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Imagine a parallellogram with sides ABCD. The sides AD and DC are each divided into three equally large distances. Show that BE, BD and BF divides AC into four equally large distances.

What I have done: Tried to setup some relationsships between the 90 degree angle in the middle and the given distance one the sides. The there are two different angles from point B which are different. I dont know what to do. Any tips?

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    $\begingroup$ What are $E$ and $F$? $\endgroup$ – Ernie060 Sep 21 '18 at 20:53
  • $\begingroup$ Can you define $E$ and $F$? $\endgroup$ – Max0815 Sep 21 '18 at 21:00
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    $\begingroup$ Presumably $E$ and $F$ are points on $\overline{CD}$ and $\overline{AD}$, respectively, such that $|CE|:|ED|=1:3=|AF|:|FD|$. It is known that the diagonals of a parallelogram bisect each other; let their intersection be $X$. The question becomes: Show that $\overline{BE}$ and $\overline{BF}$ bisect $\overline{CX}$ and $\overline{AX}$. But then the task reduces to showing this property of triangles: In $\triangle ABD$, "side-trisector" $\overline{AF}$ bisects median $\overline{BD}$. $\endgroup$ – Blue Sep 21 '18 at 21:20
  • $\begingroup$ As for "the 90 degree angle in the middle" ... (If my interpretation of the problem is correct) You may be reading too much into whatever diagram you're seeing. There need not be any right angles in the figure for the result to hold. $\endgroup$ – Blue Sep 21 '18 at 21:23
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We have points $E$ and $F$ on sides $AD$ and $CD$ of a parallelogram, such that $AE:ED=CF:FD=1:2$. We want to show that chords $BE$, $BD$, $BF$ divide diagonal $AC$ into four equal parts: $AG=GM=MH=HC$ (see diagram).

As pointed out by Blue in a comment, we already know that diagonal $BD$ bisects $AC$, hence we need only prove that $AG=GM$ and $MH=HC$.

Join $DG$ and consider the areas of the five triangles into which $ABD$ is divided. If $area_{AGE}=x$ and $BG/GE=k$, then $area_{ABG}=kx$, because triangles $AGE$ and $ABG$ have vertex $A$ in common and bases $GE$, $BG$ on the same line.

By the same argument we also have $area_{DGE}=2x$ and $area_{DBG}=2kx$. But then $$ area_{MBG}=area_{MDG}={1\over2}area_{DBG}=kx. $$ It follows that $area_{MBG}=area_{ABG}$ and as a consequence $MG=AG$, as it was to be proved.

By an analogous argument one can prove that $MH=CH$.

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