University plane geometry question about splitting the diagonal of a parallelogram into 3 equal parts 
$ABCD$ is a parallelogram. If the two sides $\overline{AB}$ and $\overline{AD}$ are bisected in $E$ and $F$, respectively, show that $\overline{CE}$ and $\overline{CF}$ when joined cut the diagonal $\overline{BD}$ in three equal parts.

I have no idea how to do this question, any help would be appreciated.
 A: Hint:
loock at the figure.

the first step is to show that the triangles $FDG$, $AFE$ and $EBH$ are congruent, so that $GF=FE=EH$. Than use the Thales intercept theorem.
A: Suppose $ABCD$ is a square as follows:
y ^
  |
1 D---C
  |\  |
  F \ |
  |  \|
0 A-E-B-->x
  0   1

The line through $BD$ has equation $y=1-x$, that through $CE$ $y=2x-1$ and that through $CF$ $y=(x+1)/2$. The intersection points of the latter two lines with the first can be easily shown to be $(\frac23,\frac13)$ and $(\frac13,\frac23)$ respectively – obviously trisecting $BD$.
Now note that for any choice of $A,B,D$ in the plane, the resulting parallelogram can be affinely transformed into the above square. Since affine transformations preserve ratios of lengths, $BD$ will be trisected in all parallelograms.
A: Please use Emilio's parallelogram $ABCD$. 

$F,E$ are the midpoints of sides $\overline{AD}$ and $\overline{AB}$ resp.
$\overline{BD}$ is a diagonal, and let $\overline{AC}$ be the other diagonal.
Let $M=\overline{BD}\cap\overline{AC}$
$M$ bisects each of the diagonals ( Property of diagonals in a parallelogram).
Let $Z_1=\overline{FC}\cap\overline{BD}$
$(1)$ $\Delta ACD$ 
$(a)$ $\overline{FC}$ is a median to $\overline{AD}$ 
$(b)$ $\overline{DM}$ is a median  to $\overline{AC}$.
Medians of a triangle intersect at the centroid which divides them in the ratio $2:1$.
The medians of the $\Delta ACD$, $\overline{FC}$ and $\overline{DM}$ intersect at $Z_1$
Ratio:$\frac{\left|DZ_1\right|}{\left|Z_1M\right|}=\frac{2}{1}$ 
Now, look at triangle $ABC$.
$\overline{EC}$ and $\overline{BM}$ are medians of $\Delta ABC$
Let $Z_2=\overline{EC}\cap\overline{BM}$
Same argument as before:
Ratio: $\frac{\left|BZ_2\right|}{\left|Z_2M\right|}=\frac{2}{1}$
Putting the parts together we have with
$d:=|DM|=|BM|$ ($M$ bisects $\overline{BD}$);
$\frac{2}{3}d=\left|DZ_1\right|,\frac{1}{3}d=\left|Z_1M\right|$and likewise:
$\frac{2}{3}d=\left|BZ_2\right|,\frac{1}{3}d=\left|Z_2M\right|$
$\left|Z_1M\right|+\left|Z_2M\right|=\left|Z_1Z_2\right|=\frac{2}{3}d$
$Finally$:
the three equal parts:
$\left|DZ_1\right|=\left|Z_1Z_2\right|=\left|BZ_2\right|$
$Q.E.D$
