This is a question whether a proof I will describe is palatable.
I want to prove something, esp. a question on geometry, by assuming that which is in question to be in fact true, and then prove that it is absolutely probable to construct it geometrically that way.
For example, in this problem, it asks me to prove that $T,K$ and $M$ are collinear.
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.
In the proof I submitted, I assume that $T,K,M$ are in fact collinear by drawing a line that passes through $T,K,M$ and showing that a possible geometric construction can be made, and that the circumstances and constraints of the problem are satisfied.
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.
Is this "proof by reverse construction" even allowed? Or am I just arbitrarily naming a generic process?