Show that this homothety takes $K$ to $M$ 
Let $C_1$ be a circle with centre $O$ and a chord $AB$,and consider a circle $C_2$ tangent internally to $C_1$ at $T$ and to $AB$ at $K$. Let $M$ denote the midpoint of the arc AB not containing $T$. Show that $T,K,M$ are collinear.

I found a homothety with centre at $T$, which send $P$ to $O$.So,we need to show that this homothety send $K$ to $M$. No other ideas came to my mind. Thank you!
 A: I do not know how rigorous this proof is, but here goes.

Suppose that we start by creating circle $C_1$ and it's chord $AB$. Now we find the midpoint of sector $ACB$ and label it $M$. Now we draw an arbitrary line that goes through $M$, $AB$ and circle $C_1$. 
The intersections will be $T,K$ and $M$ and thus by definition, they are and will be collinear.
Now we draw a line through $C_1$ and $T$. Then we draw a line perpendicular to line $C_1T$, and by definition, the perpendicular line will be tangent to $\bigcirc \ C_1$ at $T$. Since $\bigcirc C_2$ should be internally tangent to $\bigcirc C_1$ at $T$, then the center $C_2$ must necessarily lie on $C_1T$.
Now since $\bigcirc C_2$ must pass through, and in fact be tangent to, $T$ and $K$, then $TK$ will be a chord of $\bigcirc C_2$, and a perpendicular bisector of $TK$ at $C$ will pass through the center of $\bigcirc C_2$. Since we know the center of $\bigcirc C_2$ must be on line $C_1T$ then, the intersection of the perpendicular bisector of $TK$ and line $C_1T$ will be the center of $\bigcirc C_2$. This now begs the question as to why $\bigcirc C_2$ should be tangent to $AB$ at $K$. 

Notice that there a kite forms from the points $T,C_2,K,G$. Notice too that $C_2T \cong C_2K$, since they are the radii of $\bigcirc C_2$. Notice that $TC\cong KC$, because $C$ is the midpoint of $TK$. Because of these, $\angle C_2KG$ must be a reflection of $\angle C_2TG$. And through that, we prove that $K$ is indeed the point of tangency between $\bigcirc C_2$ and $AB$, and thus completing the proof that $T, K$ and $M$ are collinear.
Additionally, we can look at an extreme case:
Notice that when the lines of tangency are parallel to the $x$-axis, where the center of $\bigcirc C_2$ sits on the point of origin, and when $TK$ is the diameter of $\bigcirc C_2$, $T,K,M$ are collinear. All the centers and $T,K,M$ are aligned.

A: Homothety with center at $T$ and ratio $k= {R\over r}$ ($R$ is radius of $C_1$ and $r$ radius of $C$) takes circle $C$ to $C_1$.
Let $M'$ be the image of $K$. Since $K$ is on tangent $AB$ for $C$ then $A'B'$ ($X'$ is image of $X$) is also tangent on $C_1$ at $M'$. So the line $M'O$ is perpendicular to $A'B'$. But since $A'B'$ is parallel to $AB$ the line $M'O$ must be perpendicular to $AB$ also. But then it is perpendicular bisector for $AB$ and thus $M'$ halves arc $AB$. So $M'=M$ and we are done.
