We don't need the condition $ABCD$ is cyclic. Let $BC \cap AD=G$, $EF \cap BC=H$, $EF \cap AD=K$. Since $\angle APK = 180^{\circ} - \angle APB - \angle BPE$ and $\angle DPK = 180^{\circ} - \angle DPC - \angle CPE$, it follows $\angle APB = \angle DPC$ iff $\angle APK = \angle DPK$. From $(G,H;B,C)=-1$ and $\angle BPH = \angle HPC$, it follows $\angle GPH =90^{\circ} \Longrightarrow \angle GPK =90^{\circ}$. Now from $(G,K;A,D)=-1$ and $\angle GPK =90^{\circ}$, we have $\angle APK = \angle DPK$ as desired.

it suffices to prove $A,B,F,P$ and $C,D,P,F$ are concyclic. it's easy though~~~

littletush wrote: it suffices to prove $A,B,F,P$ and $C,D,P,F$ are concyclic. it's easy though~~~ not easy for me. how??

We can think that radical axis of circumcircle of APB and ABCD and CPD, ABCD meet at E. Intersection of circumcircle of APB and CPD at K, then KF pass through E, and angle BKE=angle CKE. Therefore, K=P.

Let $AD,BC$ intersect in $R$ and let $EF$ and $AD$ intersect in $S$. $E(A,D;S,R)=-1$ thus $EP\perp RP$ but again $(A,D;S,R)=-1$ so $PF$ is the angle bisector of $\angle APD$ and we are done.

Notice that $PE$ and $PF$ are isogonal in angle $BPC$ and $EB \cap FC=A$, $EC \cap FB=D$. Thus, by the isogonality lemma, $PA,PD$ are isogonal in angle $BPC$ and the result $\angle APB=\angle DPC$ holds.

nikolapavlovic wrote: Let $AD,BC$ intersect in $R$ and let $EF$ and $AD$ intersect in $S$. $E(A,D;S,R)=-1$ thus $EP\perp RP$....................... Please, why is this so obvious ? I think that first we must take one more harmonic division on segment $BCR$ . Babis

You're absolutely right must've written it in a hurry,sorry. $PF\cap BC=\{Q\}$ $\implies$ $(B,C;Q,R)=-1$ $\implies$ $EP\perp PR$ and than proceed as in above.

nikolapavlovic wrote: You're absolutely right must've written it in a hurry,sorry. $PF\cap BC=\{Q\}$ $\implies$ $(B,C;Q,R)=-1$ $\implies$ $EP\perp PR$ and than proceed as in above. Very well ! Thank you for the nice solution!!! Babis