Arbelos and its angle bisector I have recently been reading about a very interesting geometry problem and have tried to solve it. I'm now in a point, in which I don't know how to move forward and would appreciate if someone could help.
The problem is the following

Consider an arbelos $ACB$ (with diameters $\overline{AB}, \overline {AC}$ and $\overline{CB}$) as shown on the image. Let $M$ be the centre of the circle being tangent to the three semicircles and denote by $K$ the tangency point with the semicircle over $\overline{AB}$.
Prove that $\angle AKC=\angle CKB$ or equivalently, that $\overline{KC}$ is the angle bisector of $\angle AKB$.


My attempt so far:
Denote:

*

*by $\omega$ the incircle centered at $M$


*by $\tau$ the circle with the diameter $\overline{AB}$


*by $\varphi$ the circle with the diameter $\overline{AC}$


*by $\sigma$ the circle with the diameter $\overline{CB}$


*by $H$ and $J$ the tangency points between $\omega$ and $\varphi$, $\sigma$ respectively


*by $D,O,E$ the midpoints of the segments $\overline {AC},\;\overline {AB}$ and $\overline {CB}$ respectively


*by $P$ the intersection of the segment $\overline {KB}$ with $\omega$


*by $L$ and $N$ the higher points of $\varphi$ and $\sigma$ respectively (i.e. the intersection between $\varphi$, $\sigma$ and the perpendicular bisectors to $\overline {AC}$ and $\overline {CB}$ respectively)


*by $t$ the tangent line to $\omega$ and $\varphi$ throught $H$ (colored purple)


*by $Q$ and $R$ two random points on $t$ such that $Q$ lies left of $H$ and $R$ right of $H$

I've found out that the quadrilateral $HCBK$ is cyclic and has $N$ as circumcentre, which is awesome because by the theorem of the inscribed angle and Thales' theorem $$\angle CKB= \frac{\angle CNB}{2}=\frac{\pi}{4}$$
which is exactly half of the angle $\angle AKB=\frac{\pi}{2}$
The part where I'm stuck is where you have to prove that $N$ is the circumcentre of the cyclic quadrilateral $HCBK$

You can find the proof that shows that $HCBK$ is cyclic here
I'll, first of all, introduce a well know Lemma

$\mathbf {Lemma \; 1}$

This is an extract from the excellent book "Euclidean Geometry in Mathematical Olympiads" (pg. 16) from Evan Chen. I guess this is almost trivial, so I won't prove it.

Now back to the problem, note that the circle $\tau$ and the circle $\omega$ are homothetic with $K$ (the tangency point) as homothety center.
The point $P$ is mapped to $B$, which is the "rightest" point of the circle $\tau$. $P$ is thus also the "rightest" point of $\omega$.
Note, furthermore, that $\omega$ and $\varphi$ are also homothetic with $H$ as homothety center. Since $A$ is the "most left" point of $\varphi$ and the homothety coefficient regarding $H$ is negative, $A$ is mapped to $P$, which implies that $A$, $H$ and $P$ are collinear.
By Lemma 1: $$\angle HKP= \angle RHP$$
Since $A$, $H$ and $P$ are collinear: $$\angle RHP=\angle QHA$$
Again, by Lemma 1: $$\angle QHA= \angle HCA$$
which finally implies $$\angle HKP= \angle HCA \Rightarrow \angle BCH+\angle HKP= \pi$$
Since opposite angles sum up to $\pi$, $HCBK$ is cyclic.
Q.E.D.

I've read a proof which uses inversion, but since I'm not that familiarized with inversion, I would appreciate if you could find a proof that uses elementary geometry or analytic geometry.
PS: It is obvious that the circumcenter must lie on the perpendicular bisector of $\overline{CB}$ (just like $N$), the problem is proving that is has to be $N$.

EDIT regarding the bounty (10.02.2019):
Unless someone comes up with an extraordinary answer, I would like to reward @Calum Gilhooly's exceptional contribution.

 A: Here's a possible simple answer. Set first of all:
$$
AC=2a,\quad BC=2b, \quad \angle AOK=\phi,\quad KM=r.
$$
We have then:
$$
MD=a+r,\quad ME=b+r,\quad OD=b,\quad OE=a,\quad OM=a+b-r.
$$
By the cosine rule applied to triangles $OMD$ and $OME$ we obtain two equations:
$$
(a+r)^2=b^2+(a+b-r)^2-2b(a+b-r)\cos\phi \\
(b+r)^2=a^2+(a+b-r)^2+2a(a+b-r)\cos\phi \\
$$
which can be solved for $r$ and $\cos\phi$ to get:
$$
r={ab(a+b)\over a^2+b^2+ab},\quad \cos\phi={b^2-a^2\over a^2+b^2}.
$$
On the other hand:
$$
AK^2=2(a+b)^2(1-\cos\phi)={4a^2(a+b)^2\over a^2+b^2},\quad
BK^2=2(a+b)^2(1+\cos\phi)={4b^2(a+b)^2\over a^2+b^2}.
$$
Hence:
$$
{AK\over BK}={a\over b}={AC\over BC}
$$
and the thesis follows from the angle bisector theorem. 
A: This isn't a solution, just a suggestion of another possible line of attack - although it looks difficult.
Because the three circles with collinear centres, $\tau$, $\varphi$ and $\sigma$, play quite similar roles in the problem, it is natural to construct the points $H$ and $J$, and ask if $HB$ and $JA$ have similar properties to $KC$; and it looks as if they do.
Thinking this was just a side-issue, however, and not wanting to clutter up my diagram, I initially only drew light dotted lines for $HB$ and $JA$, so it was a long time before I noticed an "obvious" fact: $KC$, $HB$ and $JA$ appear to concur.
Moreover, they appear to concur also with the circle $\omega$, at its "South Pole", $S$, located on the perpendicular from $M$ to $AB$, and projected by the homothety between $\omega$ and $\tau$ to the point of $\tau$ that was significant in a previous answer (now deleted).
If this concurrency could be proved, then the equality of $\angle AKC$ and $\angle CKB$ would follow by a very similar argument to the one in the previous answer: the chords $TS$ and $SP$ are equal, and therefore subtend equal angles at the circumference of $\omega$.

The apparent concurrency with $\omega$ throws into sharp relief the different role played by this circle compared with the other three. Setting aside the difference, it is natural to wonder whether $KC$, $HB$, $JA$ remain concurrent even when $D$, $O$, $E$ are no longer supposed collinear. This also appeared to be true:

Then, searching the Internet, I found the article Soddy Circles and David Eppstein's Centers, from which I quote:

Thus, the Soddy circles have a long history, in the course of which they were under close attention of several exceptional people. This is then so much more remarkable that one interesting property of the configuration has been discovered as late as 2001. In 2001, D. Eppstein, of the Geometry Junkyard fame, published the following observation (see also a partial online version):
(1) Four touching circles, when taken two by two, define two points of tangency and, therefore, a straight line. There are three such lines. The three lines are concurrent.
This is true for both inner and outer configuration, so there are two points of note, known now as Eppstein's points.
Eppstein derives his result as a particular case of a 3D configuration of four spheres. Any four mutually tangent spheres determine six points of tangency. The six points are naturally divided into three pairs of opposite tangencies, i.e. tangencies, in which one is defined by two spheres distinct from the pair of the spheres defining the other.
Lemma
[Altshiller-Court, p. 231].
[I can't find this in the 2007 Dover reprint - CG.]
The three lines through the opposite points of tangency of any four mutually tangent spheres in R3 are concurrent. 

As even proving the mere concurrency of $KC$, $HB$, $JA$ thus appears to be non-trivial, I judged it wise to leave matters at this point!

This figure is rather cluttered (partly because I couldn't resist including the square $KFCG$, from my first attempt at a proof, which suffered a similar fate to another poster's deleted answer!), but it clearly shows the angle bisection properties of $HB$ and $JA$:

