Let ABC be a triangle and D, E and F be the points of tangency of its incircle with the sides BC, CA and AB respectively. Let P and Q be the intersectiom of DE and Df with the bisectors of the angles ACB and ABC respectively. Show that E, F, P and Q belong in the same circunference.
What I have tried so far is, I have shown that if X and Y are the intersection of the diagonal FP and EQ with the incircle , then if the triangle XYD is isoceles the proof follows.
To show this, I've tried to show that the angles ADX and BDY are equal.
This isn't true in general.
Here is why. If points $E,F,P,Q$ were concyclic, the power (see here) of point $D$ with respect to this circle could be expressed in two equal ways :
$$DP.DE=DQ.DF\tag{1}$$
As triangles $CDE$ and $BDF$ are isosceles, (1) is equivalent to :
$$2DP^2=2DQ^2$$
itself equivalent to $DP=DQ$ which isn't true in general.
Take for example triangle $ABC$ right isosceles triangle with right angle in $B$ (see figure) :
Fig. 1 : In a visible way, $E,F,P,Q$ do not belong to a same circle (this can be established by computation).