In a given triangle $ABC$ lets choose inside side $AC$ points $P$, $Q$ and inside side $BC$ points $R$,$S$ so, that $|AP|=|PQ|=|QC|$, $|BR|=|RS|=|SC|$
Next denote point $E$ as intersection of diagonals in trapezoid $ABRP$, point $F$ as intersection of diagonals in trapezoid $PRSQ$ and point $G$ as intersection of diagonals in trapezoid $ABSQ$. Points $E$, $F$ and $G$ lies on median of triangle $ABC$ from vertex $C$. Determinate the ratio $|GF|:|EF|$ for any triangle.
Triangle sketch:
First off, let's label the intersections of the median with $QS$, $PR$, and $AB$ as $I$, $H$, and $K$, respectively. Note that by the three parallels theorem, $CI = IH = HK$, so let's set the length of $HK = x$.
Note also that triangles $CQS$, $CPR$, and $CAB$ are similar, and using the appropriate ratio of similarities, we can find that $QS = \frac{1}{3} AB$ and $PR = \frac{2}{3} AB$. So $QS = \frac{1}{2} PR$.
Let's consider now trapezoid $APRB$. Using angles, we can conclude that triangles $EPR$ and $EBA$ are similar. So then using that $PR = \frac{2}{3} AB$, we find that $HE: EK = 2:3$. Thus $HE = \frac{2}{5}x$.
Using a similar method for trapezoid $PQSR$, we can conlcude that $HF = \frac{2}{3}x$ and $IF = \frac{1}{3}x$, so $EF = HE + HF = \frac{16}{15}x$. And then from trapezoid $AQSB$, we can conclude that $IG = \frac{1}{2}x$, so $GF = IG - IF = \frac{1}{6}x$.
Thus $GF:EF = \frac{1}{6} : \frac{16}{15}$, or more succinctly, $5:32$.