For part a: Let $R=PE \cap QD$. Then as $PQ \parallel DE$ and $2 \cdot DE=PQ$ it follows $R$ is the reflection of $P$ in $E$. Setting $\odot ABC$ to the unit circle we get: $$F=\frac{2a+b+c}{2} \quad \quad P=2a+c$$And therefore: $$R=2E-P=-a$$. Which is the point diametrically opposite $A$ so is on the circumcircle.

For part b: Let $S=PD \cap QE$ then $S$ is just the centroid of $RPQ$ and so: $$S=\frac{P+Q+R}{3}=\frac{3a+b+c}{3}=\frac{1}{3}H+\frac{2}{3}A$$Hence $S$ lies on $AH$

Solution without complex numbers: a)Obviously,$BHPA,CHQA,PQBC$ are all parallelograms,so $PQ=BC,PQ||BC$.Let $QD \cap PE =T$ then since $DE||BC||PQ,DE=1/2 PQ$ we deduce that $D,E$ are midpoints of $QT,PT$,which means that $TBQA,TCPA$ are parallelograms,so $BT=AQ=CH,BT||AQ||CH$ which means that $HCTB$ is also parallelogram,so $T$ is the reflection of $H$ in midpoint of $BC$ and lies on $(ABC)$ b)If $G=QE \cap PD$ then $G$ is the centroid of $PQT$,but this means that $G$ is also the centroid of $QAC$ and since $AF$ is a median of this triangle,so $G \in AH$.

Why triangles PQT and QAC have common centroid?

because $QE$ is a median of triangle $QAC$ and since $G$ is the centriod of $PQT$,$QG=2GE$ and this means that its also the centroid of $QAC$($E$ is the midpoint of both $AC$ and $PT$).

Thank you Yaghi. sbealing wrote: For part a: Let $R=PE \cap QD$. Then as $PQ \parallel DE$ and $2 \cdot DE=PQ$ it follows $R$ is the reflection of $P$ in $E$. Setting $\odot ABC$ to the unit circle we get: $$F=\frac{2a+b+c}{2} \quad \quad P=2a+c$$And therefore: $$R=2E-P=-a$$. You can also prove it this way. $2E-P=-a=2D-Q$ Hence reflection of $P$ in $E$ lies on lines $PE,QD$ and so does reflection of $Q$ in $D$. Because these lines are different, we have that $R=2E-P=-a=2D-Q$.

a) Let perpendicular from B to AB meet circumcircle of ABC at S. it's well-known BS = CH and BH = CS. FE/CH = 1/2 ---> FE/BS = 1/2 and PF/PB = 1/2 and P,F,B are collinear ---> P,E,S are collinear. DF/BH = 1/2 ---> DF/CS = 1/2 and QF/QC = 1/2 and Q,F,C are collinear ---> Q,D,S are collinear. so PE and QD meet circumcircle of ABC at S. b) Let QE and PD meet AH at X and Y. Triangles AQX and EXF are similar ---> FX/XA = EF/QA = EF/CH = 1/2 Triangles APY and DYF are similar ---> FY/YA = DF/PA = DF/BH = 1/2 so X and Y are same point so QE and PD meet at AH.