Let $DH$ be the midline of triangle $AFG$. Now, $DH=\frac{1}{2}AG$ and $AD=\frac{1}{3}b$, so points $D,H,E$ are colinear. Since $AD=\frac{1}{3}b$ and $BE=\frac{1}{3}a$ we have that $DE \parallel c$ and $DE=\frac{2}{3}c=2AG$ Since $GQ=\frac{2}{3}GH$ and $GP=\frac{2}{3}GE \implies QP \parallel DE \parallel c$ or simply $QP \parallel AG$ and $PQ=\frac{2}{3}HE$ Finally, $PQ=\frac{2}{3}HE=\frac{2}{3}(DE-DH)=\frac{2}{3}(2AG-\frac{1}{2}AG)=AG$ Now, since $PQ \parallel AG$ and $PQ=AG \implies AGPQ$ is a parallelogram

Only we need to prove that $\vec{AG} $ = $\vec{QP}$ $\vec{QP}$=$\vec{QG}$ + $\vec{GP}$=1/3. $\vec{FG}$ + 2/3. $\vec{GE}$ ==>>we only have to prove that $\vec{FG}$ + 2. $\vec{GE} $= $\vec{AB}$ now again we only have to prove that $\vec{FA}$ + $\vec{AG}$ + 2. $\vec{GB}$ + 2. $\vec{BE}$=$\vec{AB}$ Since $\vec{AG}$ + $\vec{GB}$ =$\vec{AB}$ we only have to prove that $\vec{FA}$ + $\vec{GB}$ + 2. $\vec{BE}$ =$\vec{0}$ we only need to prove that 2/3. $\vec{CA}$ + 2/3. $\vec{AB}$ + 2/3.$\vec{BC}$=$\vec{0}$ we know that is true and our proof is completed now

Let $N$ be midpoint of $AF$, Let $M$ be midpoint of $FG$ and Let $S$ be midpoint of $AB$. Let $CS$ meet $NE$ at $K$. By Thales Theorem we have $N,M,E$ are collinear and $EM || AB$. we also have $QP || ME$ so $PQ || AG$. $EM = EK + KM = 3NM$. $PQ/AG = (PQ/EM) . (EM/MN) . (MN/AG = (2/3) . (3) . (1/2) = 1$ so $PQ = AG$ so $AGPQ$ is parallelogram. we're Done.

Employ barycentrics on $\triangle ABC$. Set $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then $E=(0,2/3,1/3),F=(1/3,0,2/3),G=(2/3,1/3,0)$. And we can find $P=(2/3-4/9,1/3+2/9,0+2/9),Q=(2/3-1/9,1/3-1/9,0+2/9)$. We have the normalized coordinates of all points, so it just suffices to check that $A+P=G+Q$ where points are added by components, which is obviously true.