Consider △ABC and an inside point P. If AP, BP, CP meet the opposite sides at D, E, F, respectively. Prove that at least one of ratios AP/PD, BP/PE, CP/PF is less than or equal to 2 and one of them is greater than or equal to 2.
Source : CTPCM (Olympiad book)
My try -
Now first I apply menelaus to appropriate triangles to get these 3 ratios relation with other side so that I think I will find something useful in end . So I get
AP/PD . BP/PE . CP/PF = (BD/DC +1) (AF/FB +1) (CE/EA +1)
Now the ratios are converted into ceva but I don't know how to use this to prove given question...
Next, I think if I could prove that sum of the 3 ratios given in question = 6 then by piegon hole principle average form I will conclude... But not getting how to prove that.... Then I also relate these 3 ratios with the ratio of areas of triangles in which they are contained..but I couldn't able to conclude....
Any hints will be greatly helpful... Thankyou