Sorry guys, I need to overkill this problem, $B(0,0) , C(3a,0) , A(3b,3c)$ thus since G barycenter $G(a+b,c)$ In complex coord, we know that $a+d = g+b$ if AGDB parallelogram thus, in real we can transform the real coord into a complex number $3b+3ic+d = a+b+ci+0$ $d = a-2b-2ic $ $D = (a-2b,-2c)$ $BG \parallel CD$ iff gradient equal $\nabla{BG}=\frac{c}{a+b}$ $\nabla{CD} =\frac{-2c}{-2b-2a} = \frac{c}{a+b}$ so we are done

It is of course much easier: If $G'$ is the reflection of $G$ over the midpoint of $AB$ so that $AG'BG$ is a parallelogram, then $G'C=2G'G$ and moreover $G',B,D$ are collinear with $G'D=2G'B$. Hence not only $BG \parallel CD$ but in fact also $CD=2BG$.

see the figure:

Attachments:

$$C-D=C-(B+G-A)=A-B+C-G=2(G-B)$$