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$ABCD$ is a parallellogram. If the two sides $AB$ and $AD$ are bisected in $E$ and $F$, respectively, show that $CE$ and $CF$ when joined cut the diagonal $BD$ in three equal parts.

I have no idea how to do this question, any help would be appreciated.

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    There exists an affine transformation that maps this parallelogram to a square. If you can prove it in a square it will also hold in any parallelogram since affine transformations preserve ratios of lengths. – 2017-02-22

3 Answers 3

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Hint:

loock at the figure.

enter image description here

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.

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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.

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Please use Emilio's parallelogram ABCD. F,E are the midpoints of sides AD and AB resp.

BD is a diagonal, and let AC be the other diagonal.

Let the point of intersection of the two diagonals be M. M bisects each of the diagonals ( Property of diagonals in a parallelogram).

Let the point of intersection of FC and DB be Z$_1$.

1) Triangle ADC :

a) FC is a median to AD, and

b) DM is a median to AC.

Medians of a triangle intersect at the centroid which divides them in the ratio 2:1.

The medians of the triangle ADC, FC and DM intersect at Z$_1$.

Ratio: Length of DZ$_1$ to length of Z$_1$M is 2:1.

Now look at triangle ACB.

EC and BM are medians of triangle ACB. Let the point of intersection be Z$_2$.

Same argument as before:

Ratio: Length of BZ$_2$ to length Z$_2$M is 2:1.

Putting the parts together we have with

d : = length DM = length BM ( M bisects DB );

2/3 d = length DZ$_1$; 1/3 d = length Z$_1$M, and likewise:

2/3 d = length BZ$_2$; 1/3 d = length Z$_2$M .

Adding the lengths Z$_1$M to Z$_2$M we get length Z$_1$Z$_2$ = 2/3 d,

$Finally$:

the three equal parts:

Length DZ$_1$ = length Z$_1$Z$_2$ = length BZ$_2$,

q.e.d.