Finding $\frac{PQ}{QR}$ in a right angled $\triangle ABC$, where $AD$ is the median line dropped from the opposite vertex of the hypotenuse

Question

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.

Answer 1

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 $QR/AD$.

Answer 2

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