JBMO 2002 Shortlist - Centroid Let $ ABC$ be a triangle with centroid $ G$ and $ A_1,B_1,C_1$ midpoints of the sides $ BC,CA,AB$. A paralel through $ A_1$ to $ BB_1$ intersects $ B_1C_1$ at $ F$. Prove that triangles $ ABC$ and $ FA_1A$ are similar if and only if quadrilateral $ AB_1GC_1$ is cyclic.
I have so far managed to prove that $\angle AA_1F$ and $\angle ABC$, having assumed that $ AB_1GC_1$ is cyclic, but struggle to show that $C_1C||AF$.
 A: 
First of all $BB_1FA_1$ is a parallelogram and then $BA_1=B_1F$ but $C_1B_1=BC/2=B_1F$ and then $C_1AFC$ is a parallelogram because it has diagonal cutting in the mid point: that was already pointed in the previous parcial solution.
$1)$ If $AB_1GC_1$ is cyclic $\Rightarrow$ $\Delta ABC$ and $\Delta FA_1A$ are similars:
$a)$ The line $C_1B_1F$ is parallel to $BC$ so $\angle AC_1B_1=\angle ABC$;
$b)$ $AB_1GC_1$ is cyclic then $\angle AC_1B_1=\angle AGB_1$;
$c)$ $BB_1$ is parallel to $A_1F$ then $\angle AGB_1=\angle AA_1F$;
then $\angle ABC=\angle AA_1F$.
$d)$ $AB_1GC_1$ is cyclic then $\angle B_1AG=\angle GC_1B_1$;
$e)$ $C_1AFC$ is a parallelogram then $\angle GC_1B_1=\angle AFC_1$
$f)$ $AB_1GC_1$ is cyclic then $\angle C_1AG=\angle C_1B_1G$;
$g)$ $BB_1$ is parallel to $A_1F$ then $\angle C_1B_1G=\angle C_1FA_1$
then $\angle A=\angle C_1AG+\angle B_1AG=\angle C_1FA_1+\angle AFC_1=\angle AFA_1$ and finally  $\Delta ABC$ and $\Delta AFA_1$ are similars.
$2)$ If $\Delta ABC$ and $\Delta FA_1A$ are similars $\Rightarrow$ $AB_1GC_1$ is cyclic:
$a)$ $BC$ and $FC_1$ are parallels so $\angle ABC=\angle AC_1B_1$
$b)$ $\Delta ABC$ and $\Delta FA_1A$ are similars then $\angle ABC=\angle AA_1F$ (prove it);
$c)$ $BB_1$ and $A_1F$ are parallels so $\angle AA_1F=\angle AGB_1$
then $\angle AGB_1=\angle AC_1B_1$ and $AB_1GC_1$ is cyclic.
A: You can have a look at this to see if it helps you:
http://www.artofproblemsolving.com/community/c6h239185p1316351
To answer your question:
Notice that $FA_1BB_1$ is a parallelogram and thus $FB_1=\frac{BC}{2}$ 
But also $B_1C_1=\frac{BC}{2}$ thus the diagonals of $FCC_1A $ bisect each other and thus $FCC_1A $ is a parallelogram
