# For △ABC and point P, prove that at least one of ratios AP/PD, BP/PE, CP/PF is less than or equal to 2 and one is greater than 2

Question:

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.

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

According to the Ceva's theorem,

$$\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$$

which holds automatically with $$\frac{AF}{FB}=\frac yx$$, $$\frac{BD}{DC}= \frac zy$$, $$\frac{CE}{EA}= \frac xz$$, where $$x$$, $$y$$, $$z>0$$. Then, $$\frac{AP}{PD}= \frac{Area_{ABE}}{Area_{DBE}} = \frac{\frac{EA}{CA}Area_{ABC}}{\frac{CE}{CA}\cdot \frac{BD}{BC}Area_{ABC}} =\frac{EA}{CE}\cdot\frac{BD+DC}{BD}=\frac zx\left(1+\frac yz\right)=\frac{y+z}{x}$$

Similarly, $$\frac{BP}{PE}=\frac{z+x}{y},\>\>\>\>\>\frac{CP}{PF}=\frac{x+y}{z}$$

1) To prove that at least one of the ratios is greater or equal to 2, evaluate their product with the inequality $$\frac{a+b}{\sqrt{ab}}\ge 2$$,

$$\frac{AP}{PD}\cdot\frac{BP}{PE}\cdot\frac{CP}{PF} = \frac{(y+z)(z+x)(x+y)}{xyz}\ge 2\cdot2\cdot2=8$$

It then follows that one of the ratios on the LHS is greater or equal to 2.

2) To prove that at least one of the ratios is less or equal to 2, evaluate the sum of their reciprocals with the Nebitt's inequality,

$$\frac{PD}{AP}+\frac{PE}{BP}+\frac{PF}{CP}=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge \frac32$$

It then follows that one of the ratios $$\frac{PD}{AP},\>\frac{PE}{BP},\>\frac{PF}{CP}$$ is greater or equal to $$\frac12$$, hence one of the reciprocals is less or equal to 2.

Similar to Quanto's answer, but without Ceva's:

$$\frac{AP}{PD} = \frac{\triangle APB}{\triangle DPB} = \frac{\triangle APC}{ \triangle DPC} = \frac{\triangle APB + \triangle APC}{\triangle BPC}.$$ Similarly $$\frac{BP}{PE} = \frac{\triangle BPA + \triangle BPC}{\triangle APC}$$ and $$\frac{CP}{PF} = \frac{\triangle CPA + \triangle CPB}{\triangle APB}.$$

Here I'm using lazy notation that a triangle also denotes its area.