Show that $MC$ is the bisector of the angle $\widehat{AMN}$ .

Question

Let $\alpha $ be a semicircle by diameter $AB $ and $\beta $ a circle tangent to $AB $ in $C$ and tangent to $\alpha $ in $T $.

Let $M\in\alpha $ and $N\in CB $ such that $MN\perp AB $ and $MN$ is tangent to $\beta $.

Show that $MC$ is the bisector of the angle $\widehat{AMN}$ .

I tried to apply the reciprocal bisector theorem and the power of a point.

Answer 1

Let $Q$ be a touching point of $MN$ and $\beta$. First show that $Q$, $B$ and $T$ are colinear (observe a homothety at $T$ which takes $\alpha$ to $\beta$ and thus it must take tangent at $B$ to a tangent on $\beta$ so they are parallel and thus $B$ goes to $Q$) and that $A,N,Q,T$ are concyclic. Then by the power of the point of $B$ we have:

$$BC^2 = BQ\cdot BT = BN\cdot BA = BM^2$$ The last equation is POP of $B$ with respect to the circle with diameter $AM$. So $BC =BM$ and should be an easy angle calculation....

Answer 2

Assuming that the semicircle $\alpha$ is the graph of $f(x)=\sqrt{1-x^2}$ over $[-1,1]$, the center of $\beta$ has to lie on the graph of the parabola $g(x)=\frac{1-x^2}{2}$. Assuming $C=(x,0)$ we have $N=\left(x+\frac{1-x^2}{2},0\right)$. Now you may easily compute the $y$-coordinate of $M$ and check that $$\frac{AM}{MN}=\frac{AC}{CN}=\frac{2x+2}{1-x^2}=\frac{2}{1-x} $$ holds, proving that $MC$ bisects $\widehat{AMN}$. Analytic geometry is not always the wrong tool.