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

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 $$QR$$.

A small help will be necessary. Thanks in advance.

• 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 $\overline{AQ}: \overline{QD} = 2:1$. – Hw Chu Feb 26 at 17:13
• 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. – Hw Chu Feb 26 at 17:57

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 $$QR/AD$$.
• 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. – Anirban Niloy Feb 27 at 14:25
• Of course the same theorem also applies to the other cevian, just switch left and right sides: $${EF\over BF}={EC\over EA}\cdot{BD\over BC}.$$ – Aretino Feb 27 at 14:29