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In $\triangle ABC$, $L$ and $M$ are two points on $AB$ and $AC$ such that $AL = \frac{2AB}{5}$ and $AM = \frac{3AC}{4}$. $BM$ and $CL$ intersect at the point $P$ and the extension line of $AP$ and the side $BC$ intersect $BC$ at the point $N$. What is the value of $\frac{BN}{BC}$?

SOURCE: BANGLADESH MATH OLYMPIAD

My Attempt:

Let denote the area of $\triangle ALP$, $[ALP] = x$ and $[APM] = y$.

So, from $\triangle APB$, showing the relation of the base of both the triangle $\triangle APL$ and $\triangle LPB$, I got $[LPB] = \frac{3x}{2}$.

Similarly, from $\triangle APC$, doing the above likewise approach, I got $[MPC] = \frac{y}{3}$. After that, getting two triangle $\triangle ACL$ and $\triangle BCL$ and showing their relation of area with their particular base, I got

$[BPC] = \frac{5x}{6}$....(given that $AL:AB =2:5$)

Again from $\triangle ABM$ and $\triangle CBM$,

$[BPC] = 2y$.....($AM:AC = 3:4$)

So, $2y = \frac{5x}{6}$ $\implies$ $x =\frac{12y}{5}$

And then expressing the area of all the triangle by $y$, I got $[ABC] = \frac{28y}{3}$

But, I can't anyhow relate the area of $\triangle BPN$ with $y$. Here, I got stuck. What should I do to find out the area of $\triangle BPN$. I need really some help Thank you.

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    $\begingroup$ I think you can directly get your result using Ceva's Theorem. Have you tried? $\endgroup$ – Matteo Feb 20 at 18:20
  • $\begingroup$ @Matteo Nope. I haven't tried. How can I do that? $\endgroup$ – Anirban Niloy Feb 20 at 18:29
  • $\begingroup$ have a look here: en.wikipedia.org/wiki/Ceva%27s_theorem $\endgroup$ – Matteo Feb 20 at 18:31
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    $\begingroup$ Good! By the way, if I remember correctly, the simplest demonstration of the theorem proceeds, as you were doing, equating areas. $\endgroup$ – Matteo Feb 20 at 18:47
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    $\begingroup$ Yes, it could be intersting to get to the result independently of Ceva's Theorem, just for didactic purposes... $\endgroup$ – Matteo Feb 20 at 19:22
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$$\frac{BN}{NC}\cdot\frac{CM}{MA}\cdot\frac{AL}{LB}=\frac{S_{\Delta PBN}}{S_{\Delta PCN}}\cdot\frac{S_{\Delta PCM}}{S_{\Delta PAM}}\cdot\frac{S_{\Delta PAL}}{S_{\Delta PBL}}=$$ $$=\frac{\frac{1}{2}PB\cdot PN\sin\measuredangle BPN}{\frac{1}{2}PC\cdot PN\sin\measuredangle CPN}\cdot\frac{\frac{1}{2}PC\cdot PM\sin\measuredangle CPM}{\frac{1}{2}PA\cdot PM\sin\measuredangle APM}\cdot\frac{\frac{1}{2}PA\cdot PL\sin\measuredangle APL}{\frac{1}{2}PB\cdot PL\sin\measuredangle BPL}=1.$$ Thus, $$\frac{BN}{NC}\cdot\frac{1}{3}\cdot\frac{2}{3}=1.$$ Can you end it now?

I got $$\frac{BN}{BC}=\frac{9}{11}.$$

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Hint: Consider using Ceva's Theorem. Since the lines $AN, CL$ and $BM$ concur at $P$ $$\frac{CM}{MA}·\frac{AL}{LB}·\frac{BN}{NC}=\frac{1}{3}·\frac{2}{3}·\frac{BN}{NC}=1$$ Thus

$$\frac{BN}{NC}=4,5\iff \frac{BN}{NC}=\frac{4,5}{4,5+1}=\frac{9}{11}$$

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