How is AC an angle bisector of $\angle PAB$?

Question

Here is a problem involving tangent circles.

Let $\omega$ be a circle with a diameter $PQ$. Another circle $t$ is tangent to $\omega$ at $M$ and also tangent to $PQ$ at $C$. Let $AB$ be a segment such that $AB\perp PQ$. Also, $AB$ is tangent to $t$. Also $A\in\omega$ and $B\in PQ$. Prove that $AC$ bisects $\measuredangle PAB$.

Here is my approach:

Let $AB$ tangent to $t$ at $J$. According to the Archimedies' lemma, $M$, $J$, $Q$ are collinear. Also, $MQ\times QJ=QA^2=QN^2$.

Now consider an inversion $\Psi$ centered at $Q$ with radius $QA=QN$. Note that $t$ gets maps to itself. So $\Psi(C)=C$.

Now it is clear that $QA=QN=QC$. So $C$ is the incenter of $\triangle PAN$. So, it implies that $AC$ is the angle bisector of $\measuredangle PAB$.

Answer 1

As you see in figure $\angle PBC\neq\angle QBC$ but $\angle PDC=\angle CDQ=45^o$ which is easy to prove. angle CDQ is opposite to arc QE which equal to arc EP so angles opposite to this arcs are equal .

Answer 2

I will try to give better details to some claimed properties in the figure, and provide arguments where the (implicit) posted question from my point of view did not come with a solid argument.

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Here is my picture with some more ingredients for easy reference.

So let us recall. The points in the picture are constructed in a natural way in the following order. Start with $P,Q$. Let $O$ be the mid point of the segment $PQ$. Draw $(\omega)$, the circle centered in $O$ with radius $OP=OQ$. Pick some point $M\in(\omega)$ and draw the tangent in $M$ to $(\omega)$. It intersects $POQ$ in a point $S$. The angle bisector of $\measuredangle MSP$ intersects the radius $OM$ in a point $T$. We have $OTM\perp MS$. Let $C$ be the projection of $T$ on $POQS$. We draw the circle $(t)$ centered in $T$ with radius $TM=TC$. We draw the square $CTJB$ so that $B$ is between $C$ and $Q$. Let $A$ and $N$ be the two points of intersection of $JC$ with $(\omega)$.

We have to show that $AC$ bisects $\measuredangle PAB$.

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Along the lines of the OP, we can argue as follows.

The two triangles $\Delta TMJ$ and $\Delta OMQ$ are isosceles in $T$, respectively $O$, and their angles in $T$, $O$ are equal since $TJ\| OQ$ (since both $TJ$, $OQ$ are perpendicular on $ABN$). So they are similar, so their angles in $M$ coincide, so $M$, $J$, $Q$ are colinear.

The power of the point $Q$ w.r.t. the circle $(t)$ is then $$ QC^2=QJ\cdot QM\ , $$ and we consider the inversion $\Psi$ centered in $Q$ with this power, i.e. fixing $C$. For a handy notation of the inversion of a point $X\ne Q$, we alternatively denote the point $\Psi(X)$ by $X^*$.

Which is the image by this inversion of the (projective) line $AJBN$? It is a circle through the center of inversion $Q$. Since $QOP$ is perpendicular on this line, the center of this circle is on $QOP$. Since $J^*=M$ is on this circle, this circle is determined now, it is $(\omega)$.

In particular $B^*=P$ and $P^*=B$.

Where is the point $A^*$? From $A= QA\cap AJBN$ we get $A^*= (QA)^*\cap(AJBN)^*=QA\cap (\omega)=A$. So $A$ is also fixed by the inversion. This gives $QA=QC$. So the triangle $\Delta QAC$ is isosceles. From here: $$ \begin{aligned} \widehat{CAB} &= \widehat{CAQ} - \widehat{BAQ}\\ &= \widehat{QCA} - \widehat{QPA}\\ &= \widehat{QAC} \ . \end{aligned} $$ $\square$

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Observation: From here we see that the incircle $(\kappa)$ of $\Delta PAN$ is centered in $C$, has the same size / radius as $(t)$, passes through $T$ and $B$.