are you sure? $AD\cdot BE=DP^2$

Let $E_1$ be on $AC$ such that $BE_1||ED$. Then $AD*BE=AD*DE_1=PD^2$ hence $\angle{E_1BA}=\angle{E_1PA}=\frac{\angle{B}-\angle{A}}{2}$ hence $AE_1PB$ is cyclic ,then $\angle{PBE_1}=\angle{PAE_1}=\frac{\angle{A}}{2}$ hence $\angle{PBC}=\frac{\angle{B}}{2}$ which means $I=P$ so it is easy to see $FC^2-IF^2=IF^2=FB*FC$. SO WE ARE DONE.

By applying Menelaus' Theorem, point $D,E,F$ are collinear then \[\frac{DC}{DA} \cdot \frac{FA}{FB} \cdot \frac{EB}{EC}=1.\] We also have $CE=DE$ so $\frac{BE}{BF}= \frac{DA}{FA}$. Since $AP$ is the bisector of angle $BAC$ so \[\frac{BE \cdot DA}{BF \cdot FA}= \left( \frac{DA}{FA} \right)^2= \frac{PD^2}{PF^2}.\] It follows that $PD^2=BE \cdot AD \iff FP^2=FA \cdot FB$. Let $I$ be the midpoint of $DE$ then $\angle ICE= \angle ICD$ and $\angle CIE=90^{\circ}$. Therefore $I$ is the incenter of triangle $ABC$. Hence \[\angle CIE+ \angle BIE= 90^{\circ}+ \angle BIE=\angle BIC= 90^{\circ}+ \frac{\angle A}{2}.\] We obtain $\angle BIE= \frac{ \angle A}{2}= \angle IAB$. It follows that $\triangle FIA \sim \triangle FBI \; ( \text{A.A})$. Hence $FI^2=FB \cdot FA$. Thus, \[PD^2=BE \cdot AD \Leftrightarrow PF^2=FA \cdot FB \Leftrightarrow PF=FI \\ \Leftrightarrow P \equiv I \Leftrightarrow \angle CPF=90^{\circ} \Leftrightarrow FP^2+PC^2=FC^2 \\ \Leftrightarrow FA \cdot FB+CP^2=CF^2.\]

mathuz wrote: are you sure? $AD\cdot BE=DP^2$ Let $P$ be any point on the segments $ED$. Then \[ FB\cdot FA+CP^2 = CF^2\:\Longleftrightarrow\: AD\cdot BE = PD\cdot PE . \] The our problem will be particular case.

shinichiman wrote: By applying Menelaus' Theorem, point $D,E,F$ are collinear then \[\frac{DC}{DA} \cdot \frac{FA}{FB} \cdot \frac{EB}{EC}=1.\] We also have $CE=DE$ so $\frac{BE}{BF}= \frac{DA}{FA}$. Since $AP$ is the bisector of angle $BAC$ so \[\frac{BE \cdot DA}{BF \cdot FA}= \left( \frac{DA}{FA} \right)^2= \frac{PD^2}{PF^2}.\] It follows that $PD^2=BE \cdot AD \iff FP^2=FA \cdot FB$. Let $I$ be the midpoint of $DE$ then $\angle ICE= \angle ICD$ and $\angle CIE=90^{\circ}$. Therefore $I$ is the incenter of triangle $ABC$. Hence \[\angle CIE+ \angle BIE= 90^{\circ}+ \angle BIE=\angle BIC= 90^{\circ}+ \frac{\angle A}{2}.\] We obtain $\angle BIE= \frac{ \angle A}{2}= \angle IAB$. It follows that $\triangle FIA \sim \triangle FBI \; ( \text{A.A})$. Hence $FI^2=FB \cdot FA$. Thus, \[PD^2=BE \cdot AD \Leftrightarrow PF^2=FA \cdot FB \Leftrightarrow PF=FI \\ \Leftrightarrow P \equiv I \Leftrightarrow \angle CPF=90^{\circ} \Leftrightarrow FP^2+PC^2=FC^2 \\ \Leftrightarrow FA \cdot FB+CP^2=CF^2.\] You say $I$ is incenter of triangle $ABC$ . But if you know that we can immediately conclude that $AI$ is angle bisector which means $P=I$ . Your reasons are not enough to conclude that $I$ is incenter.

Can someone clarify why P is the incenter of the triangle ABC?