In $\triangle ABC$, $AD, BE$ and $CF$ are concurrent lines. $P, Q, R$ are points on $EF, FD, DE$ such that $DP, EQ$ and $FR$ are concurrent. Prove that $AP, BQ,$ and $CR$ are also concurrent.
(To be done with Menelaus theorem, Ceva's theorem, Stewart's theorem and elementary geometry)
Source:CTPCM
I am unable to first of all draw a good figure but with my rough figure I tried using Ceva's Theorem and then linking it with areas. But I was just getting stuck so I want some kind of hint and drawing to visualize.
Best Answer
Proof from EGMO, as this is quite well known as Cevian Nest.
Proof : Swap $D,E,F$ with $X,Y,Z$ and $P,Q,R$ with $D,F,E$ for convenience.
By sine rule, $\frac{\sin BAD}{\sin CAD}=\frac{\frac{ZD}{ZA}\sin ADZ}{\frac{YD}{YA}\sin ADY}=\frac{ZD}{YD}\frac{YA}{ZA}$, use trigonometric form of Ceva now to get that it is enough to prove $\frac{ZD×YA×XE×ZB×YF×XC}{YD×ZA×ZE×XB×XF×YC}=1$, which follows from Ceva's Theorem on $\Delta XYZ$ and $\Delta ABC$