University plane geometry question about splitting the diagonal of a parallelogram into 3 equal parts

Question

$ABCD$ 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.

Answer 1

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.

Answer 2

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.

Answer 3

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:

$\left|DZ_1\right|=\left|Z_1Z_2\right|=\left|BZ_2\right|$

$Q.E.D$