A problem relating to Circumscribed Quadrilaterals and Triangles. Given a triangle ABC. For an arbitrary interior point X of the triangle denote by A1(X) the point intersection of the lines AX and BC, denote by B1(X) the point intersection of the lines BX and CA, and denote by C1(X) the point intersection of the lines CX and AB.
Determine all points P in the interior of the triangle for which each of the quadrilaterals AC1(P)PB1(P), BA1(P)PC1(P) and CB1(P)PA1(P) has an inscribed circle.

What I have been able to prove includes-
i)  B3A1P = A4B1P, likewise for other corresponding pairs
ii) AB1P-AC1P = PB1P-PC1P=B2C1=DA1P , likewise for other corresponding pairs
iii)    AC1P+BA1P+CB1P = s       (semi perimeter of ∆ABC)
I know that this point must be the Inner Soddy Center of ∆ABC but am finding it difficult to prove it.
Please help..

Finally cracked it!!
To show that point P is the inner soddy centre, we show that P lies on a hyperbola with focii $B,C$ and passing through $D$.
That means that we effectively have to show that
$$CP-BP=CD-BD$$
Notice that $CB_1=CB_3=$ Length of Tangent from $C$ to the circle $O_B$.
Hence,
$$\begin{align}CP+B_1P =CB_3 \\
\implies CP=CB_3-B_1P 
\end{align}$$
Similarly,$$BP=BC_4-PC_2$$
So,
$$\begin{align}
CP-BP
&=CB_3-BC_4+(PC_2-PB_1) &(1)\\
&=(CC_4+B_3C_4)-(BB_3+B_3C_4)+(PC_1-PB_2) &(2) \\
&=CC_4-BB_3+B_2C_1 &(3) \\
&=CA_{1P}-BD+DA_{1P} &(4) \\
&=CD-BD &(5)
\end{align}$$
Also, $$BP-AP=BF-AF$$
$$AP-CP=AE-CE$$
Therefore P lies on the intersection point of all three Hyperbolae that i mentioned earlier-This must be a unique point.
Now, the inner Soddy Centre satisfies these properties.  Thus, it is the required point P.
