<s>[INST] Finding the ratio of a side of $\triangle ABC$ and its segment where one cevian line from the opposite vertex intersect the side in any point 
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.
 [/INST] $$\frac{BN}{NC}\cdot\frac{CM}{MA}\cdot\frac{AL}{LB}=\frac{S_{\Delta PBN}}{S_{\Delta PCN}}\cdot\frac{S_{\Delta PCM}}{S_{\Delta PAM}}\cdot\frac{S_{\Delta PAL}}{S_{\Delta PBL}}=$$
$$=\frac{\frac{1}{2}PB\cdot PN\sin\measuredangle BPN}{\frac{1}{2}PC\cdot PN\sin\measuredangle CPN}\cdot\frac{\frac{1}{2}PC\cdot PM\sin\measuredangle CPM}{\frac{1}{2}PA\cdot PM\sin\measuredangle APM}\cdot\frac{\frac{1}{2}PA\cdot PL\sin\measuredangle APL}{\frac{1}{2}PB\cdot PL\sin\measuredangle BPL}=1.$$
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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