Let $R$ on segment $BD$ such that $PB=PR$ hence $PRQC$ is parallelogram. Now let $M=PQ\cap CR$ and $G=AM\cap RC$. We have $AG=2GM$ and the problem is done.

tuanh208 wrote: Let $R$ on segment $BD$ such that $PB=PR$ hence $PRQC$ is parallelogram. Now let $M=PQ\cap CR$ and $G=AM\cap RC$. We have $AG=2GM$ and the problem is done. you mean $G=AM\cap BD$

Let $N$ be the midpoint of $PQ$. Then the projection from $N$ to $AC$ will be equidistant to $BD$ and $C$. Let $M=AN\cap BD$ So $AM=2MN$

I tossed it on cartesian coordinates with inclined axes $\omega$ and equal $x$ and $y$ coordinates for the rhombus. The problem then becomes trivial.

Let $R,M$ like in first proof; $AM$ is common median of $\triangle APQ,\triangle ARC$, hence they share the same centroid, which lies onto the median $BD$ of $\triangle ARC$, done. Best regards, sunken rock