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Let $\triangle ABC$ be a right angled triangle where $\angle A = 90^\circ$. $D, F, E $ and $G$ are the midpoints of $BC, AB, AF$ and $FB$ respectively. $AD$ interesect the lines $CE, CF$ and $CG$ at point $P, Q$ and $R$ respectively. Find out $\frac {PQ}{QR}$

By 'Apollonius's Theorem', I was only able to show the relation of $AD$ with the base and height of the right-angled $\triangle ABC$. But I couldn't anyhow measure its segments such as $PQ$ and $QR$.

SOURCE: Bangladesh Math Olympiad

A small help will be necessary. Thanks in advance.

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    $\begingroup$ Well, the simplest way (in the sense of no geometry is needed) is to set up coordinates: Say, let $A = (0,0)$, $B = (0,a)$ and $C = (b,0)$ and try to solve the coordinates of everything. For some geometric proof, try to mimic the proof of the property $\overline{AQ}: \overline{QD} = 2:1$. $\endgroup$ – Hw Chu Feb 26 at 17:13
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    $\begingroup$ Two auxiliary segments I was thinking about are $\overline{GD}$ and a $E'$ on $\overline{BC}$ such that $\overline{EE'} // \overline{AC}$. But I realized that the solution in my answer below is better. $\endgroup$ – Hw Chu Feb 26 at 17:57
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As a consequence of Menelaus's theorem, if two cevians $AD$ and $BE$ of triangle $ABC$ meet at $F$, then: $$ {DF\over AF}={DC\over DB}\cdot{AE\over AC}. $$

You can use this to compute $PD/PA$ and $RD/RA$, and from them $AP/AD$ and $AR/AD$. Combining these results with $AQ/AD=2/3$ ($Q$ is the centroid of $ABC$) you can then find $PQ/AD$ and $QR/AD$.

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  • $\begingroup$ I'm afraid I don't understand what is this proportion of two segments of cevian line 𝐴𝐷 which I didn't use. Could you explain what you mean? $\endgroup$ – Aretino Feb 27 at 14:15
  • $\begingroup$ Haha. I meant to say like you could have used the ratio $\frac{EF}{BF}$. And also I wanted to say about the cevian line $BE$, not $AD$. My mistake!!😒😒😒 I hope, I didn't offer so much fear to an Italian high school teacher. $\endgroup$ – Anirban Niloy Feb 27 at 14:25
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    $\begingroup$ Of course the same theorem also applies to the other cevian, just switch left and right sides: $${EF\over BF}={EC\over EA}\cdot{BD\over BC}.$$ $\endgroup$ – Aretino Feb 27 at 14:29
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Added and erased some objects. Did you see anything?

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