Concyclic incenters of triangles concerning Apollonius circle Problem:

$ABCD$ is a tangential quadrilateral and $P$ is a point such that $$\dfrac{PB}{PD}=\dfrac{AB}{AD}=\dfrac{CB}{CD}.$$
  Let $I_1$, $I_2$, $I_3$, $I_4$ be the incenters of $\triangle PAB$, $\triangle PBC$, $\triangle PCD$, $\triangle PDA$, respectively.
Proof that $I_1$, $I_2$, $I_3$ and $I_4$ are concyclic.


I have read this problem Prove that $ I_1, I_2, I_3, I_4 $ are concyclic, but it is not actually similar to my problem.
I suspect the problem has something to do with Apollonius circles (i.e., circle $PAC$), but I don't know how to use this information in any way. I cannot understand how Apollonius circles are connected to incenters.
Edit:
My analysis is that the problem clearly implies two cases: either $AC$ is the perpendicular bisector of $BD$, or $AB = CB$ and $AD = CD$ (a kite). The first case is trivial, since $I_1$, $I_2$, $I_3$, $I_4$ here forms a rectangular. The real difficult part of this problem is the second case.
Based on observations, I have made out the following results, but I cannot prove them. They may provide some clues for the original problem. I would appreciate it if anyone could help with these conjectures too:


Assume the outer and inner bisectors of $\angle BAD$ intersect $BD$ at point $M$ and $N$ respectively (so $M$ and $N$ lie on the Apollonius circle), then:
  
  
*
  
*$M$, $I_1$ and $I_4$ are colinear, likewise are $M$, $I_2$ and $I_3$;
  
*$N$, $P$, $I_1$ and $I_2$ are concyclic, likewise are $N$, $P$, $I_3$ and $I_4$.
  

 A: Partial solution:
Well say $a=AB$, $b=BC$, $c=CD$ and $d=DA$, then we have $${a\over d} = {b\over c} =:t \implies a= dt \;\;\wedge \;\; b=ct$$ 
Since $ABCD$ is tangential we have $$a+c = b+d\implies d (t-1)= c(t-1)$$
From here we have two options:


*

*If $t=1$ then $a=d$ and $b=c$ and thus also $PB =PD$. So $A,C,P$ lies on a perpendicular bisector for $BD$ and $I_1I_2I_3I_4$ make rectangle and we are done.

*If $t\ne 1$ then $d=c$ and $a=b$...

A: COMMENT: 
Let angles A and B considered as $A_1$for left part and $A_2$ for right part; similarly  $C_1$for left and $C_2$ for right part of C. For B and D, $B_1$ for the top of line BD and $B_2$ for the below. Similarly $D_1$ for the top and $D_2$ for bellow for angle D.We can write:
$\angle I_1=180-(\frac{A_1}{2}+\frac{B_1}{2})$
$\angle I_2=180-(\frac{B_2}{2}+\frac{C_1}{2})$
$\angle I_3=180-(\frac{C_2}{2}+\frac{D_2}{2})$
$\angle I_4=180-(\frac{D_1}{2}+\frac{A_2}{2})$
Summing these relations we get:
$720-(\frac{A}{2}+\frac{B}{2}+\frac{C}{2}+\frac{D}{2})=720-\frac{360}{2}=540$
Also:
$\angle I_2+\angle I_4=180-(\frac{B_2}{2}+\frac{C_1}{2})+180-(\frac{D_1}{2}+\frac{A_2}{2})=360-(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})$
The sides of these angles cross two by two mutually and construct an octagon.
Now when P is on BD, we have:
$(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})=\frac{360}{4}=90$
⇒$\angle I_2+\angle I_4=360-90=270$
similarly:
⇒$\angle I_1+\angle I_3=360-90=270$
Now we must prove two points:
1: for all positions of P we have:
$(\frac{B_2}{2}+\frac{C_1}{2}+\frac{D_1}{2}+\frac{A_2}{2})=\frac{360}{4}=90$
2: In an octagon if the sum of two opposite angle is $270^o$ the octagon is cyclic.
