1
$\begingroup$

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: Triangle sketch

$\endgroup$
3
$\begingroup$

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, $5:32$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.