Angle chase: $\angle{FPI} = \angle{FDI} = \frac{\angle{C}}{2} = \angle{ECI} = \angle{EPI}$. Since $ID = IE = IF$, this implies that $P$ lies on the perpendicular bisector of $EF$, and so does $Q$, and $EF \parallel BC$ since $\frac{FB}{EC} = \frac{AB}{AC}$, so we are done. $\square$

Nice problem! The first, we have $\frac{AB}{BF}=\frac{AB}{BD}=\frac{AC}{CD}=\frac{AC}{CE}$, so $FE \parallel BC$ Let $D \not = X = (CEI) \cap BC$. Since $\angle XCI = \angle ECI$ we have $IX = IE = (\text{from } CD=CE, BD=BF)=ID=IF$. So, $XDEF$ is inscribed quadrilateral with $FE \parallel XD$, so $FD = XE$ and $\Delta XIE = \Delta DIF$ and $\angle XPI = \angle DPI$. From this and $IX=ID$ we have $PI \perp BC$. By analogy, $QI \perp BC$, so $QP \perp BC$

Consider that $\dfrac{BF}{AB}=\dfrac{BD}{CD}=\dfrac{CE}{AC}$ Which menas that $EF//BC$ Notice $I$ is the circumcenter of $\triangle ABC$ So $\angle IPD=\angle AFD=90^{\circ}-\angle FED=\angle ICD=\angle ICE=\angle EPI$ Which means that $P,D,E$ are collinear So $\angle BDP=\angle EDC=90^{\circ}-\angle ICB$ Hence $IP\bot BC$ Similarly $IQ\bot BC$ So $PQ\bot BC$

Quite easy problem. Call $D'$ tangent point of incircle with $BC$.Let $(CEI) \cap ID'$ at $P$ from angle chasing we get $P$ is on $(DIF)$.Let $ID' \cap FD$ at $Q$ from angle chasing we get $Q$ is on $(BFI),(DIF)$. And we are done.