Are $M, P, N $ collinear? Let $\alpha $ a circle of diameters  $AB $ and $\beta $ a circle tangent to $AB $ in $ C$ and tangent to $\alpha$ in $T $. 
Let  $M\in \alpha $ and  $N\in CB $ s.t. $MN\perp AB $ and $MN $ is tangent to $\beta $. 
Show that $\angle AMC=\angle CMN $.
My idea:  Let $BT\cap \beta=$ {P}. 
I need to prove that $MN\cap \beta =${P}. In GeoGebra seems to be right. 
If this is true, then:
$\triangle BNP$ ~ $\triangle BTA $ $\Rightarrow BN\cdot AB=BP\cdot BT \Rightarrow BM ^2=BC^2$.
So, $\angle BMC=\angle BCM\Rightarrow \angle AMC=\angle CMN $.
 A: Let O and Q be the centers of $\alpha$ and $\beta$ respectively. Note that CNPQ is a square and therefore if PQ is extended to cut $\beta$ at X, then, PQX // AOCNB.

Since $\angle TXP = 0.5 \angle TQP = 0.5\angle TOB = \angle TAB$, TXA is a straight line because the two red shaded angles are on the corresponding angle positions of the two parallel lines.
Similarly, since $\angle STP = \angle TXP = \angle TAB = \angle STB$, therefore T, P, and B are collinear.
A: The points $P$, $M$, and $N$ are indeed collinear.  Consider the homothety $h$ about the point $T$ which maps the circle $\beta$ to the circle $\alpha$.  Then, $h(P)=B$.
Now, let $\ell$ be the tangent to $\beta$ at $P$ and $L$ the tangent to $\alpha$ at $B$.  Clearly, $h(\ell)=L$.  Since $L$ is perpendicular to $AB$, $L$ is parallel to $MN$.  That means $\ell$ is parallel to $MN$.  Note that $MN$ is tangent to $\beta$ at some point $Q$. This shows that either $P=Q$ (so the line $MN$ and the line $\ell$ are the same), or $PQ$ is a diameter of $\beta$.  You can rule out the latter case easily.
