See the figure below:
Let $I_1F = r_1$, $I_2G = r_2$, and $I_1E = r_3$.
Note that $GD =r_2$, $DF = r_1$, and $I_2E = r_3$.
Note also that $\triangle I_1DI_2$ is a right triangle.
So using Pytagoras' Theorem in triangles $\triangle DGI_2$, $\triangle DFI_1$, and $\triangle I_1DI_2$ we get:
$$DI_1= r_1 \sqrt{2},$$
$$DI_2= r_2 \sqrt{2},$$
$$I_1I_2= \sqrt{2(r_1^2+r_2^2)} \tag1$$
Let $m(\angle I_1AI_2)= \alpha$, as $AI_2$ is bisector of $\angle CAD$ and $AI_1$ is bisector of $\angle BAD$, we can conclude that
$$\alpha = 45 ^{\circ}$$
But if $\alpha = 45 ^{\circ}$ then $\angle I_1EI_2$ is a right angle (central angle), and using Pytagoras' Theorem in $\triangle I_1EI_2$ and equation $(1)$ we get:
$$r_3=\sqrt{r_1^2+r_2^2} \tag2$$
Now let's calculate $r$, the inscribed circle radius of $\triangle ABC$, and compare it with $r_3$.
See the picture below:
We know that $\triangle ADC$, $\triangle BDA$, and $\triangle BAC$ are similar, then
$$\frac{r_1}{c}=\frac{r_2}{b}=\frac{r}{a}=k \tag3$$
So using relation $(3)$ and Pytagoras' Theorem again ($\triangle ABC$) we get:
$$a^2=b^2+c^2 \Rightarrow \left(\frac{r}{k}\right)^2=\left(\frac{r_1}{k}\right)^2 + \left(\frac{r_2}{k}\right)^2 \Rightarrow $$
$$\Rightarrow r^2=r_1^2+r_2^2 \Rightarrow r=\sqrt{r_1^2+r_2^2} \tag4$$
Therefore comparing $(2)$ and $(4)$ we can conclude finally that
$$r=r_3.$$
Based on your picture, let's assume $M$ is the center of the circumscribed circle of $PCQ$.
Now consider the line segment $MC$.
We first show that $MC$ is the bisector. Notice that:
$$\angle HCA=90^{\circ}-\angle A \implies \angle PCA=\angle HCP=\angle 45^{\circ}-\angle \frac{A}{2}\implies \angle HPC= 45^{\circ}+\angle \frac{A}{2};$$
hence: $\angle QCM=90^{\circ}-\angle QPC=45^{\circ}-\angle \frac{A}{2}$ (because we assumed $M$ is the center of the circumscribed circle). As a result, we have: $\angle QCM=\angle PCA$.
Now, realize that $\angle QCP=45^{\circ}$. Therefore:
$$45^{\circ}= \angle QCP=\angle QCM +\angle MCP=\angle PCA +\angle MCP=\angle MCA\implies \angle MCA=45^{\circ}.$$
Since we assumed $M$ is the center of the circumscribed circle and $\angle QCP=45^{\circ}$, we easily conclude $\angle AQM=45^{\circ}$. By using the law of sine in $AQM$ and $AMC$, we get:
$$\frac {\sin \angle QAM}{\sin 45^{\circ}}=\frac {QM}{AM}=\frac {MC}{AM}=\frac {\sin \angle MAC}{\sin 45^{\circ}};$$
hence $\angle QAM=\angle MAC$ (simply because $\angle A<90^{\circ}$), which means $AM$ is the bisector as well.
Best Answer
This may be also a picture proof, the reader may want to ignore below everything but the pictures and come with her or his own proof. But for didactic purposes, since we have a NMO problem and want to collect all points, there is also the story around the pics.
I need a lemma first, some points from the posted question are involved, notations are kept. I will introduce some more points, showing how the picture was realized.
Proof: Draw parallels and perpendiculars through the circumcenters $O$ and $\Omega$ to the sides $BA$ and $BC$ of $\Delta ABC$. We obtain as in the picture two orange triangles with same hypotenuse $O\Omega$, and one of the legs is $m/2$ in each, the distance from the mid points of $AB$ and $MB$ for one triangle, respectively mid points of $CB$ and $NB$ for the other one.
These triangles are thus congruent, so $\Omega O$ is an angle bisector for an angle delimited by sides parallel to $BA$ and $BC$. So $\Omega O$ is parallel to the angle bisector $BS$ of $\widehat{ABC}$. Since $\Omega O\|BS\perp BR$ we obtain that $\Omega O$ is also the side bisector of $BR$, thus $\Omega B=\Omega R$, so $R$ is also on the circle $\odot(\Omega)=\odot(BMN)$.
To see that $RM=RN$, compare $\Delta RNC$ and $\Delta RMA$. They have $RC=RA$, $NC=m=NA$, and $\widehat{NCR}=\widehat{BCR}=\widehat{BAR}=\widehat{MAR}$. So the two triangles are congruent. (And a rotation around $R$ with angle $\widehat{NRM}=\widehat{NBM}=\widehat{CBA}=\hat B$ brings one triangle into the other.)
$\square$
We are now in position to show the claimed result. Here there is a slightly changed way to introduce points, done in an equivalent manner. Since some other points, useful for the proof, are also introduced let us explicitly restate:
Proof: For $\bbox[yellow]{(1)}$ - using the above Lemma: $$ \widehat{RNK} = \widehat{RNM} = \widehat{RBM} = \widehat{RBA} = \widehat{RCA} \ . $$ So $CNRK$ cyclic. Similarly $AMRK$ cyclic.
$\bbox[yellow]{(2)}$ take first from the picture only the piece involving the points $K,C,N;Q,Q',Q*$ and the angle bisectors in $C,K$. Denote by $x,y$ half of the angle $\hat C$, respectively $\hat K$ in $\Delta CKN$. Then $$ \begin{aligned} \widehat{Q^*CQ'} &= \widehat{Q^*CN} + \widehat{NCQ'} = \widehat{Q^*KN} + \widehat{NCQ'} =x+y=\widehat{Q'CK} + \widehat{Q'KC} \\ &=\widehat{Q^*Q'C}\ . \end{aligned} $$ So $\Delta Q^*Q'C$ isosceles, $Q^*Q'=Q^*C$. The angle bisectors $CQ$, $CQ'$ are perpendicular, so $\Delta QCQ'$ has a right angle in $C$. Its mid point of the hypotenuse is the only point of this side at same distance from $C$ and $Q'$, so it is $Q^*$. We obtain thus $Q^*Q=Q^*C=Q^*Q'$. Since $Q^*$ is the mid point of the arc $\overset\frown{CN}$ of $\odot(CNRK)$, we can add also $Q^*N$ to the chain of equalities. So $Q^*$ is the center of the circle $\odot(CQNQ')$. The same happens also for $\odot(APMP')$.
We can attack $\bbox[yellow]{(3)}$. The obvious $R$-rotation moves $\Delta RNC\to\Delta RMA$, so take in the movement also corresponding objects, circles $\odot(RNC)\to\odot(RMA)$, arcs of them $\overset\frown{NC}\to\overset\frown{MA}$, and their mid points $Q^*\to P^*$. So $$ RQ^*=RP^*\ , $$ so the perpendicular bisector of $P^*Q^*$ goes through $R$, and w.r.t. this bisector we have reflected pairs of points $(P^*,Q^*)$, and $(P,Q)$, and $(P',Q')$, the last two pairs since we know the distances on the line $\kappa=KPP^*P'QQ^*Q$.
$\square$