $Z$ is a real point interior to segment $XY$. Then we have also $M=0$ and
$$M_2=\frac{t-1}{2}+\frac{t+1}{2}i,$$
$$M_3=\frac{t+1}{2}+\frac{1-t}{2}i.$$
The point $T$  may with some work (based on its being on both $|z|=1$ and on the circle $C$ through the three $M_j$), be shown to be
$$T=\frac{2t}{t^2+1}+\frac{1-t^2}{t^2+1}i.$$
We'll say something about how this needs more work after the cross-ratios are determined.
We now compute cross-ratios, and find
$$R(M,M_3,M_1,M_2)=\frac{2}{t+1},$$ a real number, so that these four are on a circle $C$, and also
$$R(M,Z,M_3,M_1)=\frac{-t^2-1}{t-1},$$ real and putting $Z$ also on $C$, and finally
$$R(M,Z,T,M_1)=\frac{t^2+1}{1-t},$$ again real and putting $T$ on circle $C.$ Thus all six points $M,Z,M_1,M_2,M_3,T$ lie on the circle $C$.
NOTE: I found the expression for the complex number $T$ by assuming it lies on $C$ and on the unit circle. Further work is necessary in order to show that this $T$ is the point of tangency of the circle centered at $O$ in the diagram with the unit circle. This may be involved; all I can think of now is that we can find the distances from $O$ to the centers of the two smaller circles, and thus determine an expression for $O$; then further computation would be needed to show the circle centered at $O$ of appropriate radius is actually tangent to the three semicircles. So really this answer only shows that five of the six points (excluding $T$) lie on the circle $C$. I found the equations not to be simple which would correctly show $T$ has the required tangency property w.r.t. the circle centered at $O$.
A: I'd like to start the other way round:


*

*Show that the lines $M_2M_3$ and $M_1Z$ bisect (by using the coordinates), thus the four points $M_1, M_2, M_3, Z$ lie on the same circle.

*By symmetry, point $M$ lies on the circle centered at $I$.

*Define the point $T$ at the intersection between the largest semicircle and the circle centered at $I$. 


Now the challenge is to show that $T$ is also on the circle centered at $O$.

